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Module 1: Measuring the world: units, sizes and estimates

Module 2: Motion in a line

Module 3: Vectors and motion in two dimensions

Module 4: Forces and Newton's laws

Module 5: Friction, springs, slopes and circles

Module 6: Work and energy

Module 7: Momentum and collisions

Module 8: Rotation and torque

Module 9: Gravity and orbits

Module 10: Oscillations

Module 11: Waves and sound

Module 12: Fluids

Module 13: Heat, temperature and thermodynamics

Module 14: Charge and electric fields

Module 15: Current and circuits

Module 16: Magnetism and induction

Module 17: Light and optics

Module 18: Special relativity

Module 19: Quantum physics

Module 20: Physics like an expert

Answer key

  1. 1. Measuring the world: units, sizes and estimates · Physics looks for the few rules behind everything · QuizA model treats a thrown object as a point with no air resistance. For which object does it predict the flight best?A. A 7 kg shot put thrown 15 mAir pushes on everything, but what matters is how big that push is compared with the object’s weight. A heavy, compact shot put barely notices the air over 15 m, so the simple model is excellent. A feather or a sheet of paper is light with a large surface, so the air decides how it moves and the model fails completely. A model is not right or wrong in general: it is good enough for some cases and not for others.
  2. 1. Measuring the world: units, sizes and estimates · Seven base units measure everything · QuizWhich of these is an SI base unit?C. The kilogramThe kilogram is the base unit of mass, oddly the only one with a prefix already in its name. The gram is a thousandth of it, not a base unit in its own right. The newton is derived (kg·m/s2), and kilometres per hour is a unit people use, but it is not SI at all: the SI unit of speed is the metre per second.
  3. 1. Measuring the world: units, sizes and estimates · Powers of ten and prefixes · Fill the blanksGive each prefix its power of ten, in metres per unit.1: 3, 2: -3, 3: -6, 4: -9Kilo is a thousand, 103; milli a thousandth, 10−3; micro a millionth, 10−6; nano a billionth, 10−9. The classic slip is confusing the two "m"s: capital M is mega, a million times bigger, and small m is milli, a thousand times smaller. A 5 MW turbine and a 5 mW laser pointer differ by a factor of a billion.
  4. 1. Measuring the world: units, sizes and estimates · Convert units by multiplying by one · Code challengeWrite kmhToMs(kmh), which turns a speed in kilometres per hour into metres per second, and msToKmh(ms), which goes back.One solution:
    Solution
    double kmhToMs(double kmh) {  return kmh * 1000 / 3600;}
    double msToKmh(double ms) {  return ms * 3600 / 1000;}
    One kilometre is 1000 m and one hour is 3600 s, so km/h to m/s is ×1000÷3600, which is ÷3.6. Going back flips both factors. The usual mistake is multiplying by 3.6 in the wrong direction: a quick sense check catches it, because a number in m/s is always smaller than the same speed in km/h: an hour is 3600 times longer than a second, but a kilometre is only 1000 times longer than a metre. The parameters are doubles on purpose: with an int speed, kmh / 3600 * 1000 would be 0, since an int divided by an int drops the fraction. The tests use EXPECT_NEAR with a small tolerance, never EXPECT_EQ, since 72×1000/3600 need not come out as exactly 20 in binary.
  5. 1. Measuring the world: units, sizes and estimates · Convert units by multiplying by one · ProblemA cyclist rides 18km in 45min. What is her average speed in metres per second, and in kilometres per hour?In m/s 20/3 ≈ 6.66667 m/s; In km/h 24 km/h
    1. Convert both quantities to SI units, writing each conversion factor out:
    2. 18km×1000m1km=18000m,45min×60s1min=2700s
    3. v=18000m2700s=203m/s≈6.67m/s
    4. In kilometres per hour, 45 minutes is 0.75h:
    5. v=18km0.75h=24km/h
    6. Check: 24÷3.6=6.67, the same speed. The average speed is 6.67m/s, or 24km/h.
  6. 1. Measuring the world: units, sizes and estimates · Dimensional analysis checks a formula before you trust it · QuizA stone is dropped from rest. Which formula for the distance d it falls in time t could be right? (g is an acceleration.)B. d=12gt2gt2 has dimension [L][T]−2×[T]2=[L], a length, as a distance must be. gt is [L][T]−1, a speed, and g2t is [L]2[T]−3, which is nothing useful. Dimensions cannot check the number in front, 12: a formula with the right dimensions can still be wrong by a factor, but one with the wrong dimensions is always wrong.
  7. 1. Measuring the world: units, sizes and estimates · Dimensional analysis checks a formula before you trust it · ProblemA mass m bobs on a spring of stiffness k, measured in N/m, which is kg/s2: its dimension is [M][T]−2. Suppose the period is T=Cmakb, where C is a number with no dimension. Match the dimensions of the two sides to find the powers a and b.a= 1/2 ≈ 0.5; b= -1/2 ≈ −0.5
    1. Write the dimensions of the right side:
    2. [makb]=[M]a([M][T]−2)b=[M]a+b[T]−2b
    3. The left side, a period, is [M]0[T]1. Match the powers:
    4. [T]:−2b=1⇒b=−12[M]:a+b=0⇒a=12
    5. So T=Cm/k: a heavier mass is slower, a stiffer spring faster. Module 10 shows that C=2π. The answer is a=12, b=−12.
  8. 1. Measuring the world: units, sizes and estimates · Significant figures say how precise a number is · QuizA floor measures 4.25m by 2.1m. What is its area, to the right number of significant figures?C. 8.9m24.25×2.1=8.925, but 2.1 has only 2 significant figures, so the answer gets 2: 8.9m2. Writing 8.925 claims you know the area to a thousandth of a square metre when one side was only measured to a tenth of a metre. Calculators print every digit; deciding how many mean something is your job.
  9. 1. Measuring the world: units, sizes and estimates · Every measurement comes with an uncertainty · Code challengeWrite mean(values), the sum of the measurements divided by how many there are, and halfRange(values), half the difference between the largest and the smallest. Both get a non-empty vector.One solution:
    Solution
    #include <algorithm>#include <vector>
    double mean(const std::vector<double>& values) {  double sum = 0;  for (double v : values) sum += v;  return sum / values.size();}
    double halfRange(const std::vector<double>& values) {  double lo = std::ranges::min(values);  double hi = std::ranges::max(values);  return (hi - lo) / 2;}
    The mean adds every value and divides by the count, values.size(). That division is safe because sum is a double; an int sum of whole-number readings would be cut to a whole number. The half-range uses std::ranges::min and std::ranges::max, from <algorithm>, which find the smallest and largest in the vector; a loop that keeps the smallest and largest so far does the same. Forgetting the / 2 gives the full range, twice the uncertainty you should quote. The half-range is crude: one wild timing inflates it. Careful experiments and any statistics course use the standard deviation instead, which weighs every value.
  10. 1. Measuring the world: units, sizes and estimates · Every measurement comes with an uncertainty · ProblemA rectangular tile measures 12.0±0.1cm by 5.0±0.1cm. Find its area, the relative uncertainty in the area as a percentage, and the absolute uncertainty in the area. Use the rule that relative uncertainties add when you multiply.Area 60 cm²; Relative uncertainty 17/6 ≈ 2.83333 %; Absolute uncertainty 1.7 cm²
    1. The area is length times width: A=12.0×5.0=60.0cm2.
    2. Each side’s relative uncertainty, then their sum:
    3. 0.112.0≈0.83%,0.15.0=2.0%,0.83%+2.0%≈2.8%
    4. Turn it back into an absolute uncertainty:
    5. ΔA=0.0283×60.0cm2=1.7cm2
    6. So the area is 60.0±1.7cm2, a relative uncertainty of about 2.8%. The short side dominates: the same 0.1cm is a bigger share of 5 cm than of 12 cm.
  11. 1. Measuring the world: units, sizes and estimates · Fermi estimates get within a factor of ten in a minute · Put in orderPut the steps of a Fermi estimate in order.1. Say exactly what number you want, with its unit 2. Break it into factors you can each guess 3. Guess each factor to one significant figure 4. Multiply, keeping track of the units and powers of ten 5. Check the answer against anything you know, and the units against what you wantedA clear question comes first, or you cannot tell which factors you need. Then split, guess, multiply. The last step is the one people skip: if the units left over are not the unit you asked for, a factor is missing, and if the answer contradicts something you know (ten times more heartbeats than seconds in a life, say, when a heart beats about once a second), a guess is badly off.
  12. 1. Measuring the world: units, sizes and estimates · From protons to the observable universe · QuizAn atom is about 10−10m across and a grain of sand about 1mm. Roughly how many atoms side by side span the grain?C. About ten million1mm=10−3m, so the count is 10−3/10−10=107, ten million. Dividing powers of ten subtracts the exponents: −3−(−10)=7. "About 7" is the tempting slip, the number of orders of magnitude rather than the count itself.
  13. 1. Measuring the world: units, sizes and estimates · Practice problems · ProblemThe Sun is 1.50×1011m from the Earth, and light travels at 3.00×108m/s. How long does sunlight take to reach us, in seconds and in minutes?In seconds 500 s; In minutes 25/3 ≈ 8.33333 min
    1. Time is distance over speed:
    2. t=1.50×1011m3.00×108m/s=0.500×1011−8s=0.500×103s=500s
    3. 500s×1min60s≈8.33min
    4. Sunlight is about 8 minutes 20 seconds old when it arrives: 500s, or 8.33min.
  14. 1. Measuring the world: units, sizes and estimates · Practice problems · ProblemDensity is mass per unit volume. Gold has a density of 19.3g/cm3. Convert that to kg/m3. Then find the mass, in grams, of a solid gold cube 2.00cm on each side.Density 19300 kg/m³; Mass of the cube 154.4 g
    1. Convert grams to kilograms and cubic centimetres to cubic metres, with 1m3=(100cm)3=106cm3:
    2. 19.3gcm3×1kg1000g×106cm31m3=19300kgm3
    3. The cube’s volume is (2.00cm)3=8.00cm3, and mass is density (ρ, rho) times volume:
    4. m=ρV=19.3gcm3×8.00cm3=154.4g≈154g
    5. Gold is 19300kg/m3, and a 2 cm cube has a mass of about 154g. Forgetting to cube the 100 gives 193kg/m3, lighter than water: a sense check catches it.
  15. 1. Measuring the world: units, sizes and estimates · Practice problems · ProblemA leaking tap drips once a second, and each drop is 0.050mL. How many litres does it waste in a year of 365 days, and how many cubic metres is that? (1L=1000mL and 1m3=1000L.)Water wasted 1576.8 L; In cubic metres 1.5768 m³
    1. One drop a second, so the number of drops is the number of seconds in a year:
    2. 60×60×24×365=31536000≈3.15×107 drops
    3. Multiply by the volume of one drop, and convert:
    4. 3.15×107×0.050mL=1.58×106mL=1580L=1.58m3
    5. About 1580L, or 1.58m3, a year: about ten bathtubs, from one drip a second.
  16. 1. Measuring the world: units, sizes and estimates · Practice problems · ProblemA human hair is about 80μm wide, and an atom about 0.10nm across. Roughly how many atoms side by side span the width of the hair?800000
    1. In metres, the hair is 80×10−6m=8.0×10−5m and the atom 0.10×10−9m=1.0×10−10m.
    2. N=8.0×10−5m1.0×10−10m=8.0×10−5−(−10)=8.0×105
    3. About 8×105, nearly a million atoms across one hair. The metres cancel, leaving a plain count, as they should.
  17. 2. Motion in a line · Position is measured from an origin you choose · QuizYou walk 40 m east, then 60 m west. Taking east as positive, what are the distance and the displacement?B. Distance 100 m, displacement −20 mDistance counts every metre walked: 40+60=100m, and is never negative. Displacement is where you ended minus where you started: +40−60=−20m, 20 m west of the start. The sign carries the direction, which is exactly what distance throws away.
  18. 2. Motion in a line · Speed has no direction; velocity does · QuizA runner does one full lap of a 400 m track in 50 s and finishes where she started. What are her average speed and average velocity?B. Speed 8 m/s, velocity 0 m/sAverage speed is 400/50=8m/s. Her displacement is zero, because she ends where she began, so her average velocity is 0/50=0. It sounds odd, but it is what the definition says: average velocity only cares about where you started and ended, not the route.
  19. 2. Motion in a line · On a position–time graph, the slope is the velocity · QuizOn a position–time graph, a straight line sloping down means the object is:A. Moving in the negative direction at a constant speedA straight line has one slope, so the velocity is constant, and a downward slope makes it negative. The graph is not a picture of the path: "down" on the graph is not "down" in space. Slowing down would show as a curve that gets flatter, and standing still is a horizontal line, wherever it sits.
  20. 2. Motion in a line · Acceleration is how fast the velocity changes · QuizPositive is to the right. A car drives to the left and brakes, slowing from 15 m/s to 5 m/s in 2 s. What is its acceleration?B. +5m/s2Moving left, its velocities are negative: −15 then −5m/s. So a=−5−(−15)2=102=+5m/s2. It is braking, yet the acceleration is positive, because it points right, against the leftward motion. Writing every velocity with its sign before subtracting is what avoids the tempting −5.
  21. 2. Motion in a line · Acceleration is how fast the velocity changes · ProblemA ball is rolled up a ramp at 6.0m/s. 4.0s later it is rolling back down at 2.0m/s. Taking up the ramp as positive, what was its average acceleration? Give it with its sign.a= -2 m/s²
    1. With up the ramp positive, the velocities are vi=+6.0m/s and vf=−2.0m/s.
    2. a=vf−viΔt=−2.0−(+6.0)4.0=−8.04.0=−2.0m/s2
    3. The acceleration is −2.0m/s2, down the ramp the whole time: it slows the ball on the way up and speeds it up on the way down. Using speeds without signs, (2−6)/4=−1, is the trap.
  22. 2. Motion in a line · On a velocity–time graph, the area is the displacement · QuizA cyclist’s velocity rises steadily from 2 m/s to 8 m/s over 4 s. How far does she travel in that time?B. 20 mThe area under a straight line from 2 to 8 over 4 s is a trapezium: the average of the two velocities times the time, 2+82×4=20m. Taking the final speed for the whole time gives 32 m, too much, because she was slower for most of it. 24 m is 6×4, the change of velocity times the time, which has the units of a distance but is not the one travelled.
  23. 2. Motion in a line · Four equations for constant acceleration · Fill the blanksPut the same example in code: the final velocity, then the displacement.1: a * t, 2: t * tv=v0+at is v0 + a * t, and 12at2 is 0.5 * a * t * t: C++ has no power operator, so a square is t * t or std::pow(t, 2). A common slip is 0.5 * a * t, forgetting to square the time: that gives 13.5, and the units say why it is wrong, since m/s2×s is a speed, not a distance. Another is writing the half as 1 / 2: both are ints, so it is 0, and so is the distance.
  24. 2. Motion in a line · Four equations for constant acceleration · ProblemA plane starts from rest and accelerates steadily at 2.5m/s2 along the runway. It needs 75m/s to take off. How long must the runway be, at least, and how long does the take-off run last?Runway length 1125 m; Time 30 s
    1. With v0=0, v=75m/s and a=2.5m/s2, use the equation without time:
    2. v2=v02+2aΔx⇒Δx=v2−v022a=752−02×2.5=56255=1125m
    3. Then the time, from v=v0+at:
    4. t=v−v0a=75−02.5=30s
    5. Check with the fourth equation: 12(0+75)(30)=1125m. The runway must be at least 1125m long, and the run takes 30s.
  25. 2. Motion in a line · Stopping distance grows with the square of speed · Code challengeWrite stoppingDistance(speed, reactionTime, braking): the distance in metres a car at speed (m/s) covers while the driver reacts for reactionTime (s) and then brakes with a deceleration of size braking (m/s², a positive number).One solution:
    Solution
    double stoppingDistance(double speed, double reactionTime,                        double braking) {  double thinking = speed * reactionTime;  double brakingDistance = speed * speed / (2 * braking);  return thinking + brakingDistance;}
    Thinking distance is 20×1=20m, braking distance 202/(2×5)=40m, 60 m in all. The usual mistake is v/(2a) instead of v2/(2a); checking units catches it, since m/sm/s2 is seconds. Also watch the brackets: speed * speed / 2 * braking divides by 2 and then multiplies by braking.
  26. 2. Motion in a line · Free fall is constant acceleration downward · QuizA ball is thrown straight up. At the very top of its flight, what are its velocity and acceleration?B. Velocity zero, acceleration 9.8 m/s² downwardAt the top the ball stops for an instant, so v=0. But gravity has not switched off: the velocity is changing from upward to downward, and a changing velocity is an acceleration, still 9.8m/s2 down. If the acceleration were zero there, the ball would stay hanging in the air. Zero velocity and zero acceleration are different things.
  27. 2. Motion in a line · Step motion through time, the way simulations do · Put in orderPut one step of semi-implicit Euler in order.1. Work out the acceleration from the current state 2. Add acceleration × Δt to the velocity 3. Add the new velocity × Δt to the position 4. Add Δt to the timeThe acceleration depends on where things are now, so it comes first. Then velocity, then position using the updated velocity: that order is what makes it semi-implicit. Updating position with the old velocity (plain Euler) works for a falling stone but slowly gains energy in anything that oscillates, like a spring, so orbits spiral outwards in a game.
  28. 2. Motion in a line · Step motion through time, the way simulations do · Code challengeWrite simulateFall(v0, dt, steps): start at height y = 0 with velocity v0 (up is positive), take steps steps of semi-implicit Euler with a = -9.8, and return the final State, its y and v.One solution:
    Solution
    struct State {  double y;  // height, m  double v;  // velocity, m/s};
    State simulateFall(double v0, double dt, int steps) {  double y = 0;  double v = v0;  for (int i = 0; i < steps; i++) {    v += -9.8 * dt;    y += v * dt;  }  return {y, v};}
    Each step subtracts 9.8Δt from v and adds the new v * dt to y. The velocity comes out exact, since its change is constant, but the position does not: with 0.1 s steps the stone falls 5.39 m in a second instead of 4.9 m, because each step uses the speed at the end of the step for all of it. Shrinking dt shrinks the error in proportion; that trade between speed and accuracy is the heart of every simulation. A function returns one value, so the small State struct carries both, and return {y, v}; fills its fields in order.
  29. 2. Motion in a line · Practice problems · ProblemA ball is thrown straight up at 14.7m/s. Use g=9.8m/s2 and ignore air. How high above the hand does it rise, how long does it take to reach the top, and how long until it is back at the hand?Maximum height 11.025 m; Time to the top 1.5 s; Time back to the hand 3 s
    1. Up is positive: v0=14.7m/s, a=−9.8m/s2, and v=0 at the top. The equation without t gives the height:
    2. 0=14.72−2(9.8)Δx⇒Δx=216.0919.6≈11.0m
    3. 0=14.7−9.8t⇒t=1.5s
    4. Back at the hand, Δx=0:
    5. 0=14.7t−4.9t2=t(14.7−4.9t)⇒t=0 or t=3.0s
    6. It rises 11.0m, reaches the top after 1.5s and is back after 3.0s: the way down takes exactly as long as the way up.
  30. 2. Motion in a line · Practice problems · ProblemYou drop a stone into a well, and it hits the water 2.5s later. Use g=9.8m/s2, and ignore air and the time the sound takes to come back up. How far below you is the water, and how fast is the stone moving when it hits?Depth 30.625 m; Speed 24.5 m/s
    1. Down is positive, v0=0, a=9.8m/s2, t=2.5s:
    2. Δx=v0t+12at2=0+12(9.8)(2.5)2=4.9×6.25≈30.6m
    3. v=v0+at=0+9.8×2.5=24.5m/s
    4. The water is about 30.6m down, and the stone hits it at 24.5m/s. Sound, at about 340 m/s, takes some 0.09 s to come back up, so a real timing would make the well seem slightly deeper than it is.
  31. 2. Motion in a line · Practice problems · ProblemThe graph shows a tram’s velocity between two stops: it speeds up steadily from rest to 8.0m/s in the first 4.0s, cruises until t=10s, then slows steadily to rest at t=14s. Find its acceleration while speeding up, the distance between the stops, and its average velocity over the trip.Acceleration at first 2 m/s²; Distance 80 m; Average velocity 40/7 ≈ 5.71429 m/s
    1. The slope of the first part is the acceleration: a=8.0−04.0=2.0m/s2.
    2. The area under the graph is the displacement, in three pieces:
    3. Δx=12(4.0)(8.0)+(10−4.0)(8.0)+12(14−10)(8.0)=16+48+16=80m
    4. v‾=ΔxΔt=80m14s≈5.71m/s
    5. The tram accelerates at 2.0m/s2, the stops are 80m apart, and its average velocity is 5.71m/s, less than its cruising speed because of the slow start and finish.
  32. 2. Motion in a line · Practice problems · ProblemA car waits at a red light. As the light turns green, the car sets off with a constant acceleration of 2.0m/s2, and at that same moment a truck passes it at a steady 12m/s. How long does the car take to catch the truck, how far from the light does it catch it, and how fast is the car going then?Time 12 s; Distance 144 m; Car’s speed 24 m/s
    1. From the light, with t=0 when it turns green:
    2. xcar=12(2.0)t2=t2,xtruck=12t
    3. The car catches up when the positions match:
    4. t2=12t⇒t(t−12)=0⇒t=12s
    5. Then x=12×12=144m, and the car’s speed is v=2.0×12=24m/s.
    6. It catches the truck after 12s, 144m from the light, going 24m/s: twice the truck’s speed, so that its average speed over those 12 s matches the truck’s 12 m/s.
  33. 3. Vectors and motion in two dimensions · A vector is a size and a direction · QuizWhich of these is a vector?C. A wind of 30 km/h from the north-westThe wind has a size (30 km/h) and a direction (from the north-west), so it is a velocity, a vector. Speed is the magnitude of a velocity with the direction thrown away, so it is a scalar, as are temperature and mass: there is no sense in which 12 kg points anywhere.
  34. 3. Vectors and motion in two dimensions · Break a vector into components · ProblemA plane flies at 250m/s, heading 40° north of east. Taking east as x and north as y, find the components of its velocity.vx= 191.511 m/s; vy= 160.697 m/s
    1. The angle is 40° above the x axis, so:
    2. vx=vcos⁡θ=250cos⁡40°=250×0.766≈192m/svy=vsin⁡θ=250sin⁡40°=250×0.643≈161m/s
    3. Check with Pythagoras: 191.52+160.72=250m/s. The plane moves about 192m/s east and 161m/s north.
  35. 3. Vectors and motion in two dimensions · Add vectors tip to tail, or by components · Code challengeWrite components(size, angleDeg), returning a Vec with the x and y of a vector of that magnitude at angleDeg degrees above the x axis, and magnitude(x, y). std::cos and std::sin take radians, and std::numbers::pi is π.One solution:
    Solution
    #include <cmath>#include <numbers>
    struct Vec {  double x;  double y;};
    Vec components(double size, double angleDeg) {  double angle = angleDeg * std::numbers::pi / 180;  return {size * std::cos(angle), size * std::sin(angle)};}
    double magnitude(double x, double y) {  return std::sqrt(x * x + y * y);}
    Degrees become radians by ×π/180; skipping that is the classic mistake, and std::sin(30) happily returns −0.988, the sine of 30 radians. std::hypot(x, y) also works for the magnitude. The last test is the useful check: going to components and back must give the size you started with, whatever the angle.
  36. 3. Vectors and motion in two dimensions · Add vectors tip to tail, or by components · ProblemA hiker walks 3.0km east, then 5.0km in a direction 60° north of west. Taking east as x and north as y, how far is she from where she started, and in what direction, in degrees north of east?Distance sqrt(19) ≈ 4.3589 km; Direction 83.4132 ° north of east
    1. The two walks as components, in km:
    2. A→=(3.0, 0),B→=(−5.0cos⁡60°, 5.0sin⁡60°)=(−2.5, 4.33)
    3. R→=A→+B→=(0.5, 4.33)
    4. |R→|=0.52+4.332=19.0≈4.36km,θ=tan−1⁡4.330.5≈83.4°
    5. She ends up 4.36km from the start, 83.4° north of east: almost due north, though she walked 8 km.
  37. 3. Vectors and motion in two dimensions · Velocity depends on who is watching · QuizThat boat, steering straight across at 4 m/s in a 3 m/s current, crosses a river 60 m wide. How long does the crossing take?B. 15 sOnly the across component of the velocity carries the boat across: 60/4=15s. The current pushes it sideways, 3×15=45m downstream, but does not speed up the crossing. Dividing by the 5 m/s resultant gives 12 s, but the boat then also covers a longer, slanted path, 5×15=75m, so the time is the same 15 s.
