The 100 toughest problems in the subject.
Problems built to be harder than the paper, in the tradition of the classic hard-problem books — original statements and original numbers, not a reproduction of anyone's set. Each one names the bench where you can play with the situation before you commit to algebra, and each is solved more than one way, because the choice of route is the skill being trained.
Statements, hints and answers are free to read. The worked routes are part of the pass.
The chain that starts to slide
A uniform chain of mass M = 2.0 kg and length L = 1.0 m lies on a smooth horizontal table with a length a = 0.20 m hanging over the edge. It is released from rest. Take g = 10 m s⁻².
Find the speed of the chain at the instant the last link leaves the table.
Hint 1
The whole chain moves at one speed, but the force driving it grows as more of it hangs. Write that force when a length x hangs.
Hint 2
For the energy route you need the hanging part's centre of mass. It sits at depth x2 below the edge, not x.
Answer, without the method
v² = g(L² − a²)L = 9.6 m² s⁻², so v ≈ 3.1 m s⁻¹
3 routes to it
Energy route
Newton route (v dv/dx)
Bench route (numeric)
Afterthought. With a → 0 the chain would never start, yet the limit v → √(gL) = 3.16 m s⁻¹ is only 2 % above the answer. A 20 % overhang has already bought you almost all the speed there is.
The wedge that will not stay still
A block of mass m = 1.0 kg is released from rest at the top of a smooth wedge of mass M = 3.0 kg, face angle θ = 30° and height h = 0.45 m. The wedge is free to slide on a frictionless floor.
Find the wedge's speed when the block reaches the floor, and how far the wedge has moved by then.
Hint 1
Nothing horizontal acts on the block-plus-wedge pair, and the pair started at rest. That is one equation before you draw anything.
Hint 2
In the ground frame the block does not move along the incline — only its velocity relative to the wedge does. Write the block's velocity as the wedge's plus the relative one.
Answer, without the method
V = 0.72 m s⁻¹; the wedge slides 0.19 m backwards.
3 routes to it
Momentum + energy route
Centre-of-mass route (the displacement, with no dynamics at all)
Wedge-frame route
Afterthought. Let M → ∞ and u² → 2gh: the fixed-wedge answer. Every free-wedge formula should collapse to the textbook one in that limit — a five-second check worth doing on every line you write.
The sphere meets a step
A uniform solid sphere of radius R = 0.10 m rolls without slipping along a floor at v = 2.0 m s⁻¹ and meets a step. It pivots about the step's corner without slipping there and without bouncing.
Find the tallest step it can mount.
Hint 1
The impact is violent: kinetic energy is not conserved through it. Ask instead what the corner's impulse cannot change.
Hint 2
Once the sphere is pivoting, energy is conserved again — and the centre must be lifted by exactly h.
Answer, without the method
hmax = 7.0 cm — that is 0.70 R, and the sphere's speed drops by a quarter in the impact.
3 routes to it
Angular-momentum route
Where the energy went
Numeric route
Afterthought. A hoop (I = mR²) climbs a taller step than a solid sphere at the same speed, and a point mass none at all. The moment of inertia that slows a body down the ramp is the same one that carries it over the kerb.
The bead on a spinning rod
A smooth horizontal rod of length 0.60 m is spun about a vertical axis through one end at a constant ω = 4.0 rad s⁻¹. A bead of mass m = 50 g is threaded on it and starts at r₀ = 0.10 m, at rest relative to the rod.
Find the bead's speed relative to the rod as it leaves the far end, the time it takes to get there, and the force the rod exerts on it at that instant.
Hint 1
A smooth rod can push the bead sideways but never along its own length. Write the radial equation with that in mind.
Hint 2
ṙ dṙ = ω² r dr gets you the speed with no exponentials anywhere.
Answer, without the method
ṙ = ω√(r² − r₀²) = 2.37 m s⁻¹; t = ω⁻¹ cosh⁻¹(r/r₀) = 0.62 s; horizontal force 2mωṙ = 0.95 N (plus mg carried vertically).