  38. 3. Vectors and motion in two dimensions · A projectile is two motions at once · QuizFrom the same height, at the same moment, one ball is dropped and another is fired horizontally at 50 m/s. Ignoring air, which lands first?C. They land at the same timeBoth start with zero vertical velocity and both fall with ay=−g, so their heights match at every instant and they land together. The fired ball also travels sideways, but horizontal motion does not change vertical motion. People expect the fast ball to "stay up" longer; that is how it looks, not how it falls.
  39. 3. Vectors and motion in two dimensions · Forty-five degrees goes farthest · QuizAt the same speed, which goes farther on flat ground: a ball launched at 30° or at 60°?C. They land the same distance awaysin⁡(2×30°)=sin⁡60° and sin⁡(2×60°)=sin⁡120° are equal, so the ranges are equal. The 60° ball goes higher and stays up longer, but moves sideways more slowly, and the two effects cancel exactly. Mass does not appear in any of the formulas: without air, a heavy ball and a light one fly the same path.
  40. 3. Vectors and motion in two dimensions · Simulate a projectile step by step · Find the bugThis gives nonsense ranges. Which line is wrong?Line 4: return speed * speed * std::sin(2 * angle) / g;Line 3 converts to radians, and line 4 then uses the degrees anyway: std::sin(90) is the sine of 90 radians, about 0.89, so a 45° kick at 20 m/s comes out as 36.5 m instead of 40.8 m, and other angles are even worse. It is wrong without crashing, the worst kind of bug. Naming variables with their unit, like angleDeg, is what makes this one visible.
  41. 3. Vectors and motion in two dimensions · Simulate a projectile step by step · Code challengeWrite simulateRange(speed, angleDeg, dt): launch from (0, 0), step with semi-implicit Euler (update vy by -9.8 * dt, then x and y by the velocities times dt) until y is no longer above 0, and return x.One solution:
    Solution
    #include <cmath>#include <numbers>
    double simulateRange(double speed, double angleDeg, double dt) {  double angle = angleDeg * std::numbers::pi / 180;  double x = 0;  double y = 0;  const double vx = speed * std::cos(angle);  double vy = speed * std::sin(angle);  do {    vy -= 9.8 * dt;    x += vx * dt;    y += vy * dt;  } while (y > 0);  return x;}
    vx never changes, so it is a const double; only vy feels gravity. A do … while takes the first step before testing, which matters because y starts at 0 and a while (y > 0) loop would never run. With 1 ms steps it lands within about a centimetre of the formula. Add a drag term to the accelerations (module 5) and the same loop handles what no formula can.
  42. 3. Vectors and motion in two dimensions · Moving in a circle means accelerating towards the centre · QuizThe same car takes the same curve at twice the speed. What happens to its centripetal acceleration?C. It becomes four times as bigThe speed is squared in v2/r, so doubling it multiplies the acceleration by 22=4. That is why taking a bend too fast is so dangerous: a little more speed asks a lot more of the tyres’ grip. Module 5 looks at the force that provides this acceleration.
  43. 3. Vectors and motion in two dimensions · Moving in a circle means accelerating towards the centre · ProblemA child sits 2.0m from the centre of a playground merry-go-round that turns steadily once every 4.0s. What is her speed, and what is her centripetal acceleration?Speed pi ≈ 3.14159 m/s; Centripetal acceleration pi^2/2 ≈ 4.9348 m/s²
    1. Speed is the circumference over the period:
    2. v=2πrT=2π×2.0m4.0s=πm/s≈3.14m/s
    3. ac=v2r=π22.0≈4.93m/s2
    4. She moves at 3.14m/s with a centripetal acceleration of 4.93m/s2 towards the centre, about half of g, even though her speed never changes.
  44. 3. Vectors and motion in two dimensions · Practice problems · ProblemA ball rolls off the edge of a 20m high cliff, moving horizontally at 15m/s. Use g=9.8m/s2 and ignore air. How long does it take to land, how far from the foot of the cliff does it land, and how fast is it moving when it lands?Time 2.02031 s; Distance from the cliff 30.3046 m; Speed on landing sqrt(617) ≈ 24.8395 m/s
    1. Vertically it starts with vy=0 and falls 20m:
    2. 20=12(9.8)t2⇒t=409.8≈2.02s
    3. Horizontally it keeps 15m/s all the way: x=15×2.02≈30.3m.
    4. On landing, vy=9.8×2.02≈19.8m/s downward, so
    5. v=vx2+vy2=152+19.82=225+392=617≈24.8m/s
    6. It lands after 2.02s, 30.3m out, at 24.8m/s. Adding the speeds, 15+19.8, is the slip: they are at right angles.
  45. 3. Vectors and motion in two dimensions · Practice problems · ProblemA football is kicked from the ground at 25m/s, 35° above the horizontal, and lands back on the ground. Use g=9.8m/s2 and ignore air. Find its time of flight, its range and its greatest height.Time of flight 2.92641 s; Range 59.9294 m; Greatest height 10.4908 m
    1. The components: vx=25cos⁡35°≈20.5m/s and vy=25sin⁡35°≈14.3m/s.
    2. T=2v0sin⁡θg=2×14.39.8≈2.93sR=vxT=20.5×2.93≈59.9mH=(v0sin⁡θ)22g=14.3219.6≈10.5m
    3. Check the range with R=v02sin⁡2θ/g=625sin⁡70°/9.8≈59.9m. The ball flies for 2.93s, lands 59.9m away and rises 10.5m.
  46. 3. Vectors and motion in two dimensions · Practice problems · ProblemA swimmer swims at 1.5m/s through still water. She wants to cross a river 60m wide that flows at 0.90m/s, and land directly opposite her start. At what angle upstream of straight across must she head, in degrees, and how long does the crossing take?Angle upstream 36.8699 °; Time 50 s
    1. Heading at angle θ upstream, the upstream part of her swimming, 1.5sin⁡θ, must cancel the current:
    2. 1.5sin⁡θ=0.90⇒sin⁡θ=0.60⇒θ≈36.9°
    3. What is left carries her across:
    4. vacross=1.5cos⁡θ=1.5×0.80=1.2m/s,t=60m1.2m/s=50s
    5. She heads 36.9° upstream and crosses in 50s. Heading straight across would take only 40 s, but land her 36 m downstream.
  47. 3. Vectors and motion in two dimensions · Practice problems · ProblemThe Moon goes round the Earth on a nearly circular path of radius 3.84×108m, once every 27.3 days. Find its speed and its centripetal acceleration.Speed 1022.9 m/s; Centripetal acceleration 0.00272482 m/s²
    1. The period in seconds: T=27.3×86400≈2.36×106s.
    2. v=2πrT=2π×3.84×1082.36×106≈1.02×103m/s
    3. ac=v2r=(1023)23.84×108≈2.72×10−3m/s2
    4. The Moon moves at about 1.02km/s, with a centripetal acceleration of 2.72×10−3m/s2. That is g/3600, and the Moon is 60 Earth radii away, with 602=3600: the clue that led Newton to gravity’s inverse-square law (module 9).
  48. 4. Forces and Newton's laws · A force is a push or a pull · QuizWhich of these forces acts without the objects touching?B. The pull of a magnet on a paperclip a few centimetres awayA magnet pulls a paperclip across a gap: a non-contact force, like gravity. Friction, the floor’s push and a rope’s pull all need contact. (Look closely enough and contact forces are electric forces between atoms that come very close, but in mechanics the split is useful.)
  49. 4. Forces and Newton's laws · With no net force, motion does not change · QuizA hockey puck slides across perfectly smooth ice at 10 m/s. Ignoring air, what keeps it moving?B. Nothing: no force is needed to keep a constant velocityOnce it leaves the stick, nothing pushes it forward, and nothing needs to. With no net force its velocity stays the same, so it slides on at 10 m/s in a straight line. The stick’s force acted only while they touched. There is no "force of motion" stored inside; that is the Aristotle idea dressed up. Gravity pulls down, and the ice pushes up equally: they cancel.
  50. 4. Forces and Newton's laws · Forces add as vectors into one net force · QuizIn a tug of war, one team pulls left with 1200 N and the other right with 1150 N. What is the net force on the rope?B. 50 N to the leftTake right as positive: −1200+1150=−50N, so 50 N to the left, and the rope (and both teams) start to accelerate left. Adding the sizes, 2350 N, ignores direction. The net force is zero only when the pulls match exactly.
  51. 4. Forces and Newton's laws · The second law: force equals mass times acceleration · Code challengeForces along a line are numbers (positive one way, negative the other). Write netForce(forces), their sum, and acceleration(forces, mass), the net force divided by the mass.One solution:
    Solution
    #include <vector>
    double netForce(const std::vector<double>& forces) {  double sum = 0;  for (double f : forces) sum += f;  return sum;}
    double acceleration(const std::vector<double>& forces, double mass) {  return netForce(forces) / mass;}
    Add every force with its sign, then divide by the mass. Reusing netForce inside acceleration keeps the physics in one place. The mistake the second law guards against is using one force, the engine’s 4000 N, instead of the net force: that would give 3.3m/s2 and ignore everything pushing back.
  52. 4. Forces and Newton's laws · The second law: force equals mass times acceleration · ProblemYou push a 25kg shopping trolley forward with 60N, while friction pushes back on it with 15N. What is its acceleration, and how fast is it moving after 4.0s if it starts from rest?Acceleration 1.8 m/s²; Speed after 4.0 s 7.2 m/s
    1. With forward positive, the net force and the acceleration are
    2. Fnet=60−15=45N,a=Fnetm=45N25kg=1.8m/s2
    3. The acceleration is constant, so module 2’s equation gives the speed:
    4. v=v0+at=0+1.8×4.0=7.2m/s
    5. The trolley accelerates at 1.8m/s2 and reaches 7.2m/s, a fast run. Using the 60 N push alone would give 2.4m/s2.
  53. 4. Forces and Newton's laws · Mass is how much stuff; weight is gravity’s pull on it · QuizAn astronaut has a mass of 80 kg on Earth. On the Moon, where g=1.62N/kg, what are her mass and weight?A. Mass 80 kg, weight about 130 NMass does not change with place: still 80 kg, and still as hard to push sideways. Weight is mg=80×1.62≈130N, a sixth of her 784 N on Earth. Everyday language says "I weigh 80 kilos", which mixes the two; in physics, kilograms are mass and newtons are weight.
  54. 4. Forces and Newton's laws · Free-body diagrams show every force on one object · Put in orderPut the steps for solving a force problem in order.1. Choose the one object you are asking about 2. Draw it alone and add an arrow for every force acting on it 3. Choose axes, ideally one along the acceleration 4. Add the components of the forces along each axis 5. Set each sum equal to mass times the acceleration along that axis, and solvePick the object first: the forces on a book and on the table are different lists. Then draw every force on it, choose axes, and only then write ∑Fx=max and ∑Fy=may. Choosing an axis along the acceleration makes the other sum zero, which is often the easiest equation to start with.
  55. 4. Forces and Newton's laws · The normal force is whatever stops you going through the floor · QuizThe same 60 kg person stands on scales in a lift that accelerates downward at 2.0m/s2. What do the scales read?A. 468 NWith up positive, a=−2.0m/s2, so N=m(g+a)=60×7.8=468N. The person feels lighter. Their weight is still 588 N; only the floor’s push has changed. If the cable snapped, a=−g and the scales would read 0: the "weightlessness" of free fall, which astronauts in orbit feel all the time.
  56. 4. Forces and Newton's laws · The normal force is whatever stops you going through the floor · ProblemA 75kg person stands on bathroom scales in a lift, and the scales read 660N. Use g=9.8m/s2. Taking up as positive, what is the lift’s acceleration? Give it with its sign.a= -1 m/s²
    1. Two forces act on the person: the normal force N up and the weight mg=75×9.8=735N down.
    2. N−mg=ma⇒a=N−mgm=660−73575=−7575=−1.0m/s2
    3. The acceleration is −1.0m/s2, downward. The lift is either starting to go down or slowing on its way up: the scales cannot tell which, only the acceleration.
  57. 4. Forces and Newton's laws · Tension pulls along a rope · Code challengeWrite pulledBlocks(force, front, back): a force pulls a block of mass front, roped to a block of mass back behind it, on a frictionless floor. Return a Result holding a, their shared acceleration, and tension, the rope’s tension.One solution:
    Solution
    struct Result {  double a;        // m/s²  double tension;  // N};
    Result pulledBlocks(double force, double front, double back) {  double a = force / (front + back);  double tension = back * a;  return {a, tension};}
    The pull accelerates both blocks, so a=F/(m1+m2); dividing by the front mass alone is the starter’s mistake, and it forgets the block being dragged. The back block’s only horizontal force is the rope, so T=m2a. Check with the front block: F−T=m1a, 30−10=4×5. The tension is always less than the pull, since the pull also accelerates the front block.
  58. 4. Forces and Newton's laws · Forces come in pairs on different objects · QuizA truck hits a fly on the motorway. Which feels the bigger force?C. Both feel forces of exactly the same sizeThe third law says the forces are equal and opposite, always. What differs is the effect: by the second law, the same force on the fly’s tiny mass is a colossal acceleration, and on the truck’s an unnoticeable one. "Same force, very different acceleration" is the whole answer.
  59. 4. Forces and Newton's laws · Equilibrium: when the forces balance · Fill the blanksA 10 kg sign hangs from two ropes at 30° above the horizontal. Finish the code for the tension.1: g, 2: sinThe weight is mg=98N, and the two vertical components Tsin⁡30°=T/2 must add to it, so T=98N: each rope carries as much as the whole sign weighs. It is sin because the angle is measured from the horizontal and we want the vertical part; with cos, you would get the horizontal part and 56.6 N, too little.
  60. 4. Forces and Newton's laws · Practice problems · ProblemTwo horizontal forces act on a 4.0kg box on smooth ice: 12N east and 9.0N north. Find the size of the net force, the size of the box’s acceleration, and its direction, in degrees north of east.Net force 15 N; Acceleration 3.75 m/s²; Direction 36.8699 ° north of east
    1. The net force has components (12,9.0)N:
    2. Fnet=122+9.02=225=15N,a=15N4.0kg=3.75m/s2
    3. θ=tan−1⁡9.012≈36.9°
    4. The box accelerates at 3.75m/s2, 36.9° north of east, along the net force. Adding the sizes, 12+9=21N, ignores their directions.
  61. 4. Forces and Newton's laws · Practice problems · ProblemThree blocks sit in a row on a frictionless floor, joined by light ropes: 5.0kg at the front, 3.0kg in the middle and 2.0kg at the back. A 40N pull on the front block drags them all forward. Find the acceleration, the tension in the rope between the front and middle blocks, and the tension in the rope between the middle and back blocks.Acceleration 4 m/s²; Front rope 20 N; Back rope 8 N
    1. As one object of 10kg: a=405.0+3.0+2.0=4.0m/s2.
    2. The front rope is the only horizontal force on the middle and back blocks together (5.0kg):
    3. T1=(3.0+2.0)×4.0=20N
    4. The back rope pulls only the back block:
    5. T2=2.0×4.0=8.0N
    6. Check the middle block: T1−T2=3.0×4.0, and 20−8=12. So a=4.0m/s2, with tensions 20N and 8.0N: less the further back the rope.
  62. 4. Forces and Newton's laws · Practice problems · ProblemA 12kg street lamp hangs from the middle of a cable, each half of which is at 40° above the horizontal. Use g=9.8m/s2. What is the tension in each half of the cable?T= 91.4766 N
    1. Vertically, the two upward components balance the weight:
    2. 2Tsin⁡40°=mg⇒T=12×9.82sin⁡40°=117.61.286≈91.5N
    3. Horizontally, the two halves pull equally in opposite directions, so that sum is zero by symmetry.
    4. Each half carries about 91.5N. Flatten the cable and the tension climbs: at 10° it would be about 339N.
  63. 4. Forces and Newton's laws · Practice problems · ProblemA 0.145kg baseball arrives at 40m/s, and the catcher’s glove moves back 0.20m as it stops the ball. Assume a constant force, and ignore the ball’s weight during the short catch. What is the size of the ball’s deceleration, and of the force on it from the glove?Deceleration 4000 m/s²; Force 580 N
    1. From v2=v02+2aΔx, with v=0:
    2. 0=402+2a(0.20)⇒a=−16000.40=−4000m/s2
    3. F=m|a|=0.145×4000=580N
    4. The ball decelerates at 4000m/s2, about 400 g, under a 580N force. Letting the glove give more, over a longer distance, makes the force smaller: catchers pull their hands back for exactly this reason.
  64. 5. Friction, springs, slopes and circles · Friction holds until it can’t, then slides · QuizA 10 kg box rests on a floor with μs=0.5. You push it sideways with 20 N and it does not move. How big is the friction force?B. 20 NThe box is in equilibrium, so friction exactly balances your push: 20 N. The 49 N, μsN=0.5×98, is only the largest static friction can be, reached just before the box moves. Reaching for μsN whenever you see static friction is the usual mistake; it is a ceiling, not a value.
  65. 5. Friction, springs, slopes and circles · Pushing a box: friction in the second law · Code challengeWrite boxAcceleration(push, mass, muS, muK): a horizontal push (N) on a box of mass (kg) on a level floor, with g = 9.8. If the push is not more than the largest static friction, return 0; otherwise return its acceleration with kinetic friction.One solution:
    Solution
    double boxAcceleration(double push, double mass, double muS,                       double muK) {  double normal = mass * 9.8;  if (push <= muS * normal) return 0;  return (push - muK * normal) / mass;}
    Two regimes, two formulas: below the static limit, friction cancels the push and a=0; above it, kinetic friction is μkN and the net force is the push minus that. A common bug is using μs for the moving box, or subtracting friction even when the box is still, which gives a negative acceleration for a gentle push: a box that pushes back on its own.
  66. 5. Friction, springs, slopes and circles · Pushing a box: friction in the second law · ProblemYou pull a 15kg sled across level snow with a horizontal rope, pulling with 40N. The coefficient of kinetic friction is 0.12. Use g=9.8m/s2. Once the sled is sliding, what is the friction force on it, and what is its acceleration?Friction 17.64 N; Acceleration 1.49067 m/s²
    1. The normal force is the weight, and kinetic friction is μkN:
    2. N=mg=15×9.8=147N,fk=μkN=0.12×147≈17.6N
    3. a=40−17.615≈1.49m/s2
    4. Friction is 17.6N, and the sled accelerates at 1.49m/s2.
  67. 5. Friction, springs, slopes and circles · On a slope, split gravity along and across it · QuizWith no friction, what is the acceleration down a slope of 90°, a vertical wall?B. g, free fallgsin⁡90°=g: a vertical "slope" is just falling, which is reassuring. At 0° it gives gsin⁡0°=0, a flat floor. Checking extreme cases like these is a quick test of any formula: if gcos⁡θ had sneaked in by mistake, the wall would give zero.
  68. 5. Friction, springs, slopes and circles · On a slope, split gravity along and across it · ProblemA 5.0kg block slides down a ramp tilted at 25°, with a coefficient of kinetic friction of 0.30. Use g=9.8m/s2. Find the normal force on the block and its acceleration down the ramp.Normal force 44.4091 N; Acceleration 1.47711 m/s²
    1. Across the ramp nothing accelerates, so the normal force balances the weight’s component into the ramp:
    2. N=mgcos⁡25°=5.0×9.8×0.906≈44.4N
    3. Along the ramp, the pull down it minus friction:
    4. ma=mgsin⁡25°−μkN=20.7−0.30×44.4=20.7−13.3=7.39Na=7.395.0≈1.48m/s2
    5. The normal force is 44.4N, less than the weight of 49N, and the block accelerates at 1.48m/s2. The shortcut a=g(sin⁡θ−μkcos⁡θ) gives the same, without the mass.
  69. 5. Friction, springs, slopes and circles · Tilt until it slips to measure friction · QuizA phone on a tilted book starts to slide when the book is at 25°. What is μs between them?B. About 0.47μs=tan⁡25°≈0.47. sin⁡25°≈0.42 is the tempting wrong function. The phone’s mass does not matter, because both the pull down the slope and the friction are proportional to it.
  70. 5. Friction, springs, slopes and circles · An Atwood machine: two masses and a pulley · Code challengeWrite atwood(m1, m2) for an Atwood machine with g = 9.8: return a Result holding the acceleration a, positive when m1 falls, and the rope’s tension.One solution:
    Solution
    struct Result {  double a;        // m/s², positive when m1 falls  double tension;  // N};
    Result atwood(double m1, double m2) {  double a = (m1 - m2) * 9.8 / (m1 + m2);  double tension = 2 * m1 * m2 * 9.8 / (m1 + m2);  return {a, tension};}
    The difference in weights drives both masses, which is why the total mass is on the bottom. A frequent mistake is assuming the tension equals the heavier weight; then the heavy mass would not accelerate at all. The tension sits between the two weights, and equals each when they balance. Atwood used this in 1784 to slow free fall down enough to measure g with a clock.
  71. 5. Friction, springs, slopes and circles · Hooke’s law: a spring pulls back in proportion · Fill the blanksA 2 N weight stretches a spring by 4 cm. Find k, then how far a 5 N weight stretches it.1: force / stretch, 2: 5 / kk=F/x=2/0.04=50N/m, then x=F/k=5/50=0.1m=10cm. Leaving the stretch in centimetres, 2/4, gives k=0.5 in N/cm, which is fine on its own but wrong by 100 as soon as it meets SI units. Hooke’s law holds only up to the elastic limit: stretch too far and the spring stays bent.
  72. 5. Friction, springs, slopes and circles · Drag grows with speed until it balances weight · QuizA skydiver is falling at terminal velocity. Which statement is true?B. The drag equals the weight, so the net force and the acceleration are zeroConstant velocity means zero acceleration, and by the second law zero net force: drag up exactly equals weight down. Both forces are still large (about 784 N each). If drag were bigger, the skydiver would slow down, which is what happens for a moment when the parachute opens.
  73. 5. Friction, springs, slopes and circles · Simulate a fall with drag · Code challengeWrite fallWithDrag(mass, k, dt, steps): start from rest, take steps steps of size dt, each with acceleration 9.8 - (k / mass) * v * v (down is positive), and return the final speed v.One solution:
    Solution
    double fallWithDrag(double mass, double k, double dt, int steps) {  double v = 0;  for (int i = 0; i < steps; i++) {    double a = 9.8 - (k / mass) * v * v;    v += a * dt;  }  return v;}
    The acceleration must be recalculated inside the loop, from the current speed; computing it once before the loop is the classic mistake, and gives free fall forever. As v approaches mg/k, the acceleration shrinks to zero by itself: the simulation finds terminal velocity without being told it. For a body moving up, drag points down, and you would write -k * v * std::abs(v) to keep its direction right: std::abs(v) keeps the size of v2 and v brings its sign.
  74. 5. Friction, springs, slopes and circles · Something must supply the centripetal force · QuizA car takes a flat, dry bend at a steady speed. What force points towards the centre of the bend?B. Static friction between the tyres and the roadThe tyres grip the road sideways without sliding over it, so it is static friction, even though the car moves. The engine’s force points forward, along the motion. "Centrifugal force" is what passengers seem to feel, thrown outward, but from the road there is no such force: their bodies are just carrying on in a straight line (the first law) while the car turns under them.
  75. 5. Friction, springs, slopes and circles · Going over the top of a loop · QuizWhat is the slowest speed at which a car can pass the top of a loop of radius 10 m without leaving the track?B. About 9.9 m/svmin=gr=9.8×10=98≈9.9m/s. Forgetting the square root gives 98, and 9.8 alone gives 3.1. Any slower and gravity pulls the car down faster than the circle curves away: it falls off the track. Riders at exactly this speed feel weightless at the top, since N=0.
  76. 5. Friction, springs, slopes and circles · Practice problems · ProblemA 6.0kg mass and a 4.0kg mass hang on either side of a light, frictionless pulley, joined by one rope, and are let go from rest. Use g=9.8m/s2. Find their acceleration, the tension in the rope, and how far the heavier mass falls in the first 1.5s.Acceleration 1.96 m/s²; Tension 47.04 N; Distance fallen 2.205 m
    1. For the 6.0 kg mass (down positive) and the 4.0 kg mass (up positive):
    2. 6.0g−T=6.0aT−4.0g=4.0a
    3. Adding them removes T:
    4. 2.0g=10a⇒a=2.0×9.810=1.96m/s2
    5. Then T=4.0(9.8+1.96)=47.04N, between the two weights, 39.2 N and 58.8 N.
    6. Δx=12at2=12(1.96)(1.5)2≈2.21m
    7. The masses accelerate at 1.96m/s2, the rope carries 47.0N, and the heavier mass falls 2.21m in 1.5 s.
  77. 5. Friction, springs, slopes and circles · Practice problems · ProblemA spring hangs from a hook. Hanging a 0.60kg mass on it stretches it by 3.0cm. Use g=9.8m/s2. What is its spring constant, and how far, in centimetres, does a 1.5kg mass stretch it?k= 196 N/m; Stretch 7.5 cm
    1. At rest, the spring force equals the weight. With x=0.030m:
    2. k=mgx=0.60×9.80.030=5.880.030=196N/m
    3. x=mgk=1.5×9.8196=14.7196=0.075m=7.5cm
    4. The spring constant is 196N/m, and the heavier mass stretches it 7.5cm: 2.5 times the mass, 2.5 times the stretch, as Hooke’s law says.
  78. 5. Friction, springs, slopes and circles · Practice problems · ProblemA 1400kg car takes a flat curve of radius 80m at a steady 20m/s. Use g=9.8m/s2. What sideways friction force must the road supply, and what is the smallest coefficient of static friction that can provide it?Friction force 7000 N; Smallest μs 0.510204
    1. Friction must supply all of the centripetal force:
    2. f=mv2r=1400×20280=7000N
    3. Static friction can reach μsmg, so it must be that large at least:
    4. μs≥fmg=70001400×9.8≈0.51
    5. The road must supply 7000N, which needs μs≥0.51: fine on dry tarmac, impossible on ice.
  79. 5. Friction, springs, slopes and circles · Practice problems · ProblemA curve of radius 120m is banked so that cars going round at 25m/s need no friction at all. Use g=9.8m/s2. At what angle, in degrees, is the road banked?θ= 27.989 °
    1. Square v=rgtan⁡θ and solve for the angle:
    2. tan⁡θ=v2rg=252120×9.8=6251176≈0.531
    3. θ=tan−1⁡0.531≈28.0°
    4. The road is banked at about 28°. Slower cars would slide down the bank without friction, and faster ones up it.
  80. 5. Friction, springs, slopes and circles · More practice: drag and loops · ProblemA 75kg skydiver has a terminal velocity of 55m/s. Take the drag as Fd=kv2, with k fixed, and g=9.8m/s2. Find k, then the drag force and the skydiver’s acceleration at the moment they are falling at 30m/s.k= 0.242975 kg/m; Drag at 30 m/s 218.678 N; Acceleration at 30 m/s 6.8843 m/s²
    1. At terminal velocity the forces balance:
    2. k=mgvt2=75×9.8552=7353025≈0.243kg/m
    3. At 30 m/s, drag is smaller than the weight:
    4. Fd=0.243×302≈219N,a=735−21975≈6.88m/s2
    5. So k≈0.243kg/m, and at 30 m/s the drag is 219N and the skydiver still accelerates at 6.88m/s2, less than g and shrinking as the speed climbs.
  81. 5. Friction, springs, slopes and circles · More practice: drag and loops · ProblemA roller-coaster car passes the top of a circular loop of radius 12m at 15m/s, its riders upside down. Use g=9.8m/s2. What is the size of the normal force from the seat on a 60kg rider there, and what is the slowest speed at which the car could pass the top without leaving the track?Normal force 537 N; Slowest speed sqrt(117.6) ≈ 10.8444 m/s
    1. Towards the centre (down), the two forces together supply the centripetal force:
    2. N+mg=mv2r⇒N=60×15212−60×9.8=1125−588=537N
    3. With N=0, gravity alone bends the path:
    4. vmin=gr=9.8×12=117.6≈10.8m/s
    5. The seat pushes on the rider with 537N, a little less than their weight, and the car must pass the top at 10.8m/s or more.
  82. 6. Work and energy · Energy is never made or destroyed, only moved · QuizA car brakes to a stop. Where does its energy of motion go?B. Mostly into thermal energy in the brakes, tyres and roadEnergy is never destroyed: friction in the brake pads turns the motion into thermal energy, which is why brakes get hot (and some into sound). An ordinary car cannot get it back. Electric and hybrid cars use regenerative braking: the motor runs as a generator and stores part of that energy in the battery instead.
  83. 6. Work and energy · Work is force times distance along the force · QuizYou carry a heavy bag across a level room at constant velocity. How much work does the upward force of your hand do on the bag?B. Zero, because the force is at right angles to the motionYour hand pushes straight up and the bag moves sideways: θ=90° and cos⁡90°=0, so the work is zero. You get tired because muscles burn energy just to hold a force (their fibres keep twitching), but none of it goes into the bag. In physics, "work" means energy transferred to the object, not effort.
  84. 6. Work and energy · Work is force times distance along the force · ProblemYou pull a sled 15m across level snow with a rope, pulling with 80N at 25° above the horizontal. Friction on the sled is a steady 50N. Find the work done by your pull, the work done by friction (with its sign), and the net work done on the sled.By your pull 1087.57 J; By friction -750 J; Net work 337.569 J
    1. Your pull is at 25° to the motion:
    2. Wpull=Fdcos⁡θ=80×15×cos⁡25°≈1088J
    3. Friction is at 180° to the motion, and cos⁡180°=−1:
    4. Wfriction=50×15×(−1)=−750J
    5. The weight and the normal force are at right angles to the motion, so they do no work. The net work is the sum:
    6. Wnet=1088−750≈338J
    7. Your pull does 1088J, friction −750J, and the net 338J is what speeds the sled up (next slide).
  85. 6. Work and energy · Kinetic energy, and the work–energy theorem · QuizThe same car, with the same brakes, travels at 40 m/s instead of 20 m/s. How far does it need to stop?C. 100 mKinetic energy goes as v2, so twice the speed is four times the energy, 800 kJ, and the same braking force needs four times the distance: 100 m. Expecting 50 m, twice the distance, is the intuition the square law breaks, and why speed limits near schools are so low.
  86. 6. Work and energy · Kinetic energy, and the work–energy theorem · ProblemA 1200kg car slows from 25m/s to 10m/s. How much work do the brakes do on it? Give it with its sign. If the braking force is a steady 6000N, over what distance does the car slow down?Work by the brakes -315000 J; Distance 52.5 m
    1. The work done is the change in kinetic energy:
    2. W=12mvf2−12mvi2=12(1200)(102−252)=600×(−525)=−315000J
    3. −Fd=−315000J⇒d=3150006000=52.5m
    4. The brakes do −315kJ of work over 52.5m. Slowing from 25 to 10 m/s takes away 84% of the kinetic energy, though the speed only fell by 60%.
  87. 6. Work and energy · Potential energy is stored work · Fill the blanksA toy launcher’s spring (k=200N/m) is squashed by 10 cm and fires a 50 g ball. Find the stored energy and the launch speed.1: x * x, 2: 2 * stored / massUs=12(200)(0.1)2=1J. Setting 12mv2=1J gives v=2×1/0.05=40≈6.32m/s. The two usual slips: forgetting to square x (giving 10 J, ten times too much), and writing the mass as 50, in grams, which gives a speed of 0.2 m/s.
  88. 6. Work and energy · Without friction, kinetic plus potential stays constant · Code challengeWrite speedAfterDrop(startSpeed, drop): something moving at startSpeed (m/s) descends a height drop (m) with no friction, g = 9.8. Return its speed at the bottom. (It does not matter which way it was moving.)One solution:
    Solution
    #include <cmath>
    double speedAfterDrop(double startSpeed, double drop) {  // 0.5 v² = 0.5 v0² + g h; the mass cancels  return std::sqrt(startSpeed * startSpeed + 2 * 9.8 * drop);}
    Energy at the top, 12mv02+mgh, equals energy at the bottom, 12mv2; divide by m and solve: v=v02+2gh. Adding speeds instead, v0+2gh, is the common mistake: it is the energies that add, not the speeds. For the coaster, 25+588=613≈24.8m/s, not 5+24.2.
  89. 6. Work and energy · Energy ignores the shape of the path · QuizTwo children start from rest at the top of two frictionless water slides, both 6 m high. One is straight and steep; the other is long and winding. Which child is faster at the bottom?C. Both have the same speedBoth lose mgh of potential energy and gain the same 12mv2, so v=2×9.8×6≈10.8m/s for both, whatever they weigh. The steep one gets there sooner, but speed and time are different questions. The normal force from the slide does no work, because it always acts at right angles to the motion.
  90. 6. Work and energy · Friction turns mechanical energy into heat · QuizA ball dropped from 2.0 m bounces back to 1.5 m. What fraction of its mechanical energy did the bounce lose?A. 25%At the tops of both flights the ball is still, so all its mechanical energy is mgh. It went from mg(2.0) to mg(1.5): a quarter less, 25%. The lost energy is not gone; it warmed the ball and the floor and made the sound of the bounce. Total energy is conserved; mechanical energy is not, once friction or squashing is involved.
  91. 6. Work and energy · Power is how fast energy is moved · Code challengeWrite inputPower(mass, height, seconds, efficiency): the power (W) that must be supplied to lift mass (kg) by height (m) in seconds, by a machine that turns only the fraction efficiency (between 0 and 1) of its input into lifting. Use g = 9.8.One solution:
    Solution
    double inputPower(double mass, double height, double seconds,                  double efficiency) {  double useful = mass * 9.8 * height / seconds;  return useful / efficiency;}
    The useful power is the energy mgh over the time; a machine that wastes some must take in more, so divide by the efficiency. Multiplying by it instead is the slip to watch for: then a worse machine would need less power, which no real one does. Your muscles are about 25% efficient, which is why running up stairs makes you hot: three-quarters of the energy becomes heat.
  92. 6. Work and energy · Power is how fast energy is moved · ProblemA pump lifts 500kg of water every minute through a height of 12m. Use g=9.8m/s2. What useful power does it deliver, and how much electrical power does it draw if it is 70% efficient?Useful power 980 W; Power drawn 1400 W
    1. Each minute, the water gains mgh=500×9.8×12=58800J. Per second:
    2. Puseful=58800J60s=980W
    3. Only 70% of the power drawn does the lifting, so
    4. Pin=Puseful0.70=9800.70=1400W
    5. The pump delivers 980W and draws 1400W; the other 420W becomes heat and noise.
  93. 6. Work and energy · Efficiency: how much of the energy does the job · QuizAn electric winch uses 500 J of electrical energy to lift a 10 kg box by 4.0 m. What is its efficiency?B. About 78%The useful energy is mgh=10×9.8×4.0=392J, so η=392/500≈0.78, 78%. Dividing the wrong way round, 500/392, gives 128%, which is impossible and a sure sign of the slip. 8% comes from forgetting g (40/500): mass times height is not an energy.
  94. 6. Work and energy · Practice problems · ProblemA roller-coaster car starts from rest at the top of a 45m hill. Ignore friction and use g=9.8m/s2. How fast is it moving at ground level, and at the top of the next hill, 30m high?At ground level 29.6985 m/s; On the 30 m hill 17.1464 m/s
    1. Kinetic energy gained equals potential energy lost, and the mass cancels: v=2gΔh.
    2. Ground, Δh=45m:v=2×9.8×45=882≈29.7m/sNext hill, Δh=15m:v=2×9.8×15=294≈17.1m/s
    3. It reaches 29.7m/s at the bottom and still has 17.1m/s on the 30 m hill, whatever the shape of the track in between.
  95. 6. Work and energy · Practice problems · ProblemA toy’s spring, with k=800N/m, is squashed by 0.15m and then fires a 0.20kg ball straight up. Ignore air and use g=9.8m/s2. Find the energy stored in the spring, and how high above its starting point the ball rises. How high would a 0.40kg ball rise?Energy stored 9 J; Height, 0.20 kg 4.59184 m; Height, 0.40 kg 2.29592 m
    1. Us=12kx2=12(800)(0.15)2=9.0J
    2. At the top, all of it is mgh, measured from where the ball started:
    3. h=Usmg=9.00.20×9.8≈4.59m
    4. The same 9.0 J lifts a ball twice as heavy half as high: h=9.0/(0.40×9.8)≈2.30m.
    5. The spring stores 9.0J; the 0.20 kg ball rises 4.59m, and a 0.40 kg ball would rise 2.30m.
  96. 6. Work and energy · Practice problems · ProblemA 40kg child slides from rest down a playground slide that is 3.0m high and 5.0m long, and reaches the bottom at 6.0m/s. Use g=9.8m/s2. How much mechanical energy did friction turn into heat, and what was the average friction force along the slide?Energy lost 456 J; Average friction 91.2 N
    1. Etop=mgh=40×9.8×3.0=1176J,Ebottom=12(40)(6.0)2=720J
    2. The difference went to heat: 1176−720=456J, so Wnc=−456J.
    3. f×5.0m=456J⇒f=91.2N
    4. Friction turned 456J into heat, with an average force of 91.2N along the slide. The length is what matters for friction’s work, not the height.
  97. 6. Work and energy · Practice problems · ProblemA 2.0kg block slides from rest down a frictionless ramp 1.8m high, then out onto a rough level floor, where the coefficient of kinetic friction is 0.25. Use g=9.8m/s2. How fast is it moving at the bottom of the ramp, and how far does it slide across the floor before it stops?Speed at the bottom 5.9397 m/s; Distance on the floor 7.2 m
    1. Down the frictionless ramp, mgh becomes kinetic energy:
    2. v=2gh=2×9.8×1.8=35.28≈5.94m/s
    3. On the floor, friction μkmg does work −μkmgd, which must remove all of that energy, mgh:
    4. μkmgd=mgh⇒d=hμk=1.80.25=7.2m
    5. The block reaches 5.94m/s and slides 7.2m. Its mass and g cancel: any block from 1.8 m stops 7.2 m out on this floor.
  98. 7. Momentum and collisions · Momentum is mass times velocity · QuizWhich has more momentum: a 1500 kg car rolling at 1 m/s, or a 10 g bullet flying at 400 m/s?A. The car: 1500 kg·m/s against 4 kg·m/sThe car has 1500×1=1500kg·m/s; the bullet 0.010×400=4kg·m/s (10 g is 0.010 kg, the slip behind 4000). Yet the bullet has more kinetic energy: 12(0.010)(400)2=800J against the car’s 750 J. Momentum grows with v, energy with v2, so fast light things carry energy and slow heavy things carry momentum.
  99. 7. Momentum and collisions · Impulse: a force over time changes momentum · QuizIn a crash, how does an airbag protect your head?B. It makes the stop take longer, so the same impulse needs a smaller forceYour head goes from moving to stopped either way, so the change in momentum, and the impulse, are fixed. The airbag stretches the stop from a few milliseconds against the dashboard to tens of milliseconds, and since F=Δp/Δt, the force falls by the same factor. Crumple zones, bike helmets and bending your knees on landing all work the same way.
  100. 7. Momentum and collisions · Impulse: a force over time changes momentum · ProblemA 58g tennis ball arrives at a racket at 30m/s and leaves at 40m/s, straight back the way it came, after 5.0ms of contact. Find the size of the impulse on the ball and the average force of the racket on it.Impulse 4.06 kg·m/s; Average force 812 N
    1. With the outgoing direction positive, vi=−30m/s and vf=+40m/s:
    2. J=Δp=m(vf−vi)=0.058×(40−(−30))=0.058×70=4.06kg·m/s
    3. Favg=ΔpΔt=4.060.0050s=812N
    4. The impulse is 4.06kg·m/s and the average force 812N, over a thousand times the ball’s weight. Forgetting the reversal, 0.058×10, gives a seventh of the right answer.
  101. 7. Momentum and collisions · In an isolated system, total momentum never changes · QuizA 500 kg cannon on frictionless wheels fires a 5 kg ball forward at 200 m/s. How fast does the cannon roll backwards?B. 2 m/sBefore firing, the total momentum is zero, so afterwards 500v+5×200=0, giving v=−2m/s. The ball’s 1000 kg·m/s forward is matched by the cannon’s 1000 kg·m/s backward. The explosion is an internal force, so it cannot change the total. Kinetic energy is another story: the ball gets 100 kJ and the cannon only 1 kJ.
  102. 7. Momentum and collisions · In an isolated system, total momentum never changes · ProblemA 70kg skater stands at rest on smooth ice and throws a 3.0kg ball forward at 8.0m/s. How fast does she slide backwards?Her speed 12/35 ≈ 0.342857 m/s
    1. Take forward as positive. The total momentum stays zero:
    2. 0=70v+3.0×8.0⇒v=−2470≈−0.343m/s
    3. She slides backwards at 0.343m/s: the ball’s 24kg·m/s forward is matched by hers backward, spread over a mass more than 23 times bigger.
  103. 7. Momentum and collisions · Objects that stick together lose kinetic energy · Code challengeWrite stickTogether(m1, v1, m2, v2) for two objects on a line that collide and stick. Return a Collision holding v, their shared velocity, and energyLost, the kinetic energy lost (J).One solution:
    Solution
    struct Collision {  double v;           // shared velocity afterwards, m/s  double energyLost;  // J};
    Collision stickTogether(double m1, double v1, double m2, double v2) {  double v = (m1 * v1 + m2 * v2) / (m1 + m2);  double before = 0.5 * m1 * v1 * v1 + 0.5 * m2 * v2 * v2;  double after = 0.5 * (m1 + m2) * v * v;  return {v, before - after};}
    Averaging the velocities, as the starter does, is only right for equal masses: momentum weights each velocity by its mass. Signs matter too: in the last test the momenta cancel, so the pair stops and all 18 J of kinetic energy is lost. That is the most a collision can lose while conserving momentum, and why head-on crashes are the worst.
  104. 7. Momentum and collisions · Elastic collisions keep the kinetic energy too · Code challengeWrite elastic(m1, v1, m2, v2) for an elastic collision on a line. Return the two velocities afterwards, as a Velocities with fields v1 and v2.One solution:
    Solution
    struct Velocities {  double v1;  // m/s  double v2;  // m/s};
    Velocities elastic(double m1, double v1, double m2, double v2) {  double total = m1 + m2;  return {((m1 - m2) * v1 + 2 * m2 * v2) / total,          ((m2 - m1) * v2 + 2 * m1 * v1) / total};}
    Each formula is the other with 1 and 2 swapped, which is a good check that it is typed right. The last test checks the physics rather than numbers: kinetic energy before equals after. A neat fact behind the formulas: in an elastic collision the speed at which the balls separate equals the speed at which they approached, v2′−v1′=v1−v2.
  105. 7. Momentum and collisions · Elastic, inelastic, perfectly inelastic · QuizTwo cars collide and lock together. Comparing just before with just after the crash, what is conserved?B. Momentum onlyDamage does not affect momentum: the crash forces are internal, and in the instant of the collision friction from the road is small next to them. So total momentum is the same. Kinetic energy is not, since the cars stick, which is the most inelastic case there is. Crash investigators use exactly this to work out how fast cars were going from where they ended up.
  106. 7. Momentum and collisions · In two dimensions, each component is conserved · QuizOn a pool table, a moving ball hits an identical ball at rest, off-centre, in an (almost) elastic collision. What is the angle between their paths afterwards?C. 90°With equal masses, momentum conservation says v→1=v→1′+v→2′, and energy conservation says v12=v1′2+v2′2. The only way both hold is if the two final velocities form a right angle, by Pythagoras. Pool players use it: after a clean cut, the cue ball leaves at 90° to the object ball (ignoring spin).
  107. 7. Momentum and collisions · The centre of mass moves as if all the mass were there · Code challengeWrite centreOfMass(masses, positions): given matching vectors of masses and positions on a line, return the position of their centre of mass.One solution:
    Solution
    #include <vector>
    double centreOfMass(const std::vector<double>& masses,                    const std::vector<double>& positions) {  double total = 0;  double weighted = 0;  for (std::size_t i = 0; i < masses.size(); i++) {    total += masses[i];    weighted += masses[i] * positions[i];  }  return weighted / total;}
    Multiply each position by its mass, add them up, and divide by the total mass. Dividing by the number of objects instead gives the plain average, 2 for the second test instead of 1, which ignores that the 3 kg mass matters three times as much. The loop runs an index i over both vectors at once, from 0 to masses.size() - 1. The last test allows 5 km, 5e3 m: a tolerance is in the quantity’s own units, so astronomical numbers need one to match their size.
  108. 7. Momentum and collisions · A rocket pushes itself by throwing mass backwards · QuizWhy do rockets that reach orbit use several stages?B. Dropping empty tanks and engines means the rest of the burn does not accelerate dead weightThe logarithm is cruel: doubling the fuel adds only veln⁡2 to Δv. Once a stage’s fuel is used, its tanks and engines are pure dead mass. Throwing them away gives each later stage a good mass ratio of its own, and their $\Delta v$s add up. That is how a mass ratio no single structure could survive is achieved in practice.
  109. 7. Momentum and collisions · A rocket pushes itself by throwing mass backwards · ProblemA rocket in deep space, far from any planet, has an exhaust speed of 2500m/s. It starts at 50000kg and burns 200kg of fuel per second until its mass is 20000kg. Find its thrust, how long the burn lasts, and how much its speed changes.Thrust 500000 N; Burn time 150 s; Δv 2500*ln(2.5) ≈ 2290.73 m/s
    1. thrust=ve×ΔmΔt=2500m/s×200kg/s=5.0×105N
    2. It burns 50000−20000=30000kg of fuel, at 200kg/s: 150s.
    3. Δv=veln⁡m0mf=2500ln⁡5000020000=2500ln⁡2.5≈2290m/s
    4. The thrust is 500kN for 150s, and the rocket gains about 2290m/s, less than its exhaust speed, though 60% of it was fuel.
  110. 7. Momentum and collisions · Practice problems · ProblemA 12000kg railway wagon rolling at 3.0m/s runs into a stationary 8000kg wagon, and they couple together. Find their speed just afterwards, and the kinetic energy lost in the coupling.Speed afterwards 1.8 m/s; Kinetic energy lost 21600 J
    1. v=m1v1+m2v2m1+m2=12000×3.0+020000=1.8m/s
    2. Kbefore=12(12000)(3.0)2=54000J,Kafter=12(20000)(1.8)2=32400J
    3. The wagons move off at 1.8m/s, and 54000−32400=21600J of kinetic energy goes into the couplings as heat, sound and strain: 40% of it.
  111. 7. Momentum and collisions · Practice problems · ProblemOn a smooth track, a 0.50kg cart moving at 4.0m/s hits a 1.5kg cart at rest, in an elastic collision. Taking the first cart’s direction as positive, find both velocities afterwards, with their signs.Light cart, v1′= -2 m/s; Heavy cart, v2′= 2 m/s
    1. With m1=0.50, m2=1.5, v1=4.0 and v2=0:
    2. v1′=(m1−m2)v1m1+m2=(0.50−1.5)(4.0)2.0=−2.0m/sv2′=2m1v1m1+m2=2(0.50)(4.0)2.0=2.0m/s
    3. Check momentum: 0.50×4.0=2.0 before, 0.50(−2.0)+1.5(2.0)=2.0 after. Kinetic energy: 4.0J before, 1.0+3.0=4.0J after.
    4. The light cart bounces back at 2.0m/s, and the heavy one moves off forward at 2.0m/s.
  112. 7. Momentum and collisions · Practice problems · ProblemTo measure a bullet’s speed, it is fired into a 2.0kg wooden block hanging from strings. The 10g bullet stays in the block, which swings up until it is 0.12m higher than where it started. Use g=9.8m/s2. How fast did the block (with the bullet in it) move just after the hit, and how fast was the bullet going before it?Block just after 1.53362 m/s; Bullet before 308.258 m/s
    1. During the swing, kinetic energy becomes potential energy, and the mass cancels:
    2. V=2gh=2×9.8×0.12=2.352≈1.53m/s
    3. During the hit, momentum is conserved (kinetic energy is not: the bullet tears the wood):
    4. mv=(m+M)V⇒v=0.010+2.00.010×1.53≈308m/s
    5. The block moves off at 1.53m/s, and the bullet was doing about 308m/s. Using energy conservation across the hit would be wrong: over 99% of the bullet’s kinetic energy goes into heat and torn wood.
  113. 7. Momentum and collisions · Practice problems · ProblemAt a crossroads, a 1500kg car heading east at 20m/s collides with a 2500kg van heading north at 12m/s, and they lock together. Just after the crash, how fast do they move, and in what direction, in degrees north of east?Speed 7.5*sqrt(2) ≈ 10.6066 m/s; Direction 45 ° north of east
    1. Momentum before, east and north, in kg·m/s:
    2. px=1500×20=30000,py=2500×12=30000
    3. After, the 4000kg wreck has the same components:
    4. vx=300004000=7.5m/s,vy=7.5m/s,v=7.52+7.52≈10.6m/s
    5. The components are equal, so the direction is tan−1⁡1=45°. The wreck moves off at 10.6m/s, 45° north of east: investigators work backwards from exactly this.
  114. 8. Rotation and torque · Angles, angular velocity and angular acceleration · QuizA washing machine drum spins at 1200 revolutions per minute. What is its angular velocity?B. About 126 rad/s1200 rev/min is 20 rev/s, and each revolution is 2π radians: 20×2π≈126rad/s. 20 forgets the 2π; 7540 forgets to divide by 60. The same unit-conversion habit from module 1: write the factors out, 1200revmin×2πradrev×1min60s.
  115. 8. Rotation and torque · The equations of motion, for spinning · ProblemA grinding wheel spinning at 3000 revolutions per minute is switched off and slows steadily to rest in 20s. Find its starting angular velocity in rad/s, the size of its angular deceleration, and how many turns it makes while stopping.ω0= 100*pi ≈ 314.159 rad/s; Deceleration 5*pi ≈ 15.708 rad/s²; Turns 500
    1. ω0=3000revmin×2πradrev×1min60s=100π≈314rad/s
    2. α=ω−ω0t=0−100π20=−5π≈−15.7rad/s2
    3. Δθ=ω0t+12αt2=100π(20)−12(5π)(20)2=1000πrad=1000π2π=500 turns
    4. It starts at 314rad/s, decelerates at 15.7rad/s2 and makes 500 turns: the average rate, 1500 rev/min, for a third of a minute.