3 routes to it
Rotating-frame route
Inertial-frame route (no pseudo-forces)
Bench route
Afterthought. Nothing pulls the bead outward. It leaves because the rod keeps turning away from underneath it — which is what every “centrifugal force is not real” sentence is trying to say and rarely does this cleanly.
Two balls, one drop
A ball of mass M = 0.20 kg is dropped from h = 1.00 m with a small ball of mass m = 0.05 kg resting immediately above it, the two almost touching. Every collision is elastic.
How high does the small ball rise?
Hint 1
It is two collisions in a row, not one event: the big ball with the floor, then the two balls with each other.
Hint 2
Do the second collision in the frame of the big ball, where the small one simply reverses.
Answer, without the method
It leaves at 2.2 v and reaches ≈ 4.8 m — nearly five times the drop.
3 routes to it
Sequential-collision route
Change-of-frame route
Numeric route
Afterthought. The same arithmetic, run backwards, is the gravity assist: a spacecraft bouncing off a planet's moving gravity well leaves with the planet's speed added twice.
A tunnel that misses the centre
A straight, frictionless tunnel is bored through a uniform Earth (R = 6400 km, surface g = 9.8 m s⁻²) along a chord whose closest approach to the centre is d = 3200 km. A capsule is released from rest at one end.
Find the time to reach the far end and the maximum speed — and say what happens to each if the chord is moved.
Hint 1
Inside a uniform sphere the field is not an inverse square — the shell outside your radius contributes nothing, so g(r) = g r/R.
Hint 2
Only the component along the tunnel drives the capsule. The wall quietly takes the rest.
Answer, without the method
t = π√(R/g) = 42.2 min, the same for every chord; vmax = √(g/R)·√(R² − d²) = 6.9 km s⁻¹, which does depend on d.
3 routes to it
SHM route
Energy route
Bench route
Afterthought. 42 minutes is also the period of a low Earth orbit, halved — the tunnel and the skimming satellite are the same oscillator seen along two different axes.
What the scale reads while a chain falls on it
A chain of mass M = 1.5 kg and length L = 1.2 m hangs vertically with its lower end just touching the pan of a scale. It is released.
Find the reading when a length x has landed, and at the instant the last link arrives. Take g = 10 m s⁻².
Hint 1
Two different things press on the pan: what is already lying there, and what is arriving.
Hint 2
The arriving links bring momentum that has to be destroyed. Rate of destruction of momentum is a force, and it is not a weight.
Answer, without the method
N = 3λgx with λ = ML; at x = L the scale reads 3Mg = 45 N — three times the chain's weight.
3 routes to it
Impulse route
Whole-chain momentum route
Bench route
Afterthought. Turn it around and the same factor appears in reverse: a chain being lifted off a table at constant speed needs three times its landed weight — one to hold it, two to accelerate the links that are still lying still.
How much does a spinning wire stretch?
A steel wire of mass m = 0.50 kg, length L = 1.00 m, cross-section A = 1.0 mm² and Young's modulus Y = 2.0 × 10¹¹ Pa is spun in a horizontal circle about one end at ω = 20 rad s⁻¹.
Find the tension as a function of distance from the axis, and the total extension.
Hint 1
Cut the wire at radius r and ask what force the outer piece needs to keep going round.
Hint 2
Each element stretches by T(r)dr/(AY). The total is an integral — not TmaxL/(AY).
Answer, without the method
T(r) = mω²(L² − r²)2L, so T(0) = 100 N; ΔL = mω²L²3AY = 0.33 mm.
3 routes to it
Tension route
Extension route
Bench route
Afterthought. Every rotating machine lives inside this formula: the blade tip of a turbine is held on by exactly the T(0) its own mass demands, which is why the limit on rotor speed is set by material strength, not by the motor.
The spring between unequal blocks
Block A of mass m = 1.0 kg carries an ideal spring of stiffness k = 200 N m⁻¹ on its front face and slides at v₀ = 6.0 m s⁻¹ along a frictionless floor towards a stationary block B of mass 2.0 kg. The spring is not attached to B.