  116. 8. Rotation and torque · Rolling without slipping: the bottom is still · QuizA bicycle moves along the road at 5 m/s. How fast is the top of its tyre moving, relative to the road?C. 10 m/sThe top moves forward with the centre (5 m/s) and forward with the spin (another ωR=5m/s): 10 m/s. The bottom moves forward at 5 and backward at 5, so it is still. Look at a photo of a moving bike: the top spokes are blurred and the bottom ones sharp. The size cancels, since ωR=v for any wheel that rolls.
  117. 8. Rotation and torque · Torque is the turning effect of a force · QuizA bolt needs 60 N·m to loosen. Pushing square-on, what force does that take with a 0.20 m spanner, and with a 0.50 m one?A. 300 N and 120 NF=τ/r: 60/0.20=300N and 60/0.50=120N. A longer lever arm needs less force for the same torque, which is why mechanics slip a pipe over a spanner for a stubborn bolt. 12 and 30 come from multiplying instead of dividing, and they get the trend backwards.
  118. 8. Rotation and torque · Torque is the turning effect of a force · ProblemYou push on a door at its handle, 0.80m from the hinges, with a force of 30N at 60° to the door’s surface. What torque do you apply about the hinges? What force, pushing square-on at the same place, would give the same torque?Torque 12*sqrt(3) ≈ 20.7846 N·m; Square-on force 15*sqrt(3) ≈ 25.9808 N
    1. τ=rFsin⁡θ=0.80×30×sin⁡60°≈20.8N·m
    2. Square-on, sin⁡90°=1, so the force needed is
    3. F=τr=20.80.80≈26.0N
    4. You apply 20.8N·m; a square-on push of only 26.0N would do the same. The slanted part of your push presses along the door towards the hinges and turns nothing.
  119. 8. Rotation and torque · Moment of inertia: mass, and where it sits · Code challengeWrite momentOfInertia(masses, radii): for small masses at the given distances from an axis, return I=∑mr2.One solution:
    Solution
    #include <vector>
    double momentOfInertia(const std::vector<double>& masses,                       const std::vector<double>& radii) {  double total = 0;  for (std::size_t i = 0; i < masses.size(); i++) {    total += masses[i] * radii[i] * radii[i];  }  return total;}
    Each mass contributes mr2: the radius is squared, and forgetting to square it gives 6 instead of 18 in the first test. Distances are from the axis, so a mass on the axis adds nothing at all. The table’s formulas are this sum done with calculus over a solid shape: a hoop is MR2 because all its mass is at R; a disc has mass nearer the middle, so it is less.
  120. 8. Rotation and torque · Moment of inertia: mass, and where it sits · ProblemA flywheel is a solid disc of mass 20kg and radius 0.25m, turning about its centre. A motor applies a steady torque of 5.0N·m to it, starting from rest. Ignoring friction, find its moment of inertia, its angular acceleration, and its angular velocity after 8.0s.I= 0.625 kg·m²; α= 8 rad/s²; ω after 8.0 s 64 rad/s
    1. I=12MR2=12(20)(0.25)2=0.625kg·m2
    2. α=τI=5.00.625=8.0rad/s2,ω=ω0+αt=0+8.0×8.0=64rad/s
    3. The flywheel has I=0.625kg·m2, speeds up at 8.0rad/s2, and reaches 64rad/s, about 10 turns a second, after 8 s.
  121. 8. Rotation and torque · A rolling object has two kinds of kinetic energy · QuizA hoop, a solid disc and a solid sphere are let go together at the top of the same slope and roll without slipping. Which reaches the bottom first?C. The sphereThe sphere has the smallest k, 25, so the least of its energy goes into spinning and the most into moving down the slope. The hoop, with all its mass at the rim (k=1), comes last, sharing its energy half and half. Mass and size do not matter: a marble beats a bowling ball’s hollow cousin, and ties with a solid bowling ball.
  122. 8. Rotation and torque · Try it: the speed of a rolling object · Code challengeWrite rollingSpeed(height, k): the speed (m/s) at the bottom of a slope of height (m) for an object rolling from rest without slipping, whose moment of inertia is k * m * R * R. Use g = 9.8; k = 0 is a block sliding without friction.One solution:
    Solution
    #include <cmath>
    double rollingSpeed(double height, double k) {  return std::sqrt(2 * 9.8 * height / (1 + k));}
    From mgh=12mv2(1+k), v=2gh/(1+k). The brackets matter: std::sqrt(2 * 9.8 * height / 1 + k) divides by 1 and then adds k. The starter is the sliding answer, which is right only for k=0, and too fast for anything that spins.
  123. 8. Rotation and torque · Angular momentum is conserved: spin faster by pulling in · QuizA spinning skater pulls her arms in and her moment of inertia halves. What happens to her angular velocity and her rotational kinetic energy?C. Her angular velocity doubles; her kinetic energy doublesL=Iω is conserved, so halving I doubles ω. Kinetic energy is 12Iω2=L22I, so halving I doubles it. The extra energy is not free: her muscles do work pulling her arms inward against their tendency to fly outwards. Angular momentum is conserved; kinetic energy is not.
  124. 8. Rotation and torque · Balance needs zero net force and zero net torque · Fill the blanksA 30 kg child sits 2.0 m from a seesaw’s pivot. Where must a 40 kg child sit, on the other side, to balance it?1: childDistance, 2: otherMass * gTorque about the pivot is weight times distance: 30×9.8×2.0=588N·m. The other child needs the same, so d=588/(40×9.8)=1.5m. The g cancels, so 30×2.0=40×d would do. Without the brackets, torque / otherMass * g divides by the mass and then multiplies by g, giving 144 m.
  125. 8. Rotation and torque · A spinning top turns sideways instead of falling · QuizYou hold a fast-spinning bicycle wheel by one end of its axle, axle horizontal, and let go of the other end. What happens?B. It stays horizontal and slowly swings round, sideways, about your handGravity’s torque points horizontally, at right angles to the wheel’s angular momentum. So each moment it nudges L→ sideways, and the axle swings round in a horizontal circle instead of dropping. A wheel that is not spinning has L=0, and the same torque simply tips it over. As friction slows the spin, L shrinks and the precession speeds up until it finally droops.
  126. 8. Rotation and torque · Practice problems · ProblemA solid ball rolls from rest, without slipping, down a ramp 2.5m high. Use g=9.8m/s2. How fast is it moving at the bottom? How fast would a hoop be moving, rolling down the same ramp?Solid ball sqrt(35) ≈ 5.91608 m/s; Hoop sqrt(24.5) ≈ 4.94975 m/s
    1. With I=kmR2 and ω=v/R, energy conservation gives v=2gh/(1+k).
    2. Ball, k=25:v=2×9.8×2.51.4=35≈5.92m/sHoop, k=1:v=2×9.8×2.52=24.5≈4.95m/s
    3. The ball reaches 5.92m/s and the hoop only 4.95m/s, because half the hoop’s energy goes into spinning. A frictionless sliding block would reach 7.0m/s.
  127. 8. Rotation and torque · Practice problems · ProblemA skater spins at 1.5 revolutions per second with her arms out, when her moment of inertia is 3.6kg·m2. She pulls her arms in, and it drops to 1.2kg·m2. Ignoring friction with the ice, how fast does she now spin, in revolutions per second, and by what factor does her rotational kinetic energy grow?New spin rate 4.5 rev/s; Energy grows by a factor of 3
    1. Angular momentum is conserved:
    2. I1ω1=I2ω2⇒ω2=3.6×1.51.2=4.5 rev/s
    3. With L fixed, K=L22I, so the energy grows by the factor I shrank by:
    4. K2K1=I1I2=3.61.2=3
    5. She spins at 4.5 revolutions per second, with three times the kinetic energy, all of it supplied by her arms pulling in.
  128. 8. Rotation and torque · Practice problems · ProblemA uniform plank 4.0m long, of mass 20kg, rests on two supports, one under each end. A 60kg painter stands on it 1.0m from the left end. Use g=9.8m/s2. Find the upward force from each support.Left support 539 N; Right support 245 N
    1. The weights: plank 20×9.8=196N at 2.0 m, painter 60×9.8=588N at 1.0 m. Torques about the left end, anticlockwise equal to clockwise:
    2. R×4.0=196×2.0+588×1.0=980⇒R=245N
    3. The vertical forces balance:
    4. L+R=196+588=784N⇒L=784−245=539N
    5. The left support pushes up with 539N and the right with 245N: the painter is nearer the left, so it takes more of the load.
  129. 8. Rotation and torque · Practice problems · ProblemA 2.0kg mass hangs from a light string wound round a pulley, a solid disc of mass 4.0kg and radius 0.10m, free to turn on a frictionless axle. The mass is let go. Use g=9.8m/s2. Find the mass’s acceleration and the tension in the string.Acceleration 4.9 m/s²; Tension 9.8 N
    1. For the mass (down positive) and for the pulley, with I=12MR2 and α=a/R:
    2. mg−T=maTR=Iα=12MR2⋅aR⇒T=12Ma=2.0a
    3. Put the second into the first:
    4. 2.0×9.8−2.0a=2.0a⇒a=19.64.0=4.9m/s2,T=2.0×4.9=9.8N
    5. The mass accelerates at 4.9m/s2, half of g, and the string carries 9.8N. The pulley behaves like an extra 12M=2.0kg being dragged along. Notice the radius cancels.
  130. 9. Gravity and orbits · Every mass attracts every other mass · QuizA space probe moves from 1 million km to 3 million km from a planet. By what factor does the planet’s pull on it change?B. It drops to a ninthThe force goes as 1/r2: three times the distance gives 1/32=1/9 of the force. This inverse-square law is the same rule light, sound and electric forces follow from a point source, because the effect spreads over a sphere whose area grows as r2. "A third" is the linear guess the square makes wrong.
  131. 9. Gravity and orbits · Little g comes from big G · QuizThe International Space Station orbits about 400 km up, where the Earth’s gravity is about 8.7m/s2. Why do astronauts there float?B. They and the station are both in free fall, falling together around the EarthAt 400 km, g=9.82×(6371/6771)2≈8.7m/s2: nearly 90% of its value on the ground. The astronauts float because they, the station and everything in it fall with the same acceleration, so nothing presses on anything, like the lift whose cable snapped in module 4. "Zero gravity" is a misleading name for free fall.
  132. 9. Gravity and orbits · Little g comes from big G · ProblemMars has a mass of 6.42×1023kg and a radius of 3.39×106m. With G=6.674×10−11N·m2/kg2, find the gravitational field strength at its surface, and the weight there of an astronaut whose mass is 80kg.gMars= 3.72839 m/s²; Weight 298.272 N
    1. g=GMR2=(6.674×10−11)(6.42×1023)(3.39×106)2=4.28×10131.15×1013≈3.73m/s2
    2. W=mg=80×3.73≈298N
    3. Mars’s surface gravity is 3.73m/s2, about 38% of the Earth’s, so the astronaut weighs 298N there, though her mass is still 80 kg.
  133. 9. Gravity and orbits · An orbit is falling and always missing · Code challengeWith G = 6.674e-11, write orbitalSpeed(mass, r), the speed (m/s) of a circular orbit of radius r (m) around a body of mass (kg), and orbitalPeriod(mass, r), the time (s) for one orbit.One solution:
    Solution
    #include <cmath>#include <numbers>
    const double G = 6.674e-11;
    double orbitalSpeed(double mass, double r) {  return std::sqrt(G * mass / r);}
    double orbitalPeriod(double mass, double r) {  return 2 * std::numbers::pi * r / orbitalSpeed(mass, r);}
    Setting gravity equal to the centripetal force gives v=GM/r, and the period is one circumference over the speed. The radius is from the centre of the Earth: using the 400 km altitude instead gives a speed of 31.6 km/s, absurdly fast. The Moon test checks the formula far from where it was tested: its real sidereal month is 27.3 days, the small gap being the Moon’s own mass, which the simple formula ignores.
  134. 9. Gravity and orbits · An orbit is falling and always missing · ProblemGPS satellites move on circular orbits of radius 2.66×107m around the Earth, for which GM=3.986×1014m3/s2. Find their orbital speed, and their period in hours.Speed 3871.04 m/s; Period 11.9931 h
    1. v=GMr=3.986×10142.66×107=1.50×107≈3870m/s
    2. T=2πrv=2π×2.66×1073870≈4.32×104s=4.32×1043600h≈12.0h
    3. They move at about 3.87km/s, half the space station’s speed, and go round in 12.0 hours: twice a day, by design.
  135. 9. Gravity and orbits · Farther planets take much longer · QuizAn asteroid orbits the Sun with a semi-major axis of 4 AU. What is its period?B. 8 yearsT2=a3=43=64, so T=64=8 years. 64 forgets the square root and 16 squares instead of cubing. Four times as far, but eight times as long: the asteroid has a longer path and moves more slowly, since v=GM/r shrinks with distance.
  136. 9. Gravity and orbits · Gravitational energy is negative, and escape means reaching zero · QuizAt the same distance from a planet, how does the escape velocity compare with the speed of a circular orbit?B. Escape is 2 times fastervesc=2GM/r and vorbit=GM/r, so the ratio is 2≈1.41. A spacecraft already in orbit needs only 41% more speed to leave for good, which is why missions to the Moon and planets start from a parking orbit. "Twice" is the slip of forgetting the square root over the 2.
  137. 9. Gravity and orbits · Try it: escape velocity · Code challengeWith G = 6.674e-11, write escapeSpeed(mass, r): the speed (m/s) needed to escape from distance r (m) of a body of mass (kg).One solution:
    Solution
    #include <cmath>
    const double G = 6.674e-11;
    double escapeSpeed(double mass, double r) {  return std::sqrt(2 * G * mass / r);}
    The starter is the orbital speed; escape needs twice the kinetic energy, so the 2 goes inside the root. The Moon’s low escape speed is why Apollo’s small ascent stage could leave it, and why it has no air: its gas molecules, warmed by the Sun, are fast enough to leak away. A black hole is a body so compact that 2GM/r exceeds the speed of light.
  138. 9. Gravity and orbits · Geostationary orbit: one turn a day · Fill the blanksSolve T2=4π2GMr3 for r in code.1: cbrt, 2: T * TRearranged, r3=GMT24π2, so r is a cube root: std::cbrt, from <cmath>. That puts geostationary satellites 42 164 km from the centre, about 35 800 km above the ground, more than five Earth radii up. Using std::sqrt instead gives about 270 million km, nearly twice the distance to the Sun: the power and the root must match.
  139. 9. Gravity and orbits · Tides come from the difference in the Moon’s pull · QuizWhy do most coasts get two high tides a day, not one?B. There are two bulges, on the sides facing towards and away from the Moon, and the Earth turns under bothThe near side is pulled towards the Moon more than the centre is, and the centre more than the far side, so the ocean stretches both ways along the Earth–Moon line. A spot on the turning Earth passes through both bulges each day. The Moon takes a month to go round, not half a day, and the high tides come about 12 hours 25 minutes apart because it moves along its orbit meanwhile.
  140. 9. Gravity and orbits · Simulate an orbit step by step · Code challengeWrite stepOrbit(s, dt) for a body around a mass at the origin, with GM = 1: work out the acceleration -x / r³, -y / r³ from the current position, update the velocity, then the position with the new velocity (semi-implicit Euler), and return the new State.One solution:
    Solution
    #include <cmath>
    struct State {  double x, y;    // position  double vx, vy;  // velocity};
    State stepOrbit(State s, double dt) {  double r = std::hypot(s.x, s.y);  double ax = -s.x / (r * r * r);  double ay = -s.y / (r * r * r);  double vx = s.vx + ax * dt;  double vy = s.vy + ay * dt;  return {s.x + vx * dt, s.y + vy * dt, vx, vy};}
    The acceleration is GM/r2 towards the origin; multiplying the unit vector −r→/r by 1/r2 gives −r→/r3. Writing r * r instead of r * r * r leaves an extra factor of r and is the usual bug. Using the new velocity for the position is what keeps the orbit closed: plain Euler (old velocity) spirals out, ending a lap 12% too far away with these steps.
  141. 9. Gravity and orbits · Practice problems · ProblemThe Earth (5.97×1024kg) and the Moon (7.35×1022kg) are 3.84×108m apart, centre to centre. With G=6.674×10−11N·m2/kg2, find the gravitational force between them, and the acceleration it gives the Moon.Force 198603000000000000000 N; Moon’s acceleration 0.00270208 m/s²
    1. F=GMmr2=(6.674×10−11)(5.97×1024)(7.35×1022)(3.84×108)2=2.93×10371.47×1017≈1.99×1020N
    2. a=Fm=1.99×10207.35×1022≈2.70×10−3m/s2
    3. The pull is about 1.99×1020N, giving the Moon 2.70×10−3m/s2: within 1% of the centripetal acceleration its orbit needs, from its period alone (module 3). Gravity is exactly what keeps it going round.
  142. 9. Gravity and orbits · Practice problems · ProblemMars takes 1.88 years to go round the Sun. Using Kepler’s third law in astronomical units and years, find the semi-major axis of its orbit. Then find the period, in years, of an asteroid whose semi-major axis is 2.5AU.Mars, a= 1.88^(2/3) ≈ 1.52325 AU; Asteroid, T= 2.5^1.5 ≈ 3.95285 years
    1. For Mars, solve T2=a3 for a:
    2. a=T2/3=1.882/3=1.8823=3.533≈1.52AU
    3. For the asteroid, solve for T:
    4. T=a3/2=2.53=15.6≈3.95 years
    5. Mars orbits at 1.52AU; the asteroid, in the belt between Mars and Jupiter, takes 3.95 years.
  143. 9. Gravity and orbits · Practice problems · ProblemA rocky planet has twice the Earth’s mass and 1.5 times its radius. The Earth’s surface gravity is 9.8m/s2 and its escape velocity 11.2km/s. Find the planet’s surface gravity and its escape velocity.Surface gravity 9.8*2/2.25 ≈ 8.71111 m/s²; Escape velocity 11.2*sqrt(2/1.5) ≈ 12.9326 km/s
    1. Surface gravity goes as M/R2:
    2. gg⊕︎=21.52=22.25≈0.889⇒g≈0.889×9.8≈8.71m/s2
    3. Escape velocity goes as M/R:
    4. vv⊕︎=21.5≈1.155⇒v≈1.155×11.2≈12.9km/s
    5. You would weigh a little less there, 8.71m/s2, yet leaving takes more, 12.9km/s: surface gravity and escape speed depend on mass and radius differently.
  144. 9. Gravity and orbits · Practice problems · ProblemThe Earth goes round the Sun on a nearly circular orbit of radius 1.496×1011m, once every 365.25 days. With G=6.674×10−11N·m2/kg2, find the Earth’s orbital speed and the mass of the Sun.Orbital speed 29785.7 m/s; Mass of the Sun 1.98866e+30 kg
    1. The period is T=365.25×86400≈3.156×107s.
    2. v=2πrT=2π×1.496×10113.156×107≈2.98×104m/s
    3. M=4π2r3GT2=4π2(1.496×1011)3(6.674×10−11)(3.156×107)2≈1.99×1030kg
    4. The Earth moves at about 29.8km/s, and the Sun’s mass comes out as 1.99×1030kg, over 300 000 Earths: weighed with nothing but a calendar and a distance.
  145. 10. Oscillations · Periodic motion repeats itself · QuizA resting heart beats 72 times a minute. What are its frequency and period?B. 1.2 Hz and about 0.83 s72 beats per minute is 72/60=1.2 beats per second, so f=1.2Hz, and T=1/f≈0.83s. 72 Hz forgets that hertz means per second. Frequency and period are always reciprocals: if one is bigger than 1, the other is smaller.
  146. 10. Oscillations · Fastest in the middle, hardest pulled at the ends · QuizA mass bounces on a spring in SHM. Where is its speed largest, and where is its acceleration largest?B. Speed is largest at the middle; acceleration is largest at the endsAt the ends the mass stops to turn round, so v=0, but the spring is stretched most, so the force and acceleration (−ω2x) are largest. At the middle the spring is relaxed: no force, no acceleration, and it has been speeding up all the way in, so it is fastest. Zero velocity with maximum acceleration is the same lesson as the ball at the top of its throw in module 2.
  147. 10. Oscillations · A mass on a spring: stiffer is faster, heavier is slower · Code challengeWrite springPeriod(mass, k), the period (s) of a mass (kg) on a spring of constant k (N/m), and springFrequency(mass, k), its frequency (Hz).One solution:
    Solution
    #include <cmath>#include <numbers>
    double springPeriod(double mass, double k) {  return 2 * std::numbers::pi * std::sqrt(mass / k);}
    double springFrequency(double mass, double k) {  return 1 / springPeriod(mass, k);}
    T=2πm/k: mass on top, because heavier is slower. Swapping it to k/m is the common mistake, and the units catch it: kg/(N/m)=s2=s, while k/m is in 1/s, an angular frequency. The square root also means you need four times the mass to double the period.
  148. 10. Oscillations · A mass on a spring: stiffer is faster, heavier is slower · QuizYou pull the same mass on the same spring twice as far before letting go. What happens to the period?C. It stays the sameThe period 2πm/k has no amplitude in it. Twice as far means twice the force at the ends, so it moves twice as fast and covers twice the distance in the same time. This independence is the defining feature of SHM, and it holds as long as Hooke’s law does.
  149. 10. Oscillations · A mass on a spring: stiffer is faster, heavier is slower · ProblemA 0.25kg mass on a spring oscillates with a period of 0.50s. Find the spring constant. If the amplitude is 4.0cm, what is the mass’s top speed?k= 4*pi^2 ≈ 39.4784 N/m; Top speed 0.16*pi ≈ 0.502655 m/s
    1. T2=4π2mk⇒k=4π2mT2=4π2×0.250.502=4π2≈39.5N/m
    2. ω=2πT=4π≈12.6rad/s,vmax=Aω=0.040×4π≈0.503m/s
    3. The spring constant is 39.5N/m, and the mass passes through the middle at 0.503m/s.
  150. 10. Oscillations · A pendulum keeps time with its length · Code challengeMeasuring a pendulum is a classic way to find g. Write gFromPendulum(length, period): solve T=2πL/g for g and return it (m/s²).One solution:
    Solution
    #include <numbers>
    double gFromPendulum(double length, double period) {  double pi = std::numbers::pi;  return 4 * pi * pi * length / (period * period);}
    Square both sides, T2=4π2L/g, then rearrange: g=4π2L/T2. A common slip is squaring only the π or only the T. In a lab, time 20 swings and divide by 20: the timing uncertainty of your reaction (module 1) is then shared over all of them, and since g depends on T2, its relative uncertainty is twice T’s.
  151. 10. Oscillations · A pendulum keeps time with its length · ProblemUse g=9.8m/s2. What length of simple pendulum has a period of 1.0s? And what length has a period of 2.0s?For 1.0 s 9.8/(4*pi^2) ≈ 0.248237 m; For 2.0 s 9.8/pi^2 ≈ 0.992948 m
    1. T2=4π2Lg⇒L=gT24π2
    2. L1s=9.8×1.024π2≈0.248m,L2s=9.8×2.024π2≈0.993m
    3. A 1 s pendulum is 0.248m long, and a 2 s one 0.993m: twice the period needs four times the length.
  152. 10. Oscillations · Big swings take longer · QuizA pendulum clock runs exactly on time. Over the years its swing grows from 5° to 15° (perhaps its spring was over-wound). What happens?B. It runs slightly slow, because wider swings take a little longerAmplitude-independence is only approximate. A 15° swing has a period about 0.4% longer than a 5° one, which is several minutes a day: the clock falls behind. Clockmakers keep the swing small and constant for exactly this reason. "Faster bob" is true, but it also travels further, and the extra distance wins.
  153. 10. Oscillations · Energy sloshes between kinetic and potential · QuizThe same spring and mass oscillate with twice the amplitude. What happens to the total energy and the top speed?B. Energy ×4; top speed doublesTotal energy is 12kA2, so doubling A quadruples it. The top speed is Aω, and ω does not change, so it doubles; check: 12m(2v)2 is four times 12mv2, matching the energy. The unchanged period is the third piece of the same picture: twice as far at twice the speed.
  154. 10. Oscillations · Damping makes oscillations die away · QuizA car’s shock absorbers damp its springs. What damping do engineers aim for?C. Close to critical, so it settles quickly without bouncingLight damping leaves the car bobbing after every bump, which is uncomfortable and loses tyre grip; worn shock absorbers do exactly this. Very heavy damping makes the suspension so slow that it cannot follow the road, and bumps come straight through. Near-critical damping settles fastest. Door closers and measuring needles are tuned the same way.
  155. 10. Oscillations · Simulate a damped spring · Put in orderPut one step of the damped spring simulation in order.1. Work out the force from the current position and velocity: −kx − cv 2. Divide by the mass to get the acceleration 3. Add acceleration × dt to the velocity 4. Add the new velocity × dt to the positionThe force depends on the current state, so it comes first, then a=F/m. Updating the velocity before the position is semi-implicit Euler, which keeps an undamped spring’s energy steady. Updating the position first, with the old velocity, makes even an undamped spring’s amplitude grow a little every cycle, so it seems to gain energy from nowhere.
  156. 10. Oscillations · Simulate a damped spring · Code challengeWrite simulateSpring(m, k, c, x0, dt, steps): a mass m on a spring k with damping c, let go from rest at x0. Take steps steps of semi-implicit Euler with acceleration (-k * x - c * v) / m, and return the final position.One solution:
    Solution
    double simulateSpring(double m, double k, double c, double x0,                      double dt, int steps) {  double x = x0;  double v = 0;  for (int i = 0; i < steps; i++) {    double a = (-k * x - c * v) / m;    v += a * dt;    x += v * dt;  }  return x;}
    Recompute the acceleration every step from both x and v: the damping term needs the current velocity. The first two tests check against the formula for T, so they test the physics too. The last uses the envelope Ae−γt with γ=c/2m=2s−1: after a second the amplitude must be below 0.1e−2≈0.0135m. Forgetting the - c * v passes the first two and fails the last.