Find the maximum compression of the spring, the velocity of each block long after they separate, and the fraction of the initial kinetic energy the spring holds at maximum compression.
Hint 1
Two different instants are being asked about, and each has its own condition. Maximum compression is not the moment of separation.
Hint 2
The spring is fully compressed exactly when the two blocks move at the same speed — if they did not, the gap between them would still be closing.
Answer, without the method
xmax = 0.35 m; afterwards vA = −2.0 m s⁻¹ and vB = +4.0 m s⁻¹; the spring holds 2/3 of the initial kinetic energy, and that fraction does not depend on k.
3 routes to it
Lab-frame route (two conditions)
Centre-of-mass route (why the fraction is fixed)
Numeric route
Afterthought. The heavier B is, the closer vA gets to −v₀ and the more of the energy the spring can hold: against an infinite wall the fraction reaches 1. Against an equal mass it is ½, and A stops dead.
The sphere that runs away beneath you
A small block of mass m is released from rest at the top of a smooth sphere of radius R = 0.50 m and mass M. The sphere rests on a frictionless floor and is free to slide. Let θ be measured from the upward vertical, relative to the sphere.
Find the equation satisfied by cos θ at the instant the block leaves the surface, solve it for M = m, and say whether the block leaves earlier or later than it would on a fixed sphere.
Hint 1
Nothing horizontal acts on the sphere-and-block system: gravity is vertical and so is the floor’s push. One component of momentum is therefore conserved from the first instant to the last.
Hint 2
At the moment the block leaves, the normal force is zero — so the sphere has no horizontal force on it either, and its frame is momentarily inertial. That single instant is the only one at which you may write the ordinary circular-motion condition in the sphere’s frame.
Answer, without the method
k c³ − 3c + 2 = 0 with c = cos θ and k = m/(M + m). For M = m, c = √3 − 1 = 0.732, so θ ≈ 42.9° — earlier than the fixed sphere’s 48.2°.
3 routes to it
Relative-velocity route
Where the non-inertial route goes wrong
Numeric route
Afterthought. Every one of these angles is smaller than the fixed-sphere 48.2°, and none is smaller than 39.6°. A sphere that can run away shortens the ride, but only by about nine degrees however light it is.
Which way does the spool roll?
A spool has outer radius R = 0.20 m and an inner axle of radius r = 0.08 m. A thread wound on the axle comes off its underside and is pulled gently at an angle θ above the horizontal. The spool rolls without slipping.
Find the angle at which the spool does not move at all, say which way it rolls on either side of it, and decide whether the answer depends on the pull, the mass, or the moment of inertia.
Hint 1
Take torques about the contact point rather than the centre. The friction force acts there, and it is the one force you do not know — so choosing that point deletes it from the equation.
Hint 2
The spool stays put when the line of action of the pull passes through the contact point. Draw the thread, extend the line, and see where it crosses the floor.
Answer, without the method
θc = arccos(r/R) = arccos(0.40) = 66.4°. Shallower than that and it rolls towards you; steeper and it rolls away. Independent of F, of the mass and of I.
2 routes to it
Torque about the contact point
The long way, about the centre
Afterthought. At exactly θc the thread’s line of action passes through the contact point, so the pull has no moment about it — and since the spool cannot slip, no motion is possible at all. Pull harder and it still will not budge.
Spin the vessel until the bottom shows
An open cylindrical vessel of radius R = 0.10 m and height h = 0.12 m holds liquid to a depth h₀ = 0.045 m. It is spun about its own vertical axis until the liquid turns with it as a rigid body. Take g = 10 m s⁻².
Find the shape of the free surface, the angular speed at which the bottom is just exposed at the centre, and the condition on the vessel’s height for that to happen without spilling.
Hint 1
In the rotating frame each element needs a net inward force m ω² r, and the only thing available to supply it is the pressure gradient. Write ∂p/∂r and ∂p/∂z and integrate.
Hint 2
Volume is conserved. The mean height of a paraboloid of revolution is the average of its centre and its rim — so the rim rises by exactly as much as the centre falls.