  157. 10. Oscillations · Resonance: push at the natural frequency · QuizA child’s swing has a natural period of 3 s. To make it go highest with gentle pushes, how often should you push?B. Every 3 sPushing once per natural period, every 3 s, keeps each push in step with the motion, so every push adds energy. Pushing at other rates is sometimes in step and sometimes against it, and the gains cancel. (Every 6 s, every other swing, also works, more weakly.) The same matching is how a radio tunes to one station, and how a singer can shatter a wine glass.
  158. 10. Oscillations · Practice problems · ProblemA mass moves as x=0.080cos⁡(5.0t), with x in metres and t in seconds. Find its amplitude, its period, its top speed and its largest acceleration.Amplitude 0.08 m; Period 2*pi/5 ≈ 1.25664 s; Top speed 0.4 m/s; Largest acceleration 2 m/s²
    1. Matching x=Acos⁡ωt: A=0.080m and ω=5.0rad/s.
    2. T=2πω=2π5.0≈1.26s,vmax=Aω=0.40m/s,amax=Aω2=0.080×25=2.0m/s2
    3. The amplitude is 0.080m, the period 1.26s, the top speed 0.40m/s (in the middle) and the largest acceleration 2.0m/s2 (at the ends).
  159. 10. Oscillations · Practice problems · ProblemA 0.40kg mass on a 160N/m spring oscillates with an amplitude of 0.10m and no friction. Find its total energy, and its speed when it is 0.060m from equilibrium.Total energy 0.8 J; Speed at 0.060 m 1.6 m/s
    1. E=12kA2=12(160)(0.10)2=0.80J
    2. At x=0.060m the spring holds 12(160)(0.060)2=0.288J, so the rest is kinetic:
    3. 12(0.40)v2=0.80−0.288=0.512⇒v=2.56=1.6m/s
    4. Check with v=ωA2−x2, where ω=160/0.40=20rad/s: 20×0.080=1.6m/s. The total energy is 0.80J and the speed there 1.6m/s.
  160. 10. Oscillations · Practice problems · ProblemA pendulum has a period of 2.00s on the Earth, where g=9.8m/s2. What is its period on the Moon, where g=1.62m/s2? By what factor must it be shortened to swing with a 2.00s period on the Moon?Period on the Moon 2*sqrt(9.8/1.62) ≈ 4.9191 s; Shorten by a factor of 9.8/1.62 ≈ 6.04938
    1. With the same length, T∝1/g:
    2. TMoon=2.00×9.81.62=2.00×2.46≈4.92s
    3. To keep T, keep L/g the same, so L shrinks by the same factor as g:
    4. LEarthLMoon=9.81.62≈6.05
    5. On the Moon it swings with a period of 4.92s, and it must be about 6.05 times shorter to keep time there, about 16 cm instead of 99 cm.
  161. 10. Oscillations · Practice problems · ProblemA damped oscillator’s amplitude follows Ae−γt. It falls from 10cm to 5.0cm in 4.0s. Find the damping rate γ, and the amplitude 12s after it was 10cm.γ= ln(2)/4 ≈ 0.173287 1/s; Amplitude after 12 s 1.25 cm
    1. The amplitude halves in 4.0 s:
    2. e−4.0γ=12⇒γ=ln⁡24.0≈0.173s−1
    3. Twelve seconds is three halvings:
    4. 10cm×(12)3=1.25cm
    5. The damping rate is 0.173s−1, and after 12 s the amplitude is 1.25cm: exponential decay loses the same fraction in each equal time, not the same amount.
  162. 11. Waves and sound · A wave carries energy, not matter · QuizIn a stadium “wave”, each fan stands up and sits down as the wave sweeps round the stands. What kind of wave is it, and what travels round the stadium?B. Transverse; only the pattern travels round, the fans stay in their seatsThe fans move up and down while the pattern moves sideways round the stadium: motion across the direction of travel is transverse. The fans stay in their seats, just as the leaf stays put on the pond. That is the defining feature of every wave: the pattern and its energy travel, the medium does not.
  163. 11. Waves and sound · Speed is wavelength times frequency · Fill the blanksFinish the line that works out a wavelength from a speed and a frequency.1: speed / frequencyFrom v=fλ, divide both sides by f: λ=v/f. A common slip is multiplying instead; a units check catches it, since (m/s) ÷ (1/s) = m, while (m/s) × (1/s) is m/s², not a length.
  164. 11. Waves and sound · Speed is wavelength times frequency · ProblemOcean waves pass a moored buoy. Their crests are 12m apart, and the buoy makes 8 complete up-and-down cycles in 40s. Find the period, the frequency and the speed of the waves.Period T 5 s; Frequency f 0.2 Hz; Speed v 2.4 m/s
    1. One cycle takes the total time divided by the number of cycles, and the frequency is its reciprocal:
    2. T=40 s8=5.0 s,f=1T=0.20 Hz
    3. The distance between crests is the wavelength, λ=12 m, so:
    4. v=fλ=0.20×12=2.4 m/s
    5. The buoy itself goes nowhere: it only bobs, while the pattern moves past it at 2.4 m/s.
  165. 11. Waves and sound · The medium sets the speed · QuizA sound wave of 500 Hz passes from air into water, where sound is about 4.3 times faster. What happens to its frequency and wavelength?A. The frequency stays 500 Hz; the wavelength becomes about 4.3 times longerThe frequency is set by the source: each wobble of the air at the boundary pushes the water once, so the water wobbles 500 times a second too. The speed is set by the new medium, so λ=v/f must change: 4.3 times faster means 4.3 times longer waves. Frequency is what survives crossing a boundary.
  166. 11. Waves and sound · The medium sets the speed · ProblemA guitar string has a mass of 3.6g per metre and is stretched to a tension of 72N. How fast do waves travel along it? A wave of frequency 220Hz runs along it: what is its wavelength?Wave speed v 100*sqrt(2) ≈ 141.421 m/s; Wavelength λ 100*sqrt(2)/220 ≈ 0.642824 m
    1. In SI units the mass per metre is μ=3.6g/m=0.0036kg/m. Then:
    2. v=FTμ=720.0036=20000≈141 m/s
    3. The wavelength follows from the wave equation:
    4. λ=vf=141.4220≈0.643 m
    5. Leaving μ in grams per metre would make the speed 1000≈32 times too small.
  167. 11. Waves and sound · Path difference decides loud and quiet · QuizTwo speakers in step play a tone of wavelength 0.80 m. You stand 3.0 m from one and 3.4 m from the other. What do you hear?B. A quiet spot: the path difference is half a wavelengthThe path difference is 3.4 − 3.0 = 0.4 m, and 0.4 / 0.80 = 0.5 wavelengths. A crest from one speaker arrives with a trough from the other, so they cancel. Step sideways by a little and the path difference changes, and the sound comes back: rooms have loud and quiet spots for each frequency.
  168. 11. Waves and sound · Path difference decides loud and quiet · ProblemTwo speakers play the same 680Hz note in step, and sound travels at 340m/s. You stand 4.00m from the first speaker and further than that from the second. What is the wavelength, and what is the smallest distance from the second speaker that puts you at a quiet spot?Wavelength 0.5 m; Distance from the second speaker 4.25 m
    1. λ=vf=340680=0.500 m
    2. Destructive interference needs Δd=(n+12)λ. The smallest is n=0, half a wavelength:
    3. Δd=12λ=0.250 m⇒d2=4.00+0.25=4.25 m
    4. The next quiet spots are at path differences of 0.75 m, 1.25 m and so on, with loud spots in between.
  169. 11. Waves and sound · Standing waves fit a whole number of halves · Code challengeWrite harmonic(n, length, speed): the frequency in hertz of the nth harmonic of a string fixed at both ends, of length metres, with wave speed speed in m/s.One solution:
    Solution
    double harmonic(int n, double length, double speed) {  double wavelength = 2 * length / n;  return speed / wavelength;}
    The $n$th mode fits n half-wavelengths, so λ=2L/n, and f=v/λ=nv/(2L). A common mistake is λ=L/n: but the fundamental is half a wave (one hump), not a whole one. Pressing a guitar string against a fret halves L and doubles f: up an octave.
  170. 11. Waves and sound · Harmonics give an instrument its voice · QuizA pipe closed at one end has a fundamental of 200 Hz. Which of these is also one of its resonant frequencies?C. 600 HzA closed pipe has only odd harmonics: 200, 600, 1000 Hz and so on (1, 3, 5 times the fundamental). 400 and 800 Hz are even multiples: they would fit a whole number of half-wavelengths in the pipe, which needs the same kind of point at both ends, but the closed end (where the air can’t move) and the open end (where it moves most) are opposite kinds. 500 Hz is not a multiple at all. A clarinet’s hollow, woody sound comes from this lack of even harmonics.
  171. 11. Waves and sound · Sound is a pressure wave · QuizYou see lightning and hear the thunder 3.0 s later. Light takes almost no time to reach you. How far away was the strike?B. About 1 kmDistance is speed times time: 343 m/s × 3.0 s ≈ 1030 m, about 1 km. The rule of thumb “3 seconds per kilometre” comes from this. Light covers 1 km in about 3 microseconds, so treating its arrival as instant is safe.
  172. 11. Waves and sound · Loudness is counted in decibels · Code challengeWrite soundLevel(power, distance): the sound level in decibels at distance metres from a small source radiating power watts equally in all directions.One solution:
    Solution
    #include <cmath>#include <numbers>
    const double I0 = 1e-12;  // W/m², the threshold of hearing
    double soundLevel(double power, double distance) {  double area = 4 * std::numbers::pi * distance * distance;  double intensity = power / area;  return 10 * std::log10(intensity / I0);}
    First intensity, P/(4πr2), then the logarithm of its ratio to I0, times 10. The tests show the two rules of thumb: doubling intensity adds 10log10⁡2≈3 dB, and doubling distance quarters intensity, which takes off about 6 dB. A usual slip is std::log, which is the natural logarithm: decibels use base 10, std::log10.
  173. 11. Waves and sound · A moving source changes the pitch · Code challengeWrite doppler(f, sourceSpeed, listenerSpeed): the frequency heard, in air at 343 m/s. Both speeds are positive when moving towards the other, negative when moving away.One solution:
    Solution
    const double SOUND = 343;  // m/s
    double doppler(double f, double sourceSpeed, double listenerSpeed) {  return f * (SOUND + listenerSpeed) / (SOUND - sourceSpeed);}
    The listener’s speed adds to the rate crests are met (top), and the source’s speed shortens the wavelength (bottom). The usual mistake is a sign: going away is the same formula with a negative speed, so you never need a second formula. Note the asymmetry: a source at 10 m/s gives 700×343/333≈721.0 Hz, a listener at 10 m/s gives 720.4 Hz, because only the source changes the wavelength in the air.
  174. 11. Waves and sound · Practice problems · ProblemA violin’s A string is 0.325m long between its fixed ends and has a mass of 0.60g per metre. It must sound a fundamental of 440Hz. What wave speed does that need, what tension, and what is the frequency of its third harmonic?Wave speed 286 m/s; Tension 0.0006*286^2 ≈ 49.0776 N; Third harmonic 1320 Hz
    1. The fundamental has λ1=2L=0.650 m, so the waves must travel at:
    2. v=f1λ1=440×0.650=286 m/s
    3. Squaring v=FT/μ and solving for the tension, with μ=0.00060 kg/m:
    4. FT=μv2=0.00060×2862≈49.1 N
    5. Harmonics are whole-number multiples of the fundamental: f3=3×440=1320 Hz.
  175. 11. Waves and sound · Practice problems · ProblemAn organ pipe closed at one end is 0.85m long, and sound travels at 343m/s. What is its lowest resonant frequency, and the next one above it?Lowest 343/3.4 ≈ 100.882 Hz; Next 3*343/3.4 ≈ 302.647 Hz
    1. The fundamental fits a quarter of a wavelength in the pipe, so λ1=4L=3.4 m:
    2. f1=v4L=3433.4≈101 Hz
    3. A closed pipe has only odd harmonics, so the next is three times the fundamental, not twice:
    4. f3=3f1≈303 Hz
  176. 11. Waves and sound · Practice problems · ProblemA small loudspeaker radiates 0.50W of sound equally in all directions. Find the intensity, in milliwatts per square metre, and the sound level 4.0m away, with I0=10−12W/m2. How far away is the sound level 20dB lower?Intensity 500/(64*pi) ≈ 2.4868 mW/m²; Sound level 93.96 dB; Distance for 20 dB less 40 m
    1. The power spreads over a sphere of radius 4.0 m:
    2. I=P4πr2=0.504π(4.0)2≈2.49×10−3 W/m2=2.49 mW/m2
    3. β=10log10⁡2.49×10−310−12=10log10⁡(2.49×109)≈94.0 dB
    4. 20 dB less is 102=100 times less intensity. Intensity goes as 1/r2, so the distance must be 100=10 times greater: 40 m.
  177. 11. Waves and sound · Practice problems · ProblemA train’s horn sounds at 500Hz. The train comes towards you at 25m/s while you stand still beside the track, passes, and goes away at the same speed. Take the speed of sound as 343m/s. What frequency do you hear as it approaches, and as it goes away? By how much does the pitch drop as it passes?Approaching 500*343/318 ≈ 539.308 Hz; Going away 500*343/368 ≈ 466.033 Hz; Drop 500*343/318 - 500*343/368 ≈ 73.2756 Hz
    1. fin′=500×343343−25=500×343318≈539 Hz
    2. fout′=500×343343+25=500×343368≈466 Hz
    3. The drop is 539.3−466.0≈73 Hz. It isn’t symmetric about 500 Hz: the rise (39 Hz) is bigger than the fall (34 Hz).
  178. 12. Fluids · A fluid flows, and density says how heavy it is · QuizA room is 5 m by 4 m by 2.5 m. Roughly what is the mass of the air in it?C. About 60 kgThe volume is 5 × 4 × 2.5 = 50 m³, and m=ρV=1.2×50=60 kg: about the mass of a person. Air feels like nothing because you are immersed in it and it presses equally all round, but it is far from weightless.
  179. 12. Fluids · Pressure is force spread over area · QuizWhy do snowshoes stop you sinking into soft snow?B. They spread your weight over a bigger area, so the pressure is lowerYour weight, the force, is unchanged. Spreading it over a larger area lowers p=F/A, and snow gives way when the pressure, not the force, is too high. A sharp knife works the other way: a tiny area turns a modest force into a huge pressure.
  180. 12. Fluids · Pressure grows with depth · Code challengeWrite pressureAtDepth(depth, density): the absolute pressure in pascals at depth metres in a liquid of density kg/m³, with the atmosphere (101 325 Pa) on top and g = 9.81 m/s².One solution:
    Solution
    const double ATMOSPHERE = 101325;  // Paconst double g = 9.81;             // m/s²
    double pressureAtDepth(double depth, double density) {  return ATMOSPHERE + density * g * depth;}
    The surface pressure plus the weight of the column per square metre, ρgh. At the bottom of the Mariana Trench that is about 111 MPa, over a thousand atmospheres. Forgetting p0 gives the gauge pressure instead: fine for a tyre, wrong for a diver’s lungs.
  181. 12. Fluids · Pressure grows with depth · ProblemA diver swims 25m below the surface of the sea. Seawater has a density of 1025kg/m3, g=9.81m/s2, and the air presses on the surface at 101.3kPa. What are the gauge pressure and the absolute pressure on the diver?Gauge pressure 251.4 kPa; Absolute pressure 352.7 kPa
    1. ρgh=1025×9.81×25≈2.514×105 Pa=251.4 kPa
    2. p=p0+ρgh=101.3+251.4=352.7 kPa
    3. About 3.5 atmospheres: the air the diver breathes must be supplied at this pressure too.
  182. 12. Fluids · A push on a fluid reaches everywhere · QuizIn the lift above, you push the small piston down 50 cm. How far does the big piston rise?C. 1 cmThe oil pushed out of the small cylinder, 0.01×0.5=0.005 m³, fills the big one: 0.005/0.5=0.01 m, 1 cm. Fifty times the force over one fiftieth of the distance: the work in, 100 × 0.5 = 50 J, equals the work out, 5000 × 0.01 = 50 J. A hydraulic lift multiplies force, never energy.
  183. 12. Fluids · Archimedes: the push up equals the fluid pushed aside · QuizIce is 917 kg/m³ and seawater 1025 kg/m³. What fraction of an iceberg is above the water?B. About 11%The fraction below is 917/1025≈0.895, so about 89% is hidden and 11% shows: “the tip of the iceberg”. The trap is answering 89%, which is the part under water.
  184. 12. Fluids · Archimedes: the push up equals the fluid pushed aside · ProblemA block of wood 40cm×30cm across and 10cm thick, of density 600kg/m3, floats flat in fresh water (1000kg/m3). How deep does it sit in the water? What extra mass, placed on top, would push it down until its top is level with the water?Depth under water 6 cm; Extra mass 4.8 kg
    1. The fraction under water is the ratio of the densities, 600/1000=0.60, so the block sits 0.60×10=6.0 cm deep.
    2. Its volume is 0.40×0.30×0.10=0.012m3, so its mass is 600×0.012=7.2 kg.
    3. Fully under, it displaces 0.012m3 of water, a mass of 12 kg. It floats just level when the block and its load together have that mass:
    4. mextra=12−7.2=4.8 kg
    5. Notice that g cancels: floating is a balance of weights, so it can be worked out in masses.
  185. 12. Fluids · Weighing something under water · Code challengeWrite apparentWeight(mass, volume, fluidDensity): the reading in newtons of a scale holding a fully submerged object of mass kg and volume m³. Use g = 9.81 m/s².One solution:
    Solution
    const double g = 9.81;
    double apparentWeight(double mass, double volume,                      double fluidDensity) {  double buoyancy = fluidDensity * volume * g;  return mass * g - buoyancy;}
    Weight down, buoyancy up, and the scale takes the rest. The buoyancy uses the fluid’s density and the object’s volume: mixing them up is the usual slip. If the answer is negative, the object would float, and you would need to hold it under.
  186. 12. Fluids · Narrower pipe, faster flow · QuizWhy does water squirt further when you put your thumb over the end of a hose?B. A smaller opening means the same flow must leave fasterWith roughly the same flow rate Q forced through a smaller area, v=Q/A goes up, and faster water flies further. (In a real hose the flow drops a little too, because the tap can only supply so much pressure, but the speed still rises.)
  187. 12. Fluids · Bernoulli: faster flow, lower pressure · Fill the blanksWater flows through a level pipe that narrows. Fill in the pressure in the narrow part, from Bernoulli with no change in height.1: v1 * v1 - v2 * v2With h the same, p1+12ρv12=p2+12ρv22, so p2=p1+12ρ(v12−v22)=150000−7500. The pressure drops where the flow is fast. This is how a Venturi meter measures flow, and how a perfume atomiser pulls liquid up a tube.
  188. 12. Fluids · Bernoulli: faster flow, lower pressure · ProblemA water tank has a small hole of area 2.0cm2 in its side, 3.2m below the water’s surface. Take g=9.81m/s2. How fast does the water leave the hole, and how many litres flow out each second?Speed sqrt(2*9.81*3.2) ≈ 7.92364 m/s; Flow rate 0.2*sqrt(2*9.81*3.2) ≈ 1.58473 L/s
    1. v=2gh=2×9.81×3.2≈7.92 m/s
    2. Q=Av=(2.0×10−4 m2)(7.92 m/s)≈1.58×10−3 m3/s=1.58 L/s
    3. As the tank empties, h falls and so does the flow: a tank drains fast at first and slowly at the end.
  189. 12. Fluids · Viscosity is a fluid’s internal friction · QuizFatty deposits narrow an artery’s radius by half, with the same pressure difference across it. What happens to the blood flow, by Poiseuille’s law?C. It falls to a sixteenthQ∝r4, and (1/2)4=1/16. That is why small narrowings of blood vessels matter so much, and why the heart must work much harder to keep up. Continuity says the speed rises where a pipe narrows for a given flow; Poiseuille says how much flow viscosity lets through at all.
  190. 12. Fluids · Practice problems · ProblemA garage lift holds up a 1500kg car on a piston of radius 20cm. Oil connects it to a small piston of radius 2.0cm. Take g=9.81m/s2 and ignore the pistons’ own weight. What force on the small piston holds the car up, and how far must the small piston move down to raise the car by 10cm?Force 147.15 N; Distance 10 m
    1. The area ratio is (20/2.0)2=100, and the car weighs 1500×9.81=14715 N. The pressure is the same at both pistons:
    2. F1=F2A1A2=14715100≈147 N
    3. The same volume of oil leaves one cylinder and enters the other, A1d1=A2d2, so d1=100×0.10=10 m.
    4. Work in, 147×10≈1470 J, equals the car’s gain in potential energy, 14715×0.10≈1470 J. Real lifts pump the small piston in many short strokes.
  191. 12. Fluids · Practice problems · ProblemA crown has a mass of 1.20kg. Hung from a spring scale and fully under water (1000kg/m3), it reads 10.90N. Take g=9.81m/s2. Find the buoyant force, the crown’s volume in cubic centimetres, and its density. (Pure gold is 19300kg/m3.)Buoyant force 0.872 N; Volume 88.89 cm³; Density 13500 kg/m³
    1. Its weight is 1.20×9.81=11.772 N and the scale takes 10.90 N of it, so the water holds up the rest:
    2. Fb=11.772−10.90=0.872 N
    3. The buoyant force is the weight of the water displaced, and the crown displaces its own volume:
    4. V=Fbρwaterg=0.8721000×9.81≈8.89×10−5 m3=88.9 cm3
    5. ρ=mV=1.208.89×10−5≈13500 kg/m3
    6. Far below gold’s 19 300 kg/m³: the crown is gold mixed with something lighter, such as silver. No melting needed.
  192. 12. Fluids · Practice problems · ProblemWater flows along a level pipe of radius 4.0cm at 1.5m/s, where the pressure is 180kPa. The pipe then narrows to a radius of 2.0cm. Treat the water as frictionless, with density 1000kg/m3. Find the flow rate in litres per second, the speed in the narrow part, and the pressure there.Flow rate 2.4*pi ≈ 7.53982 L/s; Speed in the narrow part 6 m/s; Pressure in the narrow part 163.125 kPa
    1. Q=A1v1=π(0.040)2(1.5)≈7.54×10−3 m3/s=7.54 L/s
    2. Halving the radius quarters the area, so by continuity the water goes four times faster: v2=6.0 m/s.
    3. With no change in height, Bernoulli gives:
    4. p2=p1+12ρ(v12−v22)=180000+500(2.25−36)=163125 Pa≈163 kPa
    5. Faster flow, lower pressure: the drop of about 17 kPa is what a Venturi meter reads to measure the flow.
  193. 13. Heat, temperature and thermodynamics · Temperature measures jiggling, and has a true zero · QuizAt which temperature do the Celsius and Fahrenheit scales show the same number?B. −40Set F=C in F=95C+32: C−95C=32, so −45C=32 and C=−40. Check: 95(−40)+32=−72+32=−40. Kelvin is written “K”, not “°K”, and never goes negative.
  194. 13. Heat, temperature and thermodynamics · Most things grow when they warm up · QuizA metal lid is stuck on a glass jar. Why does running hot water over the lid help?B. The metal expands more than the glass for the same temperature riseMetals have larger expansion coefficients than glass (aluminium about 23×10−6/K, ordinary glass about 9×10−6/K), so the lid grows away from the rim. Glass doesn’t shrink when heated; it grows too, just less.
  195. 13. Heat, temperature and thermodynamics · Heat is energy on the move · Code challengeWrite mixTemperature(m1, c1, t1, m2, c2, t2): the final temperature when two bodies (masses in kg, specific heats in J/(kg·K), temperatures in °C) reach equilibrium with no heat lost.One solution:
    Solution
    double mixTemperature(double m1, double c1, double t1,                      double m2, double c2, double t2) {  double capacity1 = m1 * c1;  double capacity2 = m2 * c2;  return (capacity1 * t1 + capacity2 * t2) / (capacity1 + capacity2);}
    Set the heat lost equal to the heat gained, m1c1(t1−T)=m2c2(T−t2), and solve for T: a weighted average, weighted by each body’s heat capacity mc. The plain average only works for equal mc. The iron barely warms the water, because water’s heat capacity is so large: 2×4186 against 1×450.
  196. 13. Heat, temperature and thermodynamics · Heat is energy on the move · ProblemA 2.2kW kettle holds 1.5kg of water at 15 °C. Water’s specific heat capacity is 4186J/(kg·K). How much heat brings the water to 100 °C, and how long does that take if all the kettle’s power goes into the water?Heat 533.715 kJ; Time 533715/2200 ≈ 242.598 s
    1. Q=mcΔT=1.5×4186×(100−15)=533715 J≈534 kJ
    2. t=QP=5337152200≈243 s
    3. About four minutes. A real kettle takes a little longer, because some of the heat warms the kettle and the air.
  197. 13. Heat, temperature and thermodynamics · Melting and boiling take heat without warming · QuizWhy does a burn from steam at 100 °C hurt much more than one from water at 100 °C?B. Steam condensing on your skin releases its latent heat, 2256 kJ/kg, before it even starts to coolSteam at 100 °C turning into water at 100 °C gives up 2256 kJ for every kilogram, and that goes into your skin. Water at 100 °C cooling to skin temperature releases only about 4186×63≈264 kJ/kg. The same latent heat, taken from your skin, is why sweating cools you.
  198. 13. Heat, temperature and thermodynamics · Melting and boiling take heat without warming · ProblemYou drop 50g of ice at 0 °C into 0.40kg of water at 25 °C, in an insulated cup. Use Lf=334kJ/kg, and c=4186J/(kg·K) for water. How much heat does the ice need to melt, and what is the final temperature once it has all melted and mixed?Heat to melt the ice 16.7 kJ; Final temperature (41860 - 16700)/(0.45*4186) ≈ 13.3567 °C
    1. Qmelt=mLf=0.050×334000=16700 J=16.7 kJ
    2. Cooling the 0.40 kg of water all the way to 0 °C would give 0.40×4186×25=41860 J, more than enough, so all the ice melts.
    3. Call the final temperature T. The warm water’s loss pays for the melting and for warming the melted ice from 0 °C to T:
    4. 0.40×4186(25−T)=16700+0.050×4186T41860−1674.4T=16700+209.3TT=251601883.7≈13.4 ∘C
    5. Melting does most of the cooling: 50 g of water at 0 °C would only have brought the drink down to about 22 °C.