Answer, without the method
z = ω² r² / 2g, a paraboloid; the bottom is exposed at ω = (2/R)√(g h₀) = 13.4 rad s⁻¹; and it happens without spilling only if h ≥ 2h₀ — here 0.12 m ≥ 0.09 m, so it just works.
2 routes to it
Pressure-field route
Surface-element route (no calculus)
Afterthought. Halving the radius quadruples the required ω, because the drop at the centre goes as ω²R². A wide shallow dish exposes its base at a gentle spin; a narrow tube practically cannot be spun fast enough.
The bullet, the rod, and the impulse that vanishes
A uniform rod of mass M = 1.0 kg and length L = 1.0 m hangs at rest from a frictionless pivot at its upper end. A bullet of mass m = 10 g travelling horizontally at v = 300 m s⁻¹ strikes the rod a distance d = 0.60 m below the pivot and embeds itself.
Find the angular speed just after impact and the maximum angle of swing — then find the value of d for which the pivot delivers no horizontal impulse at all.
Hint 1
Linear momentum is not conserved here — the pivot is free to push. Angular momentum about the pivot is, because the pivot’s own force has no moment about itself.
Hint 2
Separate the impact from the swing. The impact is angular momentum, and energy is lost in it; the swing is energy, and angular momentum is not conserved once gravity has a moment. Using the wrong law in the wrong phase is the whole trap.
Answer, without the method
ω = 5.34 rad s⁻¹ and θmax ≈ 87°; the pivot feels nothing when d = 2L/3 = 0.67 m, the centre of percussion, whatever the bullet’s mass and speed.
3 routes to it
Angular-momentum route, then energy
Centre-of-percussion route
Numeric route
Afterthought. The pivot’s impulse and the swing angle peak at different d. The bat that stings least is not the bat that hits hardest — and both are different again from the point that maximises ω.
The crossing that drifts least
A river of width d = 200 m flows at v = 5.0 m s⁻¹. A boat can move at u = 3.0 m s⁻¹ relative to the water — slower than the current, so it cannot cross straight.
Find the heading that makes the downstream drift as small as possible, and find that smallest drift.
Hint 1
Because u < v the boat cannot land straight opposite, so "minimise the drift" is a genuine optimisation — not the usual "point upstream until the cross-current cancels", which has no solution here.
Hint 2
Write the drift as a function of the heading angle before differentiating anything. Time across and distance downstream both depend on it, and it is their ratio that matters.
Answer, without the method
Head upstream at sin θ = u/v = 0.60, i.e. 37° from straight across; the drift is then 267 m, and no heading does better.
2 routes to it
Calculus route
Vector-triangle route (no calculus)
Afterthought. A boat half as fast as the river drifts d√3 ≈ 1.7 d downstream at best. A boat matching the river drifts nothing, but takes forever — the crossing time goes to infinity exactly as the drift goes to zero.
The plank pulled out from under the block
A block of mass m = 1.0 kg rests on a plank of mass M = 4.0 kg, which rests on the floor. The coefficient of friction is μ₁ = 0.30 between block and plank, and μ₂ = 0.20 between plank and floor. A horizontal force F is applied to the plank alone. Take g = 10 m s⁻².
Find the least F that makes the plank slide out from under the block, and notice what the answer does not depend on.
Hint 1
Slipping between the two begins the moment the plank’s acceleration exceeds the greatest the block can be given by friction alone. Find that ceiling first.
Hint 2
Draw the plank’s free-body diagram carefully: it is rubbed on both faces, by the block above and the floor below, and the floor carries the whole weight of both.
Answer, without the method
Fmin = (M + m) g (μ₁ + μ₂) = 25 N — and remarkably it depends only on the total mass, not on how that mass is split between block and plank.
2 routes to it
Two free bodies, one threshold
Whole-system route
Afterthought. Because only the total mass matters, a heavy block on a light plank needs the same pull as a light block on a heavy one. What the split does change is how violently the plank leaves once you exceed 25 N.
How close do the two ships come?