  199. 13. Heat, temperature and thermodynamics · Heat moves by conduction, convection and radiation · QuizA star’s surface goes from 3000 K to 6000 K, its size unchanged. How much more power does it radiate?D. 16 timesRadiated power goes as T4, and 24=16. That steep fourth power is why hot objects glow so much brighter. It only works with kelvin: in Celsius the “doubling” would be meaningless, since 0 °C isn’t zero energy.
  200. 13. Heat, temperature and thermodynamics · The ideal gas law ties pressure, volume and temperature · Code challengeWrite gasPressure(moles, celsius, litres): the pressure in pascals of an ideal gas. Take the temperature in °C and the volume in litres, and convert them first.One solution:
    Solution
    const double R = 8.314;  // J/(mol·K)
    double gasPressure(double moles, double celsius, double litres) {  double kelvin = celsius + 273.15;  double cubicMetres = litres / 1000;  return moles * R * kelvin / cubicMetres;}
    The starter makes both classic mistakes: Celsius instead of kelvin (so a gas at 0 °C would have no pressure) and litres instead of cubic metres (1 L = 0.001 m³, a factor of 1000). The ideal gas law only works in SI units with absolute temperature.
  201. 13. Heat, temperature and thermodynamics · Pressure is molecules hitting the walls · Code challengeWrite rmsSpeed(kelvin, molarMass): the root-mean-square speed in m/s, with the molar mass in kg/mol.One solution:
    Solution
    #include <cmath>
    const double R = 8.314;
    double rmsSpeed(double kelvin, double molarMass) {  return std::sqrt(3 * R * kelvin / molarMass);}
    vrms=3RT/M. The air molecules hitting you now average about 500 m/s, faster than sound (which makes sense: sound is carried by them). Speed grows only as T. The usual slip is the molar mass in g/mol (28) instead of kg/mol (0.028), which gives a speed about 32 times too small.
  202. 13. Heat, temperature and thermodynamics · The first law: energy is conserved · QuizA gas takes in 500 J of heat and pushes a piston out, doing 200 J of work. What is the change in its internal energy?B. +300 JΔU=Q−W=500−200=+300 J. Of the heat that came in, 200 J left again as work and 300 J stayed, warming the gas. Adding them, +700 J, would mean the gas gained energy by doing work, which runs the wrong way.
  203. 13. Heat, temperature and thermodynamics · The first law: energy is conserved · ProblemA gas in a cylinder takes in 800J of heat while it expands at a constant pressure of 1.0×105Pa, from 2.0L to 5.0L. How much work does the gas do, and by how much does its internal energy change?Work done by the gas, W 300 J; ΔU 500 J
    1. The volume grows by 3.0L=3.0×10−3m3:
    2. W=pΔV=(1.0×105)(3.0×10−3)=300 J
    3. ΔU=Q−W=800−300=+500 J
    4. Of the 800 J that came in, 300 J left as work on the piston and 500 J stayed, warming the gas.
  204. 13. Heat, temperature and thermodynamics · A heat engine can never use all its heat · Find the bugThis gives a Carnot efficiency of 91% for steam at 300 °C and a river at 50 °C. Which line is wrong?Line 3: double cold = coldCelsius + 273.15;Both temperatures must be absolute, and line 3 leaves the cold one in Celsius. As written it computes 1−50/573.15≈0.91. With the fix, e=1−323.15/573.15≈0.44, 44%. Any formula with a ratio of temperatures needs kelvin on both sides. Real power stations turn about 35 to 45% of their heat into work, not far below their limit.
  205. 13. Heat, temperature and thermodynamics · Entropy never goes down · QuizWhich of these would break the second law of thermodynamics?B. An engine that turns all the heat from a hot source into work, with no cold sideA cycle that only takes heat and gives out work would lower the hot side’s entropy with nothing to make up for it. The others are fine: the fridge and freezer run on work and push extra heat out into the room, and a living thing lowers its own entropy while raising its surroundings’ more, by eating and giving off heat.
  206. 13. Heat, temperature and thermodynamics · Practice problems · ProblemA car tyre holds 0.025m3 of air at 20 °C and an absolute pressure of 300kPa. How many moles of air does it hold? After a long drive the air is at 45 °C, with the volume unchanged: what is its pressure now? Use R=8.314J/(mol·K).Amount 300000*0.025/(8.314*293.15) ≈ 3.07724 mol; New pressure 300*318.15/293.15 ≈ 325.584 kPa
    1. In kelvin the temperatures are 293.15 K and 318.15 K. From pV=nRT:
    2. n=pVRT=(300000)(0.025)8.314×293.15≈3.08 mol
    3. With n, R and V unchanged, the pressure is proportional to the absolute temperature:
    4. p2=p1T2T1=300×318.15293.15≈326 kPa
    5. Using Celsius, 300×45/20=675 kPa, would be badly wrong: the pressure rises by about 9%, not 125%.
  207. 13. Heat, temperature and thermodynamics · Practice problems · ProblemOxygen gas (M=0.032kg/mol) is at 27 °C. Find the average kinetic energy of one molecule, in units of 10−21J, and the root-mean-square speed of the molecules. Use kB=1.381×10−23J/K and R=8.314J/(mol·K).Average kinetic energy 1.5*0.01381*300.15 ≈ 6.21761 × 10⁻²¹ J; rms speed sqrt(3*8.314*300.15/0.032) ≈ 483.682 m/s
    1. In kelvin, T=27+273.15=300.15 K.
    2. Ek=32kBT=1.5×1.381×10−23×300.15≈6.22×10−21 J
    3. vrms=3RTM=3×8.314×300.150.032≈484 m/s
    4. Every gas at 27 °C has the same average kinetic energy per molecule; lighter molecules just move faster to have it.
  208. 13. Heat, temperature and thermodynamics · Practice problems · ProblemA power station’s boiler makes steam at 550 °C, and it gives its waste heat to a river at 20 °C. What is its Carnot limit, as a percentage? It actually turns 38% of the heat from its fuel into 600MW of electrical power: how much heat each second does it take from the fuel, and how much goes into the river?Carnot limit 100*(1 - 293.15/823.15) ≈ 64.3868 %; Heat from the fuel 600/0.38 ≈ 1578.95 MW; Heat into the river 600/0.38 - 600 ≈ 978.947 MW
    1. eCarnot=1−TcTh=1−293.15823.15≈0.644=64.4%
    2. Qh=We=600 MW0.38≈1579 MW
    3. Qc=Qh−W≈1579−600=979 MW
    4. More heat goes into the river than comes out as electricity, even though the station reaches more than half of its Carnot limit.
  209. 13. Heat, temperature and thermodynamics · Practice problems · ProblemA 1.0kg block of ice at 0 °C melts in a room at 20 °C, taking 334kJ of heat from the room’s air, which stays at 20 °C. Find the entropy change of the ice, of the room, and of the two together.Ice 334000/273.15 ≈ 1222.77 J/K; Room -334000/293.15 ≈ −1139.35 J/K; Total 334000/273.15 - 334000/293.15 ≈ 83.4229 J/K
    1. ΔSice=+334000273.15≈+1223 J/K
    2. ΔSroom=−334000293.15≈−1139 J/K
    3. ΔStotal≈1223−1139=+83 J/K
    4. The total rises, as the second law demands. Run it backwards, water freezing by giving its heat to a warmer room, and the total would fall: that never happens by itself.
  210. 14. Charge and electric fields · Charge comes in two kinds and in fixed lumps · QuizHow many extra electrons give an object a charge of −1 C?B. About 6.2 × 10¹⁸1/(1.602×10−19)≈6.24×1018 electrons. A coulomb is a lot of charge: static on a balloon is a few nanocoulombs. The tempting 6.0×1023 is Avogadro’s number, the size of a mole, which is a different count.
  211. 14. Charge and electric fields · Conductors let charge move; insulators hold it · QuizA negatively charged balloon sticks to a neutral wall. Why, if the wall has no net charge?B. The wall’s molecules polarise, with their positive sides nearer the balloon, and nearer charges pull harderThe balloon pushes the wall’s negative charges slightly away and pulls positive ones slightly nearer. Attraction and repulsion both act, but the attracted charges are closer, and the force grows sharply as distance falls (next slide), so attraction wins. Nothing is transferred: the wall stays neutral overall.
  212. 14. Charge and electric fields · Coulomb's law: an inverse square · Code challengeWrite coulombForce(q1, q2, r): the force in newtons between charges q1 and q2 (coulombs) r metres apart. Positive means they repel, negative that they attract.One solution:
    Solution
    const double k = 8.99e9;  // N·m²/C²
    double coulombForce(double q1, double q2, double r) {  return k * q1 * q2 / (r * r);}
    The distance is squared, r * r: C++ has no power operator, and r ^ 2 won’t compile for a double (^ is a bitwise operation on whole numbers, where 3 ^ 2 is 1). Leaving the sign of the charges in the product does the sign work for you: like charges give a positive product and repel. Two charges of one microcoulomb a centimetre apart push with the weight of a 9 kg mass: tiny charges, big forces.
  213. 14. Charge and electric fields · Coulomb's law: an inverse square · ProblemTwo small spheres carry charges of +4.0nC and −6.0nC, and their centres are 3.0cm apart. Use k=8.99×109N·m2/C2. How big is the force between them, in micronewtons? At what separation would it be 1.0mN?Force 239.73 μN; Separation 1.4689 cm
    1. F=k|q1q2|r2=(8.99×109)(4.0×10−9)(6.0×10−9)(0.030)2≈2.40×10−4 N=240 μN
    2. The charges are unlike, so it is an attraction.
    3. r=k|q1q2|F=(8.99×109)(2.4×10−17)1.0×10−3≈0.0147 m=1.47 cm
    4. Check with the inverse square: the force grew 1000/240≈4.2 times, so the distance shrank 4.2≈2.04 times, from 3.0 to 1.47 cm.
  214. 14. Charge and electric fields · A charge fills the space around it with a field · Code challengeCharges lie on the x axis. Write fieldAt(x, charges): the electric field (N/C) at position x, positive pointing in +x. Each Charge has a position at and charge q. Assume x isn’t at a charge.One solution:
    Solution
    #include <vector>
    const double k = 8.99e9;
    struct Charge {  double at;  // position on the x axis, m  double q;   // charge, C};
    double fieldAt(double x, const std::vector<Charge>& charges) {  double field = 0;  for (const Charge& c : charges) {    double r = x - c.at;    double size = k * c.q / (r * r);    if (r > 0) field += size;    else field -= size;  }  return field;}
    A positive charge’s field points away from it: to the right on its right (r>0), to the left on its left. The if on the sign of r gives that direction, and kq/r2 the size; a negative q flips it to point inward. Between a + and a − charge the fields both point from + to −, so they add. Forgetting the direction and adding sizes would give 17.98 in the cancelling case too.
  215. 14. Charge and electric fields · A charge fills the space around it with a field · ProblemA +2.0μC charge sits at x=0 and a +8.0μC charge at x=30cm. Where between them is the electric field zero? How strong is the field at the midpoint, x=15cm? Use k=8.99×109N·m2/C2.Zero field at x= 10 cm; Field strength at the midpoint 2397000 N/C
    1. Between the charges, the left one’s field points right and the right one’s points left. They cancel where (with x in cm):
    2. k(2.0)x2=k(8.0)(30−x)2⇒30−xx=2⇒x=10 cm
    3. The zero is nearer the smaller charge, as it must be. At the midpoint both charges are 0.15 m away:
    4. E=k(8.0−2.0)×10−6(0.15)2=(8.99×109)(6.0×10−6)0.0225≈2.40×106 N/C
    5. It points towards the smaller charge, because the bigger charge’s push wins there.
  216. 14. Charge and electric fields · Field lines draw the field · QuizAn electron is released at rest in a uniform field that points to the right. What does it do?C. It accelerates to the leftThe force is F→=qE→, and the electron’s q is negative, so the force points opposite to the field: left. A constant force gives a constant acceleration, so it speeds up steadily. The field’s direction is defined for a positive charge; negative ones go the other way.
  217. 14. Charge and electric fields · Voltage is energy per charge · QuizA 9 V battery pushes 2 C of charge round a circuit. How much energy does it give that charge?D. 18 JVoltage is joules per coulomb: ΔU=qΔV=2×9=18 J. Dividing (4.5) mixes up which is per which. The voltage tells you how much energy each coulomb carries; how many coulombs flow is the current’s job (next module).
  218. 14. Charge and electric fields · Voltage is energy per charge · ProblemAn electron gun accelerates electrons from rest through 1000V. Find the kinetic energy each one gains, in electronvolts and in units of 10−16J, and its speed. Use e=1.602×10−19C and me=9.109×10−31kg.Energy in eV 1000 eV; Energy in joules 1.602 × 10⁻¹⁶ J; Speed sqrt(2*1.602e-16/9.109e-31) ≈ 18754700 m/s
    1. Crossing 1000 V, each electron gains 1000 eV: that is what an electronvolt means. In joules:
    2. Ek=eΔV=(1.602×10−19)(1000)=1.602×10−16 J
    3. v=2Ekme=2×1.602×10−169.109×10−31≈1.88×107 m/s
    4. About 6% of the speed of light, so Newton’s 12mv2 is still good to well under 1%.
  219. 14. Charge and electric fields · The field points downhill in potential · Fill the blanksTwo parallel plates 2 mm apart are connected to a 12 V supply. Fill in the field between them.1: voltage / gap, 2: 6000In a uniform field, E=V/d=12/0.002=6000 V/m. This is why V/m and N/C are the same unit. Make the gap smaller and the field grows: air breaks down and sparks at about 3×106 V/m.
  220. 14. Charge and electric fields · A capacitor stores charge, and energy · QuizYou double the voltage across a capacitor. What happens to the energy stored?B. It is four times as muchU=12CV2, so twice the voltage stores four times the energy. Both the charge and the voltage doubled, and the energy is (half) their product. Camera flashes and defibrillators charge a capacitor slowly and dump its energy in a fraction of a second.
  221. 14. Charge and electric fields · A capacitor stores charge, and energy · ProblemA capacitor is made of two plates 20cm×20cm, 0.50mm apart, with plastic of dielectric constant κ=3.0 between them, connected to 9.0V. Use ε0=8.854×10−12F/m. Find its capacitance, the charge on each plate, and the energy stored.Capacitance 2.125 nF; Charge 19.125 nC; Energy 86.06 nJ
    1. The plates’ area is 0.20×0.20=0.040m2 and the gap is 5.0×10−4 m:
    2. C=κε0Ad=3.0×8.854×10−12×0.0405.0×10−4≈2.12×10−9 F=2.12 nF
    3. Q=CV=2.125×10−9×9.0≈1.91×10−8 C=19.1 nC
    4. U=12CV2=12×2.125×10−9×81≈8.61×10−8 J=86.1 nJ
    5. Without the plastic, all three would be three times smaller.
  222. 14. Charge and electric fields · Gauss's law: count the field lines leaving · QuizA closed surface encloses +3 nC and −3 nC, and a +5 nC charge sits just outside it. What is the total electric flux out of the surface?A. ZeroOnly charge inside counts, and +3−3=0. The outside charge’s field lines go in one side and out the other, adding nothing in total. The field on the surface is not zero (the outside charge makes sure of that); only the total flux is.
  223. 14. Charge and electric fields · Practice problems · ProblemTwo parallel plates 5.0mm apart are held at a potential difference of 150V. Find the field between them, the force on a proton there (in units of 10−15N), and the speed of a proton released at rest at the positive plate when it reaches the negative one. Use e=1.602×10−19C and mp=1.673×10−27kg.Field 30000 V/m; Force 4.806 × 10⁻¹⁵ N; Speed sqrt(2*1.602e-19*150/1.673e-27) ≈ 169490 m/s
    1. E=Vd=1500.0050=3.0×104 V/m
    2. F=eE=(1.602×10−19)(3.0×104)≈4.81×10−15 N
    3. Crossing 150 V, the proton gains e×150 V of kinetic energy:
    4. v=2eΔVmp=2×1.602×10−19×1501.673×10−27≈1.69×105 m/s
    5. An electron crossing the same gap would end about 43 times faster, 1836, because it is 1836 times lighter.
  224. 14. Charge and electric fields · Practice problems · ProblemA defibrillator charges a 150μF capacitor to 2000V. How much energy does it store, and how much charge? If it releases that energy through the patient in 5.0ms, what is the average power?Energy 300 J; Charge 0.3 C; Average power 60 kW
    1. U=12CV2=12(150×10−6)(2000)2=300 J
    2. Q=CV=(150×10−6)(2000)=0.30 C
    3. P=Ut=3000.0050=6.0×104 W=60 kW
    4. A modest energy, about what lifting a 30 kg bag by one metre takes, but delivered in milliseconds, so the power is huge.
  225. 14. Charge and electric fields · Practice problems · ProblemA point charge of +5.0nC is fixed in place. Find the electric potential 0.50m and 0.20m from it, and the work needed to bring a +2.0nC charge slowly from 0.50m to 0.20m away, in microjoules. Use k=8.99×109N·m2/C2.Potential at 0.50 m 89.9 V; Potential at 0.20 m 224.75 V; Work 0.2697 μJ
    1. V(0.50)=kQr=(8.99×109)(5.0×10−9)0.50=89.9 V
    2. V(0.20)=(8.99×109)(5.0×10−9)0.20≈224.8 V
    3. W=qΔV=(2.0×10−9)(224.75−89.9)≈2.70×10−7 J=0.270 μJ
    4. Positive work: you push one positive charge towards another against their repulsion, and the energy is stored as potential energy.
  226. 15. Current and circuits · Current is charge flowing per second · QuizElectrons in a household copper wire drift at well under a millimetre per second. Why does a lamp light the instant you flip the switch?B. The wire is already full of electrons; the field that pushes them spreads along the wire at nearly the speed of light, so they all start moving at onceLike water in a full hose, the electrons are already in the lamp. Switching on sets up the electric field along the whole circuit almost at once, and every electron starts drifting together. The signal travels fast; the charges themselves crawl.
  227. 15. Current and circuits · Resistance and Ohm's law · QuizA 230 V heater draws 10 A. What is its resistance?C. 23 ΩR=V/I=230/10=23 Ω. Multiplying gives 2300, which is the power in watts (next slide), not the resistance: check the units, V/A is ohms, V × A is watts.
  228. 15. Current and circuits · Resistance and Ohm's law · ProblemAn extension cable is 25m long. Current goes out along one copper wire and back along another, each with a cross-section of 1.5mm2. Copper’s resistivity is 1.68×10−8Ω⋅m. What is the resistance of the two wires together, and how much voltage is lost along them when 10A flows?Resistance 0.56 Ω; Voltage lost 5.6 V
    1. Out and back, the current passes through L=50 m of wire with A=1.5×10−6m2:
    2. R=ρLA=(1.68×10−8)(50)1.5×10−6=0.56 Ω
    3. V=IR=10×0.56=5.6 V
    4. On 230 V mains that is a loss of about 2.4%, and the lost energy heats the cable, which is why a coiled-up cable carrying a big load can get hot.
  229. 15. Current and circuits · Power is voltage times current · Fill the blanksA 2 kW kettle runs for 3 minutes. Fill in the energy in kilowatt-hours.1: 60, 2: *Energy is power times time: 2 kW×0.05 h=0.1 kWh, or 360 000 J. Kilowatt-hours need the time in hours: 3 minutes is 3/60 of an hour. That is why it says 3.0: with 3 / 60, both are ints, the division drops the fraction and gives 0, and the kettle would seem to use no energy at all.
  230. 15. Current and circuits · Power is voltage times current · ProblemA heater on 230V has a resistance of 26.5Ω. What current does it draw, what is its power, and how many kilowatt-hours does it use in 5.0 hours?Current 230/26.5 ≈ 8.67925 A; Power 230^2/26.5 ≈ 1996.23 W; Energy 230^2/26.5*5/1000 ≈ 9.98113 kWh
    1. I=VR=23026.5≈8.68 A
    2. P=V2R=230226.5≈1996 W≈2.0 kW
    3. E=Pt≈1.996 kW×5.0 h≈9.98 kWh
    4. Rounding the power to 2.0 kW first gives 10 kWh, close enough for a bill.
  231. 15. Current and circuits · Series adds resistance; parallel adds paths · Code challengeWrite series(rs) and parallel(rs): the total resistance in ohms of a list of resistors connected in series, and in parallel.One solution:
    Solution
    #include <vector>
    double series(const std::vector<double>& rs) {  double sum = 0;  for (double r : rs) sum += r;  return sum;}
    double parallel(const std::vector<double>& rs) {  double conductance = 0;  for (double r : rs) conductance += 1 / r;  return 1 / conductance;}
    In parallel you add the reciprocals (conductances), then take the reciprocal of the sum. The classic mistake is returning the sum of reciprocals itself: for 6 and 3 that is 0.5, which is in siemens, not ohms. 1 / r is a fraction only because r is a double: with whole-number resistances in a std::vector<int>, 1 / 6 would be 0. Remember the sanity check: a parallel total is always smaller than any single branch.
  232. 15. Current and circuits · Kirchhoff's laws solve any circuit · Put in orderPut the steps of solving that circuit in order.1. Combine the parallel pair into one equivalent resistance 2. Add the series resistance to get the total 3. Find the battery current from Ohm’s law, I = V / R 4. Find the voltage across the series resistor, then what is left for the parallel pair (loop rule) 5. Find each branch current, and check they add up (junction rule)Simplify from the inside out until one resistor is left, find the total current, then work back out, splitting voltages with the loop rule and currents with the junction rule. The final check costs nothing and catches most slips.
  233. 15. Current and circuits · Kirchhoff's laws solve any circuit · ProblemA 24V battery drives R1=6Ω in series with a parallel pair, R2=12Ω and R3=4Ω. Find the total resistance, the current from the battery, and the current through R2.Total resistance 9 Ω; Battery current 8/3 ≈ 2.66667 A; Current through R2 2/3 ≈ 0.666667 A
    1. R23=(112+14)−1=(412)−1=3 Ω,R=6+3=9 Ω
    2. I=249=83≈2.67 A
    3. The series resistor takes V1=83×6=16 V, which leaves 24−16=8 V across the pair:
    4. I2=812=23≈0.667 A,I3=84=2 A
    5. Check the junction: 23+2=83 A. The smaller resistor takes the larger share of the current.
  234. 15. Current and circuits · A capacitor charges along an exponential · Code challengeSimulate charging step by step. Write chargeRC(v0, r, c, t, dt): start with the capacitor at 0 V and repeat, t / dt times, “the voltage rises by (v0 − v) / (r·c) · dt”. Return the voltage.One solution:
    Solution
    #include <cmath>
    double chargeRC(double v0, double r, double c, double t, double dt) {  double v = 0;  int steps = std::lround(t / dt);  for (int i = 0; i < steps; i++) {    v += (v0 - v) / (r * c) * dt;  }  return v;}
    This is Euler’s method: each small step adds rate × time. It agrees with V0(1−e−t/RC) to within about 0.001 V here, because dt is a ten-thousandth of τ. Make dt close to τ and the answer goes badly wrong: the step size must be small compared with the fastest change in the system.
  235. 15. Current and circuits · A capacitor charges along an exponential · ProblemAn uncharged 470μF capacitor is connected through a 2.2kΩ resistor to a 12V supply. Find the time constant, the capacitor’s voltage after 2.0s, and the time it takes to reach 9.0V.Time constant 1.034 s; Voltage after 2.0 s 12*(1 - exp(-2/1.034)) ≈ 10.2656 V; Time to reach 9.0 V 1.034*ln(4) ≈ 1.43343 s
    1. τ=RC=(2200)(470×10−6)=1.034 s
    2. VC(2.0)=12(1−e−2.0/1.034)=12(1−0.145)≈10.3 V
    3. 9.0=12(1−e−t/τ)⇒e−t/τ=0.25t=τln⁡4=1.034×1.386≈1.43 s
    4. 75% of the way takes ln⁡4≈1.39 time constants, a little more than the one time constant that reaches 63%.
  236. 15. Current and circuits · Mains electricity, and staying safe · QuizWhy is a fuse always put in the live wire, not the neutral?B. When it blows, the appliance is cut off from the high voltage, not just from the return pathIn a simple loop the same current flows in live and neutral, so a fuse in either would stop it. But a fuse in the neutral would leave the appliance connected to the live wire after it blew: still at a dangerous voltage, ready to shock whoever opens it to see what is wrong.
  237. 15. Current and circuits · Practice problems · ProblemA battery has an EMF of 9.0V and an internal resistance of 0.50Ω: a resistance inside the battery, in series with whatever it drives. It lights a lamp of resistance 4.0Ω. Find the current, the voltage across the battery’s terminals, and the power the lamp gets.Current 2 A; Terminal voltage 8 V; Lamp power 16 W
    1. I=ℰR+r=9.04.0+0.50=2.0 A
    2. V=ℰ−Ir=9.0−2.0×0.50=8.0 V
    3. P=I2R=2.02×4.0=16 W
    4. The other 2.02×0.50=2 W warms the battery itself: 1 V of the 9 never reaches the lamp.
  238. 15. Current and circuits · Practice problems · ProblemA kettle is rated 1150W on 230V (RMS) mains. What RMS current does it draw, what is its resistance, and what is the peak voltage of the supply?Current 5 A; Resistance 46 Ω; Peak voltage 230*sqrt(2) ≈ 325.269 V
    1. I=PV=1150230=5.0 A
    2. R=VI=2305.0=46 Ω
    3. Vpeak=2×230≈325 V
    4. Ratings use RMS values because they give the right heating in P=I2R, the same as steady values would.
  239. 15. Current and circuits · Practice problems · ProblemA string of fairy lights has 40 identical bulbs in series on 230V, using 46W in all. Find the voltage across each bulb, the current, and each bulb’s resistance.Voltage per bulb 5.75 V; Current 0.2 A; Resistance per bulb 28.75 Ω
    1. Identical bulbs in series share the voltage equally: 230/40=5.75 V each.
    2. I=PV=46230=0.20 A
    3. R=VI=5.750.20=28.75 Ω
    4. Check: 40×28.75=1150Ω in all, and 2302/1150=46 W.
  240. 16. Magnetism and induction · Magnets have poles and fill space with a field · QuizYou snap a bar magnet in half, between its poles. What do you get?B. Two smaller magnets, each with a north and a south poleCut a magnet as often as you like, down to a single atom, and every piece has both poles. No one has ever found a lone pole, a magnetic monopole. Magnetism comes from moving charge (currents, and the spin of electrons), and a current loop always has two faces.
  241. 16. Magnetism and induction · A magnetic field pushes moving charges sideways · Code challengeWrite orbitRadius(mass, charge, speed, field) (metres) and orbitPeriod(mass, charge, field) (seconds) for a charged particle moving across a uniform magnetic field.One solution:
    Solution
    #include <numbers>
    double orbitRadius(double mass, double charge, double speed,                   double field) {  return mass * speed / (charge * field);}
    double orbitPeriod(double mass, double charge, double field) {  return 2 * std::numbers::pi * mass / (charge * field);}
    Set qvB=mv2/r and one v cancels: r=mv/(qB). The period is the circumference over the speed, 2πr/v=2πm/(qB), and v cancels again. Mass spectrometers use the radius to sort ions by mass; a common slip is squaring v, forgetting it cancels.
  242. 16. Magnetism and induction · A magnetic field pushes moving charges sideways · ProblemIn a mass spectrometer, sodium ions of mass 3.82×10−26kg and charge +e=1.602×10−19C move at 1.0×105m/s across a 0.50T field. Find the radius of their circle, in centimetres, and the time for one full circle, in microseconds.Radius 4.769 cm; Period 2.9965 μs
    1. r=mvqB=(3.82×10−26)(1.0×105)(1.602×10−19)(0.50)≈0.0477 m=4.77 cm
    2. T=2πmqB=2π(3.82×10−26)(1.602×10−19)(0.50)≈3.00×10−6 s=3.00 μs
    3. A heavier ion at the same speed circles wider, which is how the instrument sorts ions by mass.
  243. 16. Magnetism and induction · A current in a field feels a force: the motor · QuizA wire carrying 5 A runs exactly along the direction of a 0.2 T field. What force does it feel?C. NoneWith the current along the field, θ=0 and sin⁡0=0: no force at all. The magnetic force needs motion across the field. This is the difference from an electric field, which pushes a charge whichever way it moves, or if it is still.
  244. 16. Magnetism and induction · Currents make magnetic fields · Fill the blanksA solenoid has 1000 turns per metre and carries 2 A. Fill in its field.1: turnsPerMetre, 2: currentB=μ0nI=4π×10−7×1000×2≈2.51×10−3 T. It doesn’t depend on the solenoid’s width or length, only on how tightly it is wound and the current, as long as it is long compared with its width.
  245. 16. Magnetism and induction · Currents make magnetic fields · ProblemA long, straight power cable carries a steady 200A. Use μ0=4π×10−7T·m/A. What is its magnetic field 5.0m away, in microtesla? How close to the cable would its field equal Earth’s, 50μT?Field at 5.0 m 8 μT; Distance for 50 μT 0.8 m
    1. B=μ0I2πr=(2×10−7)(200)5.0=8.0×10−6 T=8.0 μT
    2. r=μ0I2πB=(2×10−7)(200)50×10−6=0.80 m
    3. Five metres away the cable’s field is about a sixth of Earth’s, so a compass is only thrown off close to it. (Mains cables carry AC, whose field alternates too.)
  246. 16. Magnetism and induction · Faraday: changing flux makes a voltage · Code challengeYou have the flux through a coil (webers) sampled every dt seconds. Write emfs(turns, flux, dt): the induced EMF between each pair of samples, −NΔΦ/Δt, so one fewer value than samples.One solution:
    Solution
    #include <vector>
    std::vector<double> emfs(double turns, const std::vector<double>& flux,                         double dt) {  std::vector<double> result;  for (std::size_t i = 1; i < flux.size(); i++) {    result.push_back(-turns * (flux[i] - flux[i - 1]) / dt);  }  return result;}
    Each EMF is a finite-difference derivative of the flux, times −N. A steady flux, however big, induces nothing: only change counts. The minus sign is Lenz’s law (next slide): it says which way the induced current flows.
  247. 16. Magnetism and induction · Faraday: changing flux makes a voltage · ProblemA coil of 250 turns, each of area 40cm2, sits with its axis along a magnetic field. The field falls steadily from 0.80T to 0.20T in 0.050s. What is the size of the EMF induced? The coil’s ends are joined, and its resistance is 12Ω: what current flows?EMF 12 V; Current 1 A
    1. ΔΦ=ΔBA=(0.80−0.20)(4.0×10−3)=2.4×10−3 Wb per turn
    2. |ℰ|=NΔΦΔt=250×2.4×10−30.050=12 V
    3. I=ℰR=1212=1.0 A
    4. By Lenz’s law the current flows so that its own field points along the original field, trying to keep the flux up.
  248. 16. Magnetism and induction · Lenz's law: induction opposes the change · QuizYou pull a magnet’s north pole away from a coil. What pole does the coil’s induced current make at the end nearest the magnet?B. South, to pull back on the magnet and oppose it leavingPulling the magnet away reduces the flux, so the coil opposes that by trying to keep the magnet: a south pole attracts the departing north. Whichever way you move the magnet, the coil resists. Answering north is the usual slip: Lenz opposes the change, not the field.
  249. 16. Magnetism and induction · A generator spins a coil in a field · QuizThe same coil is turned twice as fast. What happens to the peak EMF?B. It doublesThe peak is NBAω, proportional to ω. Turning faster changes the flux faster, and Faraday’s law says the EMF is the rate of change. So both the frequency and the peak double, which is why a bicycle dynamo’s lamp brightens as you speed up.
  250. 16. Magnetism and induction · Transformers change voltage, and save power on the grid · QuizA charger’s transformer turns 230 V into 5 V. Its primary has 920 turns. How many does the secondary have?B. 20Ns=NpVs/Vp=920×5/230=20 turns. Step-down means fewer secondary turns. (Modern chargers switch at high frequency so a tiny transformer will do, but the turns ratio works the same way.)
  251. 16. Magnetism and induction · Maxwell's equations unite electricity, magnetism and light · QuizWhich of Maxwell’s equations says that magnetic field lines never begin or end?B. Gauss’s law for magnetismZero total magnetic flux out of every closed surface means as many lines go in as come out: there is nowhere for a line to start, no north pole on its own. Gauss’s law for electricity is the same with charge on the right, which is why electric lines can start and end on charges.
  252. 16. Magnetism and induction · Practice problems · ProblemA straight wire 25cm long carries 4.0A through a uniform 0.60T field, at 30∘ to the field. What force does the field exert on it? At what angle to the field, in degrees, would the force be greatest, and how big would it be then?Force at 30° 0.3 N; Angle for the greatest force 90 °; Greatest force 0.6 N
    1. F=BILsin⁡θ=0.60×4.0×0.25×sin⁡30∘=0.60×0.50=0.30 N
    2. sin⁡θ is largest, 1, at θ=90∘: the wire square across the field.
    3. Fmax⁡=BIL=0.60×4.0×0.25=0.60 N
  253. 16. Magnetism and induction · Practice problems · ProblemA generator’s coil has 150 turns of area 0.020m2 and turns 50 times a second in a 0.30T field. Find the peak EMF and the RMS EMF.Peak EMF 150*0.3*0.02*2*pi*50 ≈ 282.743 V; RMS EMF 150*0.3*0.02*2*pi*50/sqrt(2) ≈ 199.93 V
    1. ω=2π×50≈314 rad/s
    2. ℰ0=NBAω=150×0.30×0.020×314.2≈283 V
    3. ℰrms=ℰ02≈200 V
    4. Close to mains voltage, from a modest coil turning at the grid’s 50 Hz.
  254. 16. Magnetism and induction · Practice problems · ProblemAn ideal transformer steps 230V mains down to 12V for a 60W lamp. Its primary coil has 1150 turns. How many turns does the secondary have, what current flows in the lamp, and what current does the transformer draw from the mains?Secondary turns 60; Lamp current 5 A; Mains current 60/230 ≈ 0.26087 A
    1. Ns=NpVsVp=1150×12230=60
    2. Is=PVs=6012=5.0 A,Ip=PVp=60230≈0.26 A
    3. The current is stepped up by the same ratio the voltage is stepped down, 230/12≈19.
  255. 16. Magnetism and induction · Practice problems · ProblemA power station sends 20MW along a line whose total resistance is 2.5Ω. How much power is lost as heat in the line if it is sent at 25kV, and how much if it is sent at 400kV?Loss at 25 kV 1.6 MW; Loss at 400 kV 6.25 kW
    1. at 25 kV: I=20×10625×103=800 A,Ploss=8002×2.5=1.6 MW
    2. at 400 kV: I=20×106400×103=50 A,Ploss=502×2.5=6250 W
    3. At 25 kV the line wastes 8% of the power; at 400 kV, 0.03%. Sixteen times the voltage cuts the loss 162=256 times.
  256. 17. Light and optics · Light is one band of the electromagnetic spectrum · QuizGreen light has a wavelength of 550 nm. What is its frequency?C. About 5.5 × 10¹⁴ Hzf=c/λ=(3.00×108)/(550×10−9)≈5.45×1014 Hz: over five hundred million million wobbles a second. Forgetting to turn nanometres into metres (×10⁻⁹) is the usual slip, and gives 5.5×105 Hz; dividing the wrong way round, λ/c, gives the first option, and multiplying gives the last.
  257. 17. Light and optics · Light is fastest in a vacuum · QuizThe Sun is 1.50×1011 m away. How old is the sunlight reaching you now?B. About 8 minutest=d/c=(1.50×1011)/(3.00×108)=500 s, about 8.3 minutes. You always see the Sun as it was eight minutes ago, and distant stars as they were years or millennia ago. Looking far away is looking back in time.
  258. 17. Light and optics · Light is fastest in a vacuum · ProblemLight of wavelength 600nm in a vacuum enters glass of refractive index 1.50. Use c=3.00×108m/s. Find its speed in the glass, its frequency, and its wavelength in the glass.Speed in the glass 200000000 m/s; Frequency 500000000000000 Hz; Wavelength in the glass 400 nm
    1. v=cn=3.00×1081.50=2.00×108 m/s
    2. f=cλ0=3.00×108600×10−9=5.00×1014 Hz
    3. λ=vf=2.00×1085.00×1014=4.00×10−7 m=400 nm
    4. Slower at the same frequency, so shorter waves: λ=λ0/n. The light doesn’t turn violet in the glass; its colour goes with its frequency, which hasn’t changed.
  259. 17. Light and optics · Reflection: the angles are equal · QuizWhat is the shortest flat wall mirror in which a 1.80 m tall person can see all of themselves?B. 0.90 mLight from your feet reflects off the mirror halfway between your eyes and feet in height, and light from the top of your head halfway between your eyes and your head. So the mirror needs half your height, 0.90 m, hung at the right place. Standing further back doesn’t help: the image recedes just as fast.
  260. 17. Light and optics · Refraction and Snell's law · Code challengeWrite refractionAngle(n1, n2, degrees): the angle of the refracted ray in degrees, or std::nullopt, no angle, if there is none (when Snell’s law would need a sine above 1).One solution:
    Solution
    #include <cmath>#include <numbers>#include <optional>
    std::optional<double> refractionAngle(double n1, double n2,                                      double degrees) {  double toRad = std::numbers::pi / 180;  double sine = n1 * std::sin(degrees * toRad) / n2;  if (sine > 1) return std::nullopt;  return std::asin(sine) / toRad;}
    std::sin and std::asin work in radians, so convert in and out. A std::optional<double> holds a double or nothing: return std::nullopt; is the nothing, returning a number fills it, and the caller checks has_value() before reading it with *. Going from glass to air at 60°, sin⁡θ2=1.5×0.866=1.30: no angle has that sine, so no light gets out at all. That case has a name, and it is the next slide.
  261. 17. Light and optics · Refraction and Snell's law · ProblemA ray of light in water (n=1.33) meets the surface at 35∘ from the normal and passes out into air (n=1.00). At what angle from the normal does it leave, in degrees?θ2= 49.72 °
    1. 1.33sin⁡35∘=1.00sin⁡θ2⇒sin⁡θ2=1.33×0.574≈0.763
    2. θ2=sin−1⁡(0.763)≈49.7∘
    3. Into a faster material it bends away from the normal: 49.7° is more than 35°.
  262. 17. Light and optics · Total internal reflection traps light · QuizA swimmer underwater looks up. Beyond what angle from straight up does the surface look like a silver mirror? (Water n=1.33.)C. About 48.8°sin⁡θc=1/1.33=0.752, so θc≈48.8°. Inside a cone of that half-angle the swimmer sees the whole sky squeezed in (Snell’s window); outside it, the surface reflects the pool. 41.8° is glass’s critical angle, not water’s.
  263. 17. Light and optics · Lenses form images: the thin lens equation · Code challengeWrite imageDistance(f, objectDistance) and magnification(f, objectDistance) from the thin lens equation. Distances are in any one unit; use the sign convention above.One solution:
    Solution
    double imageDistance(double f, double objectDistance) {  return 1 / (1 / f - 1 / objectDistance);}
    double magnification(double f, double objectDistance) {  return -imageDistance(f, objectDistance) / objectDistance;}
    Rearrange: 1/di=1/f−1/do, then take the reciprocal. Forgetting that last step (returning 1/di) is the usual mistake. The signs carry the meaning: a negative image distance is virtual, a negative magnification inverted.
  264. 17. Light and optics · Lenses form images: the thin lens equation · ProblemA projector’s lens has a focal length of 12cm and makes a sharp image of a slide on a screen 6.0m from the lens. How far from the lens is the slide, and what is the magnification?Slide distance do 600/49 ≈ 12.2449 cm; Magnification m -49
    1. 1do=1f−1di=112−1600=50−1600=49600
    2. do=60049≈12.24 cm,m=−dido=−600600/49=−49
    3. The slide sits just outside the focal point, and the image is 49 times bigger and upside down, which is why slides go into a projector upside down.
  265. 17. Light and optics · Two slits prove light is a wave · QuizIn Young’s experiment, which change spreads the fringes further apart?A. Moving the slits closer togetherΔy=λL/d: a smaller d means wider spacing. Blue light has a shorter λ, which narrows them; a closer screen (smaller L) narrows them too; brightness changes how bright the fringes are, not where.
  266. 17. Light and optics · Two slits prove light is a wave · ProblemRed laser light of wavelength 633nm passes through two slits 0.20mm apart onto a screen 3.0m away. What is the spacing of the bright fringes, in millimetres? At what angle from the centre, in degrees, is the second bright fringe on one side (m=2)?Fringe spacing 9.495 mm; Angle of the second fringe 0.3627 °
    1. Δy=λLd=(633×10−9)(3.0)0.20×10−3≈9.50×10−3 m=9.50 mm
    2. sin⁡θ=2λd=2(633×10−9)0.20×10−3=6.33×10−3⇒θ≈0.363∘
    3. The angle is tiny, which is why the small-angle formula for Δy works: Ltan⁡θ≈19.0 mm, two fringe spacings.
  267. 17. Light and optics · Polarisation shows light is transverse · QuizTwo crossed polarisers (at 90°) block all light. You slip a third between them, at 45° to each. What fraction of the original unpolarised light now gets through?B. One eighthThe first filter passes half. The 45° one passes cos2⁡45°=12 of that, and the last another 12: 12×12×12=18. Adding a filter lets more light through, because each filter re-polarises the light along its own axis rather than just removing some.
  268. 17. Light and optics · Practice problems · ProblemAn optical fibre has a glass core of refractive index 1.48 inside a cladding of index 1.46. What is the critical angle at the boundary between core and cladding, in degrees? How long does light take to travel 100km through the core, in milliseconds, ignoring its zigzag? Use c=3.00×108m/s.Critical angle 80.57 °; Time 1.48/3 ≈ 0.493333 ms
    1. sin⁡θc=n2n1=1.461.48≈0.9865⇒θc≈80.6∘
    2. Only rays meeting the wall at more than 80.6° from the normal, nearly parallel to the fibre, stay trapped.
    3. v=cn=3.00×1081.48≈2.03×108 m/s,t=1.00×1052.03×108≈4.93×10−4 s=0.493 ms
    4. About half a millisecond per 100 km, part of the delay you notice on a call to the other side of the world.
  269. 17. Light and optics · Practice problems · ProblemYou look at a stamp through a magnifying glass of focal length 8.0cm, held 6.0cm from the stamp. Where is the image, and what is the magnification? (A virtual image has a negative image distance.)Image distance di -24 cm; Magnification m 4
    1. 1di=1f−1do=18.0−16.0=3−424=−124⇒di=−24 cm
    2. m=−dido=−−246.0=+4
    3. A virtual image 24 cm from the lens, on the stamp’s side, upright and four times as big.
  270. 17. Light and optics · Practice problems · ProblemGreen light of wavelength 550nm falls straight onto a diffraction grating with 600 lines per millimetre. At what angle, in degrees, is the first-order bright line? What is the highest order that can be seen?First-order angle 19.27 °; Highest order 3
    1. d=1600 mm≈1.667×10−6 m
    2. sin⁡θ1=λd=550×10−91.667×10−6=0.330⇒θ1≈19.3∘
    3. m≤dλ=1.667×10−6550×10−9≈3.03
    4. So the third order, at about 82°, is the last; a fourth would need sin⁡θ=1.32.
  271. 17. Light and optics · Practice problems · ProblemUnpolarised light of intensity 80W/m2 passes through a polarising filter, then through a second filter whose axis is at 60∘ to the first. What is the intensity after the first filter, and after the second?After the first 40 W/m²; After the second 10 W/m²
    1. The first filter passes half of unpolarised light: 40 W/m².
    2. I=40cos2⁡60∘=40×(0.5)2=10 W/m2
    3. Using cos⁡60∘ instead of its square, the usual slip, would give 20 W/m².
  272. 18. Special relativity · Simultaneous for you is not simultaneous for me · QuizIn the train example, who is right about whether the doors opened at the same time?C. Both: simultaneity depends on the frame, and each measures correctly in their ownBy the first postulate no frame is special, so neither has a better claim. It is not a delay of signals either: each observer has already corrected for the time light takes to reach them. The disagreement is about when the events happened, and it is real.
  273. 18. Special relativity · The Lorentz factor measures how relativistic you are · Code challengeWrite lorentz(speed): the Lorentz factor for a speed in m/s. Use c = 299 792 458 m/s.One solution:
    Solution
    #include <cmath>
    const double c = 299792458;  // m/s
    double lorentz(double speed) {  double beta = speed / c;  return 1 / std::sqrt(1 - beta * beta);}
    Divide by c first, then square: γ=1/1−β2. At 0.8c, 1−0.64=0.36, whose root is 0.6, so γ=5/3. Squaring c and v separately works too but makes huge numbers. At a jet’s speed γ−1≈β2/2≈3.5×10−13: that is why Newton is right in everyday life.
  274. 18. Special relativity · The Lorentz factor measures how relativistic you are · ProblemA spaceship travels at 0.96c. Find its Lorentz factor. Then find the speed, as a fraction of c, at which γ=2.γ at 0.96c 25/7 ≈ 3.57143; Speed for γ=2 sqrt(3)/2 ≈ 0.866025 c
    1. γ=11−0.962=10.0784=10.28=257≈3.57
    2. 11−β2=2⇒1−β2=14⇒β=32≈0.866
    3. Even at 87% of light’s speed, γ is only 2: it stays near 1 for a long way, then climbs steeply.
  275. 18. Special relativity · Moving clocks run slow · QuizOne twin flies to a star at 0.8c and back; the other stays on Earth. Each sees the other’s clock running slow while they move apart. Who is younger when they meet again?C. The traveller, who changed inertial frames by turning roundThe situation isn’t symmetric: the traveller accelerates to turn round, switching from an outbound inertial frame to an inbound one, and that switch changes which distant events are simultaneous for them. Worked through, the traveller ages less; at 0.8c, 60% as much. Meeting again at one place, they can compare clocks directly, so they must agree.
  276. 18. Special relativity · Moving clocks run slow · ProblemPions live for 26ns on average, measured at rest. A beam of them moves at 0.95c through a laboratory. Use c=3.00×108m/s. How long does an average pion live, as measured in the lab, and how far does it travel in that time?Lifetime in the lab 26/sqrt(1 - 0.95^2) ≈ 83.2666 ns; Distance 0.95*0.3*26/sqrt(1 - 0.95^2) ≈ 23.731 m
    1. γ=11−0.952=10.0975≈3.20
    2. Δt=γΔt0≈3.20×26 ns≈83 ns
    3. d=vΔt=0.95×(3.00×108)×(83.3×10−9)≈23.7 m
    4. Without time dilation it would go only 0.95×3.00×108×26×10−9≈7.4 m. Particle physicists design beam lines around the longer figure, because it is the one that happens.
  277. 18. Special relativity · Moving lengths shrink, along the motion · QuizA spaceship 100 m long at rest flies past you at 0.6c. How long do you measure it to be?B. 80 mAt 0.6c, γ=1/1−0.36=1/0.8=1.25, so L=100/1.25=80 m. 125 m would be multiplying by γ: lengths contract, times dilate. And 60 m uses β directly, which isn’t how either effect goes.
  278. 18. Special relativity · Moving lengths shrink, along the motion · ProblemProxima Centauri is 4.24 light-years from Earth, measured in Earth’s frame (a light-year is the distance light travels in a year). A ship flies there at 0.80c. How long does the trip take in Earth’s frame, how long by the ship’s clocks, and how far away is the star as measured by the crew?Time in Earth’s frame 5.3 years; Time on the ship 3.18 years; Distance for the crew 2.544 light-years
    1. tEarth=4.24 light-years0.80c=5.30 years
    2. γ=11−0.64=10.6,tship=tEarthγ=5.30×0.6=3.18 years
    3. L=L0γ=4.24×0.6≈2.54 light-years
    4. The two views agree: for the crew, 2.544 light-years at 0.80c takes 2.544/0.80=3.18 years.
  279. 18. Special relativity · Speeds add in a way that never passes c · Code challengeWrite addVelocities(u, v), with both speeds as fractions of c: the speed you measure for something moving at u relative to a frame that moves at v relative to you.One solution:
    Solution
    double addVelocities(double u, double v) {  return (u + v) / (1 + u * v);}
    In units of c the c2 disappears: u=(u′+v)/(1+u′v). However close to 1 both are, the result stays below 1, and if either is exactly 1 the result is exactly 1. Speeds near light’s don’t pile up; the denominator grows to stop them.
  280. 18. Special relativity · Speeds add in a way that never passes c · ProblemA ship moving at 0.70c relative to Earth launches a probe at 0.50c relative to the ship. Find the probe’s speed relative to Earth, as a fraction of c, if it is launched straight forward, and if it is launched straight backward.Forward 8/9 ≈ 0.888889 c; Backward 4/13 ≈ 0.307692 c
    1. u=0.50+0.701+0.50×0.70c=1.201.35c=89c≈0.889c
    2. u=−0.50+0.701−0.50×0.70c=0.200.65c=413c≈0.308c
    3. Newton would say 1.2c and 0.2c. The forward answer is held below c, and even the backward one comes out a little faster than 0.2c.
  281. 18. Special relativity · Spacetime diagrams draw time and space together · QuizTwo events are 1 s apart in time and 2 light-seconds apart in space. Which is true?B. Their interval is spacelike: no signal can link them, and frames even disagree on their orders2=12−22=−3<0: spacelike. Linking them would take a signal at twice light’s speed. For such pairs some frames see A first, some B, some both at once, which is fine because neither can affect the other. Cause and effect are only ever timelike or lightlike, and every frame agrees on their order.
  282. 18. Special relativity · Mass is a form of energy · Code challengeWrite kineticEnergy(mass, speed): the relativistic kinetic energy in joules for a mass in kg and a speed in m/s.One solution:
    Solution
    #include <cmath>
    const double c = 299792458;  // m/s
    double kineticEnergy(double mass, double speed) {  double beta = speed / c;  double gamma = 1 / std::sqrt(1 - beta * beta);  return (gamma - 1) * mass * c * c;}
    Ek=(γ−1)mc2. At 0.8c, γ=5/3, so Ek=23mc2, while 12mv2 would give 0.32mc2, less than half as much. Don’t use γmc2 alone: that is the total energy, including the rest energy.
  283. 18. Special relativity · A glimpse of general relativity: gravity is curved spacetime · QuizGPS satellite clocks move fast (special relativity: about 7 μs a day slow) and sit higher in Earth’s gravity (general relativity: about 45 μs a day fast). What happens if engineers ignore both?B. The clocks gain about 38 μs a day, and positions drift by roughly 10 km a dayThe two effects partly cancel: 45−7=38 μs a day fast. GPS finds you by timing radio signals, and light goes about 300 m in a microsecond, so 38 μs is over 11 km of error per day. The satellites’ clocks are deliberately set to tick slow before launch to make up for it.
  284. 18. Special relativity · Practice problems · ProblemIn Earth’s frame, event B happens 10s after event A and 6.0 light-seconds away from it. Find the spacetime interval s between them, in light-seconds, the time shown by a clock carried at a steady speed from A to B, and that speed as a fraction of c.s 8 light-seconds; Time on the carried clock 8 s; Speed 0.6 c
    1. s2=102−6.02=64⇒s=8.0 light-seconds
    2. It is timelike, so a clock can go from A to B, and it reads the proper time s/c=8.0 s.
    3. v=ΔxΔt=6.0 light-seconds10 s=0.60c
    4. Check with time dilation: at 0.60c, γ=1.25, and 10/1.25=8.0 s.
  285. 18. Special relativity · Practice problems · ProblemA proton (m=1.673×10−27kg) moves at 0.90c. Using c=2.998×108m/s and 1MeV=1.602×10−13J, find γ, the proton’s rest energy in MeV, and its kinetic energy in MeV.γ 1/sqrt(0.19) ≈ 2.29416; Rest energy 938.6 MeV; Kinetic energy 1214.7 MeV
    1. γ=11−0.81=10.19≈2.294
    2. E0=mc2=(1.673×10−27)(2.998×108)2≈1.504×10−10 J≈939 MeV
    3. Ek=(γ−1)mc2≈1.294×938.6≈1215 MeV
    4. Its kinetic energy is more than its rest energy. Newton’s 12mv2=0.405mc2≈380 MeV would be far too low.
  286. 18. Special relativity · Practice problems · ProblemA city uses electrical power at an average of 1.0GW for a year (3.16×107s). How much energy is that, and how much mass, in grams, would have to be turned entirely into energy to supply it? Use c=3.00×108m/s.Energy 31600000000000000 J; Mass 3.16e16/9e16*1000 ≈ 351.111 g
    1. E=Pt=(1.0×109)(3.16×107)=3.16×1016 J
    2. m=Ec2=3.16×1016(3.00×108)2≈0.351 kg=351 g
    3. A year of a city’s electricity from a third of a kilogram. Burning fuel turns far less than a millionth of its mass into energy, which is why a coal power station of that size burns millions of tonnes a year.
  287. 19. Quantum physics · Hot things glow, and classical physics got it wrong · QuizA red star has a surface of 3000 K and a blue one 12 000 K. How do their peak wavelengths compare?A. The blue star’s peak is four times shorterWien’s law: λmax⁡∝1/T. Four times hotter means a peak at a quarter of the wavelength: 966 nm (infrared) against 242 nm (ultraviolet). Sixteen would come from mixing it up with the T4 of radiated power.
  288. 19. Quantum physics · Hot things glow, and classical physics got it wrong · ProblemYour skin is at about 34 °C. At what wavelength, in micrometres, does it glow most brightly? A star’s spectrum peaks at 290nm: what is its surface temperature? Use b=2.898×10−3m·K.Skin’s peak 2898/307.15 ≈ 9.43513 μm; Star’s temperature 2.898e-3/290e-9 ≈ 9993.1 K
    1. λmax⁡=bT=2.898×10−3307.15≈9.4×10−6 m=9.4 μm
    2. Deep in the infrared: thermal cameras see people by this glow.
    3. T=bλmax⁡=2.898×10−3290×10−9≈1.0×104 K
    4. A hot, blue-white star, whose glow peaks in the ultraviolet.
  289. 19. Quantum physics · The photoelectric effect: light comes in photons · Code challengeWrite maxKineticEV(wavelengthNm, workFunctionEV): the largest kinetic energy (eV) of electrons knocked out by light of that wavelength, or 0 if none escape. A photon’s energy in eV is 1239.84 divided by its wavelength in nm.One solution:
    Solution
    #include <algorithm>
    const double HC = 1239.84;  // eV·nm
    double maxKineticEV(double wavelengthNm, double workFunctionEV) {  double photon = HC / wavelengthNm;  return std::max(0.0, photon - workFunctionEV);}
    Photon energy minus work function, but never below zero: a photon that can’t free an electron frees none, and brighter red light only means more photons that each fall short. In C++ it is std::max(0.0, …), not std::max(0, …): both arguments must have the same type, and 0 is an int. The handy constant hc=1239.84 eV·nm saves converting to joules and metres.
  290. 19. Quantum physics · The photoelectric effect: light comes in photons · ProblemUltraviolet light of wavelength 250nm shines on tungsten, whose work function is 4.50eV. Use hc=1239.84eV·nm. Find each photon’s energy, the largest kinetic energy of the electrons knocked out, and the longest wavelength that can knock out any electrons at all.Photon energy 1239.84/250 ≈ 4.95936 eV; Largest kinetic energy 1239.84/250 - 4.5 ≈ 0.45936 eV; Longest wavelength 1239.84/4.5 ≈ 275.52 nm
    1. E=hcλ=1239.84250≈4.96 eV
    2. Ek,max⁡=hf−φ=4.96−4.50≈0.46 eV
    3. λ0=hcφ=1239.844.50≈276 nm
    4. Longer wavelengths, all of visible light included, knock nothing out of tungsten, however bright.
  291. 19. Quantum physics · A photon carries energy and momentum · QuizWhy does ultraviolet light cause sunburn while a much brighter red lamp doesn’t?B. Each UV photon has enough energy to break molecular bonds in skin cells; red photons don’t, however many there areDamage happens one photon at a time. A 300 nm UV photon carries about 4.1 eV, enough to break bonds in DNA; a 700 nm red photon carries 1.8 eV, which isn’t. A brighter lamp only sends more of the weak photons. The same idea is the photoelectric effect’s threshold.
  292. 19. Quantum physics · A photon carries energy and momentum · ProblemA green laser pointer gives out 5.0mW of light of wavelength 532nm. Find the energy of one photon in eV and in units of 10−19J, and the number of photons it gives out each second. Use hc=1239.84eV·nm and 1eV=1.602×10−19J.Photon energy in eV 1239.84/532 ≈ 2.33053 eV; Photon energy in joules 1239.84/532*1.602 ≈ 3.7335 × 10⁻¹⁹ J; Photons each second 5e-3/(1239.84/532*1.602e-19) ≈ 13392247946615962
    1. E=1239.84532≈2.33 eV=2.33×1.602×10−19≈3.73×10−19 J
    2. N=PE=5.0×10−3 J/s3.73×10−19 J≈1.34×1016 per second
    3. Over ten million billion photons a second: why light from everyday sources looks smooth and continuous.
  293. 19. Quantum physics · Everything is both wave and particle · Fill the blanksFill in the de Broglie wavelength of a neutron.1: mass * speedλ=h/(mv)=6.626×10−34/(1.675×10−27×2200)≈1.8×10−10 m. Neutrons that slow have wavelengths like the spacing of atoms, so they diffract off crystals: a way of seeing where atoms sit.
  294. 19. Quantum physics · Everything is both wave and particle · ProblemIn 1927 Clinton Davisson and Lester Germer fired electrons, accelerated from rest through 54V, at a nickel crystal and saw them diffract. Find the electrons’ momentum, in units of 10−24kg·m/s, and their de Broglie wavelength in nanometres. Use e=1.602×10−19C, me=9.109×10−31kg and h=6.626×10−34J·s.Momentum 3.97 × 10⁻²⁴ kg·m/s; Wavelength 0.1669 nm
    1. Ek=eΔV=(1.602×10−19)(54)≈8.65×10−18 J
    2. p=2meEk=2×9.109×10−31×8.65×10−18≈3.97×10−24 kg·m/s
    3. λ=hp=6.626×10−343.97×10−24≈1.67×10−10 m=0.167 nm
    4. Close to the spacing of the atoms in nickel, so the crystal acts as a grating. It was one of the first experiments to show de Broglie was right.
  295. 19. Quantum physics · You can’t know position and momentum exactly at once · QuizWhat does the uncertainty principle say?B. A particle doesn’t have an exact position and an exact momentum at the same time; it is a property of waves, not of measuring toolsIt is often explained as the measurement “disturbing” the particle, but the limit is deeper: it follows from the particle being described by a wave, and a wave can’t be both short and of one wavelength. No improved instrument gets round it. It is also why atoms don’t collapse: an electron squeezed onto the nucleus would need enormous momentum.
  296. 19. Quantum physics · Atoms have energy levels, and spectra are their fingerprints · Code challengeWrite hydrogenLineNm(upper, lower): the wavelength in nm of the photon given off when hydrogen’s electron drops from level upper to level lower. Use En=−13.6/n2 eV and hc=1239.84 eV·nm.One solution:
    Solution
    const double HC = 1239.84;  // eV·nm
    double hydrogenLineNm(int upper, int lower) {  auto level = [](int n) { return -13.6 / (n * n); };  double photon = level(upper) - level(lower);  return HC / photon;}
    The photon’s energy is upper minus lower: −1.511−(−3.4)=1.889 eV for 3 → 2, and λ=1239.84/1.889≈656 nm. level is a lambda, a small function made inside another, and -13.6 / (n * n) is a fraction because -13.6 is a double. Taking lower minus upper gives a negative energy and a negative wavelength: the sign check catches it. Astronomers see this red line in glowing hydrogen clouds across the universe.
  297. 19. Quantum physics · The wavefunction gives probabilities · QuizA particle in a box is in the state n = 2. Where are you least likely to find it?A. Near the walls and exactly in the middle|ψ2|2=2sin2⁡(2πx) is zero at x=0,0.5,1: the walls and the middle are nodes. The particle is most likely a quarter of the way in from either side. Yet it still “gets” from one half to the other: it isn’t a small ball moving about, so the classical question of how it crosses the node doesn’t apply.
  298. 19. Quantum physics · Particles tunnel through walls · QuizAn electron tunnels through a 0.5 nm barrier with a chance of about 0.006. Roughly what is its chance through 1.0 nm of the same barrier?C. About 0.00004Doubling the width squares the exponential: e−2κ(2d)=(e−2κd)2≈0.0062=3.6×10−5. Halving (0.003) is what you’d guess if it fell in proportion. Tunnelling’s steep dependence on width is what makes an STM sensitive to single atoms.
  299. 19. Quantum physics · The nucleus, and radioactive decay · Code challengeWrite remaining(n0, halfLife, t): how much is left after time t. Then age(fraction, halfLife): how long it takes to fall to that fraction. Both use the same time unit as halfLife.One solution:
    Solution
    #include <cmath>
    double remaining(double n0, double halfLife, double t) {  return n0 * std::pow(0.5, t / halfLife);}
    double age(double fraction, double halfLife) {  // fraction = std::pow(0.5, t / halfLife), so take logarithms  return halfLife * std::log(fraction) / std::log(0.5);}
    Decay is a power of one half, and undoing a power needs a logarithm: t=t1/2ln⁡(f)/ln⁡(12). Any base of logarithm works, as long as you use the same on top and bottom. A straight-line guess (10% left means 90% of the way to zero) is the usual mistake: decay never reaches zero; it keeps halving.
  300. 19. Quantum physics · Practice problems · ProblemHydrogen’s electron drops from level n=5 to n=2. Using En=−13.6/n2eV and hc=1239.84eV·nm, find the photon’s energy and its wavelength. How much energy would free the electron completely from n=2?Photon energy 2.856 eV; Wavelength 1239.84/2.856 ≈ 434.118 nm; Energy to free it from n = 2 3.4 eV
    1. E5=−13.625=−0.544 eV,E2=−13.64=−3.40 eV
    2. hf=E5−E2=−0.544−(−3.40)=2.856 eV,λ=1239.842.856≈434 nm
    3. A violet line, one of hydrogen’s four visible lines. Freeing the electron from n=2 takes 0−(−3.40)=3.40 eV.
  301. 19. Quantum physics · Practice problems · ProblemA proton is confined inside a nucleus, so the uncertainty in its position is about Δx=1.0×10−14m. Find the smallest possible uncertainty in its momentum, in units of 10−21kg·m/s, and in its speed. Use ℏ=1.055×10−34J·s and mp=1.673×10−27kg.Momentum 5.275 × 10⁻²¹ kg·m/s; Speed 5.275e-21/1.673e-27 ≈ 3153020 m/s
    1. Δp≥ℏ2Δx=1.055×10−342×1.0×10−14≈5.28×10−21 kg·m/s
    2. Δv=Δpmp=5.28×10−211.673×10−27≈3.2×106 m/s
    3. About 1% of the speed of light: protons and neutrons in a nucleus can’t sit still, and move fast.
  302. 19. Quantum physics · Practice problems · ProblemA wooden bowl has 30% of the carbon-14 that living wood has. Carbon-14’s half-life is 5730 years. How many half-lives have passed, and how old is the bowl?Half-lives ln(0.3)/ln(0.5) ≈ 1.73697; Age 5730*ln(0.3)/ln(0.5) ≈ 9952.81 years
    1. 0.30=(12)n⇒n=ln⁡0.30ln⁡0.5≈1.74
    2. t=nt1/2≈1.737×5730≈9950 years
    3. A sense check: 50% left is one half-life and 25% is two, so 30% lands in between, nearer two.
  303. 19. Quantum physics · Practice problems · ProblemAn electron is trapped in a box 0.50nm wide. Use En=n2h2/(8mL2) with h=6.626×10−34J·s and me=9.109×10−31kg, and 1eV=1.602×10−19J, hc=1239.84eV·nm. Find its lowest energy in eV, and the wavelength of the photon it gives off when it drops from n=2 to n=1.Lowest energy E1 1.5043 eV; Photon wavelength 274.73 nm
    1. E1=h28mL2=(6.626×10−34)28×9.109×10−31×(0.50×10−9)2≈2.41×10−19 J≈1.50 eV
    2. E2−E1=3E1≈4.51 eV,λ=1239.844.51≈275 nm
    3. Ultraviolet. A box the size of a small molecule gives energies of a few eV, the same scale as chemistry, and that is no coincidence.
  304. 20. Physics like an expert · How an expert attacks a problem · Put in orderPut the steps of solving a physics problem in order.1. Draw a sketch and write down what is known and what is wanted 2. Choose the principle: Newton’s laws, energy, momentum, a field law 3. Solve with symbols, to get a formula for the unknown 4. Check the formula’s units and its limiting cases 5. Put in the numbers and judge whether the size is sensibleSymbols first: a formula shows how the answer depends on each quantity, can be checked by units and limits, and carries fewer rounding errors. Numbers go in last, and the final sanity check (“a person can’t run at 300 m/s”) is the expert’s habit that beginners skip.
  305. 20. Physics like an expert · How an expert attacks a problem · QuizA student finds that a block sliding from rest down a frictionless ramp of height h at angle θ reaches the bottom at v=2ghsin⁡θ. What shows it is wrong without redoing the work?B. Energy conservation says the speed depends only on h, not on the angle; and at θ → 0 a very long, gentle ramp would give zero speedThe units are fine (sin θ has none), so a units check passes. But mgh=12mv2 gives v=2gh for any frictionless ramp, and the limit of a nearly flat but very long ramp still drops by h. Limits catch what units miss.
  306. 20. Physics like an expert · Simulating motion: not all steps are equal · ProblemYou simulate a spring with ω=2.0rad/s by Euler’s method with a step Δt=0.010s. Each step multiplies the energy by 1+ω2Δt2. By what factor has the energy grown after 10s of simulated time? What is the largest step, in milliseconds, that keeps the growth over those 10s under 1%?Growth factor (1 + 4e-4)^1000 ≈ 1.49171; Largest step 0.2488 ms
    1. Ten seconds is 10/0.010=1000 steps, each multiplying the energy by 1+4.0×0.0102=1.0004:
    2. (1.0004)1000≈e0.400≈1.49
    3. Almost 50% more energy than the spring started with. For the second part, 10/Δt steps must give a factor under 1.01:
    4. (1+ω2Δt2)10/Δt≈e(10/Δt)ω2Δt2=e40Δt<1.01Δt<ln⁡1.0140≈2.49×10−4 s=0.249 ms
    5. Forty times smaller steps, and forty times the work, just to hold Euler’s error to 1% over 10 s. Velocity Verlet keeps the energy within about 0.01% with the original step.
  307. 20. Physics like an expert · Write a Verlet stepper · Code challengeWrite verletStep(x, v, dt, accel): one velocity Verlet step, returning the new State, its x and v. accel(x) gives the acceleration at position x.One solution:
    Solution
    #include <functional>
    struct State {  double x;  // position  double v;  // velocity};
    State verletStep(double x, double v, double dt,                 std::function<double(double)> accel) {  double a = accel(x);  double newX = x + v * dt + 0.5 * a * dt * dt;  double newV = v + 0.5 * (a + accel(newX)) * dt;  return {newX, newV};}
    Move the position with the current velocity and acceleration, work out the acceleration at the new position, and update the velocity with the average of the two. That averaging makes the step time-reversible and symplectic. Plain Euler, the starter, ends the second test with an energy around 0.5×1.0110000: astronomically wrong.
  308. 20. Physics like an expert · Runge–Kutta: sample the slope four times · Code challengeWrite rk4Step(f, t, y, h): one RK4 step of size h for dy/dt = f(t, y), returning the new y. A Slope is any function of t and y.One solution:
    Solution
    #include <functional>
    using Slope = std::function<double(double t, double y)>;
    double rk4Step(Slope f, double t, double y, double h) {  double k1 = f(t, y);  double k2 = f(t + h / 2, y + h / 2 * k1);  double k3 = f(t + h / 2, y + h / 2 * k2);  double k4 = f(t + h, y + h * k3);  return y + h / 6 * (k1 + 2 * k2 + 2 * k3 + k4);}
    Each k is a slope; k2 and k3 peek at the midpoint using the slopes before them, and the final average weights the middle twice. With just 10 steps it reaches e to about six digits, where Euler gets 2.594. The usual slip is evaluating k2 and k3 at t instead of t+h/2, which the second test catches. using Slope = …; only gives the long std::function type a short name; the tests pass lambdas as f.
  309. 20. Physics like an expert · The Lagrangian: physics from one function · QuizA mass m on a spring of stiffness k has L=12mx˙2−12kx2. What equation of motion does the Euler–Lagrange equation give?B. mx¨=−kx∂L/∂x˙=mx˙, whose time derivative is mx¨; ∂L/∂x=−kx. So mx¨−(−kx)=0, that is mx¨=−kx: Hooke’s law and Newton’s second law, recovered. Losing the minus sign of V in L=T−V gives the first option, an exploding spring.
  310. 20. Physics like an expert · Noether: every symmetry gives a conservation law · QuizA puck slides without friction over a vast surface whose height changes along x but not along y. Which component of the puck’s momentum must be conserved?B. The y componentSliding the whole setup along y changes nothing, so by Noether the y momentum is conserved. Along x the height, and so the potential energy, changes, which means forces act along x and its momentum changes. Symmetry tells you what is conserved before you solve anything.
  311. 20. Physics like an expert · Chaos: simple rules, unpredictable motion · QuizA chaotic system’s gap grows by a factor of 10 every 2 s. You measure its start 1000 times more precisely. How much longer can you predict it?B. About 6 s longerExponential growth turns precision into time only logarithmically: a factor of 1000 is 103, three factors of ten, at 2 s each: 6 s more. That is the practical meaning of chaos: better measurements help, but very slowly.
  312. 20. Physics like an expert · Chaos: simple rules, unpredictable motion · ProblemIn a weather model, small errors grow as eλt and double every 2.0 days. Find the Lyapunov exponent λ, per day. An error starts at 0.1% of the size that makes a forecast useless: after how many days is the forecast useless?λ ln(2)/2 ≈ 0.346574 per day; Time 2*ln(1000)/ln(2) ≈ 19.9316 days
    1. e2λ=2⇒λ=ln⁡22≈0.347 per day
    2. eλt=1000⇒t=ln⁡1000λ=2ln⁡1000ln⁡2≈19.9 days
    3. About ten doublings (210≈1000), so about 20 days. Starting ten times more precisely adds only 2ln⁡10/ln⁡2≈6.6 days.
  313. 20. Physics like an expert · The Standard Model: what everything is made of · QuizA proton is made of three quarks and has charge +1. Which three?B. up, up, down23+23−13=1. A neutron is up, down, down: 23−13−13=0. Beta decay (module 19) is one down quark turning into an up, through the weak force, giving off a W boson that becomes an electron and an antineutrino.
  314. 20. Physics like an expert · What physics doesn’t know yet · QuizWhat is the main evidence that dark matter exists?B. Its gravity: galaxies’ rotation, clusters’ motion and the bending of light need more mass than we can seeDark matter has so far been detected only through its gravity, in many independent ways that agree. Cold gas and dust do exist, but they absorb and emit light at other wavelengths, and measured amounts fall far short. Experiments to catch a dark matter particle directly are running, so far without a confirmed signal.
  315. 20. Physics like an expert · Mastery questions: everything at once · QuizWhich combination of g (m/s²) and a length L (m) has the units of a speed?B. gLgL has units m²/s², and its square root m/s. L/g is a time (a pendulum’s period, up to 2π), and g/L is a rate squared. Water waves in a shallow sea really travel at gh: dimensional analysis can find formulas, not just check them.
  316. 20. Physics like an expert · Mastery questions: everything at once · QuizAn electron and a proton are both accelerated from rest through 1000 V. Which has the shorter de Broglie wavelength?B. The proton: both gain 1000 eV, but p=2mEk is larger for the heavier oneSame charge size, same voltage, so the same kinetic energy e×1000 V (module 14). Momentum is 2mEk, about 43 times bigger for the proton (1836), and λ=h/p is 43 times shorter (module 19). Being faster doesn’t matter; momentum does.
  317. 20. Physics like an expert · Mastery questions: everything at once · QuizYou simulate a capacitor discharging, dV/dt=−V/RC, with Euler’s method and a step of Δt=3RC. What does the simulation do?C. Flips sign each step and grows: V is multiplied by −2 each timeVn+1=Vn−VnΔt/RC=Vn(1−3)=−2Vn. The simulation explodes while the real voltage quietly dies away. Euler is only stable for this equation when Δt<2RC, and only accurate when Δt≪RC: always compare the step size with the system’s fastest time scale (module 15).
  318. 20. Physics like an expert · Mastery questions: everything at once · QuizA satellite in a circular orbit fires its engine forward along its motion, twice (once to climb, once to round off the orbit), and ends in a higher circular orbit. How does its new speed compare with the old?B. SlowerA circular orbit has v=GM/r (module 9): a higher orbit is slower. The burn adds energy, which goes into potential energy as it climbs, and more than the extra kinetic energy is used up on the way. Spacecraft chasing one ahead in the same orbit first slow down, dropping lower and faster.
  319. 20. Physics like an expert · Practice problems: motion and energy · ProblemA skier starts from rest and slides down a frictionless slope, dropping 45m vertically to a ramp that launches her horizontally. The ramp’s lip is 20m above flat ground. Take g=9.8m/s2 and ignore air resistance. How fast does she leave the ramp, how long is she in the air, and how far from the lip, horizontally, does she land?Launch speed sqrt(2*9.8*45) ≈ 29.6985 m/s; Time in the air sqrt(40/9.8) ≈ 2.02031 s; Horizontal distance 60 m
    1. mgh1=12mv2⇒v=2gh1=2×9.8×45≈29.7 m/s
    2. Launched horizontally, she starts with no vertical speed and falls 20 m:
    3. 20=12×9.8×t2⇒t=409.8≈2.02 s,x=vt≈29.7×2.02≈60 m
    4. In symbols, x=2gh12h2/g=2h1h2=245×20=60 m exactly: g cancels, so the jump would be just as long on the Moon (though slower).
  320. 20. Physics like an expert · Practice problems: motion and energy · ProblemA 2.0kg block slides at 3.0m/s on a frictionless floor and hits a 1.0kg block at rest, which is attached to a horizontal spring of stiffness 300N/m whose other end is fixed to a wall. The blocks stick together. Find their speed just after the collision, the spring’s greatest compression, and the period of the oscillation that follows.Speed after the collision 2 m/s; Greatest compression 0.2 m; Period 2*pi*sqrt(3/300) ≈ 0.628319 s
    1. (2.0)(3.0)=(2.0+1.0)v⇒v=2.0 m/s
    2. 12(3.0)(2.0)2=12(300)x2⇒x2=12300=0.04⇒x=0.20 m
    3. T=2πmk=2π3.0300=2π×0.10≈0.628 s
    4. The collision kept momentum but lost a third of the kinetic energy, 9 J down to 6 J. Using the 9 J would give a compression of 0.24 m, too much.
  321. 20. Physics like an expert · Practice problems: motion and energy · ProblemA space station orbits Earth in a circle 400km above the surface. Earth’s mass is 5.972×1024kg, its radius 6371km, and G=6.674×10−11N·m2/kg2. Find the station’s orbital speed, and its period in minutes.Speed 7672 m/s; Period 92.42 minutes
    1. The radius is r=6371+400=6771 km =6.771×106 m.
    2. v=GMr=(6.674×10−11)(5.972×1024)6.771×106≈7.67×103 m/s
    3. T=2πrv=2π(6.771×106)7672≈5545 s≈92.4 min
    4. About 16 orbits a day. Using 400 km as the radius, the usual slip, would give an impossible 31.6 km/s.
  322. 20. Physics like an expert · Practice problems: heat, fields and light · ProblemA kettle on 230V heats 1.2kg of water from 20 °C to 100 °C in 4.0 minutes. Assume all the electrical energy goes into the water, with c=4186J/(kg·K). Find the kettle’s power, the current it draws, and the resistance of its element.Power 1.2*4186*80/240 ≈ 1674.4 W; Current 7.28 A; Resistance 230^2/(1.2*4186*80/240) ≈ 31.5934 Ω
    1. Q=mcΔT=1.2×4186×80≈4.02×105 J,P=Qt=401856240≈1674 W
    2. I=PV=1674230≈7.28 A,R=VI=2307.28≈31.6 Ω
    3. In practice some heat warms the kettle and the air, so a real kettle that does this has a slightly more powerful element.
  323. 20. Physics like an expert · Practice problems: heat, fields and light · ProblemProtons start at rest, are accelerated through 20kV, then enter a uniform 0.50T magnetic field at right angles to it. Use e=1.602×10−19C and mp=1.673×10−27kg. Find their speed, and the radius of the circle they follow in centimetres.Speed sqrt(2*1.602e-19*2e4/1.673e-27) ≈ 1957100 m/s; Radius 4.088 cm
    1. 12mpv2=eΔV⇒v=2×1.602×10−19×2.0×1041.673×10−27≈1.96×106 m/s
    2. r=mpveB=(1.673×10−27)(1.957×106)(1.602×10−19)(0.50)≈0.0409 m=4.09 cm
    3. Under 1% of the speed of light, so Newton’s kinetic energy is fine (module 18). The magnetic field changes the direction, never the speed.
  324. 20. Physics like an expert · Practice problems: heat, fields and light · ProblemSunlight near Earth carries 1360W/m2. A solar sail of area 1000m2, facing the Sun, absorbs all the light that falls on it, and the craft’s total mass is 10kg. Use c=3.00×108m/s and ignore gravity. Find the force of the light in millinewtons, the craft’s acceleration, and the speed it gains in 30 days, starting from rest.Force 1360*1000/3e8*1000 ≈ 4.53333 mN; Acceleration 1360*1000/3e8/10 ≈ 0.000453333 m/s²; Speed after 30 days 1360*1000/3e8/10*30*86400 ≈ 1175.04 m/s
    1. The sail absorbs P=1360×1000=1.36×106 W. Momentum arrives at P/c each second, and that rate is the force:
    2. F=Pc=1.36×1063.00×108≈4.53×10−3 N=4.53 mN
    3. a=Fm=4.53×10−310≈4.53×10−4 m/s2,v=at=4.53×10−4×(30×86400)≈1.18×103 m/s
    4. A push about the weight of half a gram, yet after a month the craft is going over 1 km/s faster, with no fuel. A sail that reflects the light gets twice the push.