Ship A steams due east at 20 km h⁻¹. At the same instant ship B, which is 10 km due east of A, steams due north at 15 km h⁻¹. Both hold their courses.
Find their least separation and when it occurs — without writing the separation as a function of time and differentiating it.
Hint 1
Sit on ship A. In its frame A never moves, and B travels in a straight line at the relative velocity — so the question becomes the distance from a point to a line.
Hint 2
The relative velocity is vB − vA, not the difference of the speeds. Draw it: it points north-west.
Answer, without the method
They close to 6.0 km, after 0.32 h (19 minutes).
2 routes to it
Relative-frame route
Component route, as a check
Afterthought. B is moving away from A’s longitude the whole time, yet the ships still close for nineteen minutes. Separation is not governed by either motion alone but by the component of the relative velocity along the line joining them.
What the dimensions alone can tell you
A liquid drop held together by surface tension, disturbed slightly, wobbles. The only quantities that can matter are its radius R, its density ρ and the surface tension σ — gravity is irrelevant for a small enough drop, and the liquid is taken as inviscid.
Find how the period of the wobble must depend on those three quantities, then evaluate it for a water drop of radius 2.0 mm (ρ = 1000 kg m⁻³, σ = 0.072 N m⁻¹).
Hint 1
Surface tension is a force per unit length, so its dimensions are M T⁻², not M L T⁻². Getting that wrong is the whole difficulty.
Hint 2
Write T = ρa Rb σc and match the powers of M, L and T separately. Three equations, three unknowns, no physics required beyond the dimensions.
Answer, without the method
T ∝ √(ρR³/σ), and for the water drop T ≈ 10.5 ms — about 95 wobbles a second.
2 routes to it
Dimensional route
What the scaling is worth
Afterthought. The same three quantities give a speed √(σ/ρR) and an energy σR². Once you have the dimensional skeleton of a problem, every other quantity in it is one line away.
The boat that never quite stops
A boat of mass m = 200 kg moving at v₀ = 4.0 m s⁻¹ cuts its engine. The water resists with a force proportional to the square of the speed, F = −b v² with b = 10 kg m⁻¹.
Find the distance in which the speed halves, and then decide whether the boat ever stops — in distance, and in time. Compare with the answer you would get if the drag were proportional to v instead.
Hint 1
You are asked about distance, so trade dv/dt for v dv/dx before integrating. That substitution turns the v² into a v and makes the equation trivial.
Hint 2
Doing it in time as well is worth the extra two lines: the two questions "does it stop?" and "does it stop in finite distance?" have different answers, and that is the point of the problem.
Answer, without the method
The speed halves in 13.9 m. With v² drag the boat travels an unbounded distance and never stops. With v drag it stops after a finite 80 m — and still takes infinite time.
3 routes to it
Distance route
Time route
Linear drag, for contrast
Afterthought. Quadratic drag dominates at speed and linear drag near rest, so a real boat follows the first curve then the second — and does, in the end, stop.
The monkey and the counterweight
A rope hangs over a frictionless, massless pulley. A monkey of mass m = 10 kg holds one end; a counterweight of exactly the same mass 10 kg hangs from the other. The system is at rest. The monkey now begins to climb, accelerating at 2.0 m s⁻² relative to the rope.
Find the acceleration of the monkey and of the counterweight, as seen from the ground. Then answer the question that makes people argue: can the monkey get away from the counterweight?
Hint 1
The rope is massless and the pulley frictionless, so the tension is the same on both sides. Write Newton’s second law for each mass with that one tension and compare the two equations before doing anything else.
Hint 2
The monkey’s 2.0 m s⁻² is relative to the rope, and the rope is moving. Ground acceleration = acceleration relative to the rope + acceleration of the rope, and the rope’s own end moves opposite to the counterweight.
Answer, without the method
Both rise at 1.0 m s⁻² — exactly half the monkey’s climbing rate, and identically to one another. The gap between them never changes, so no, the monkey can never get away, however it climbs.
2 routes to it
Two equations, one tension
Centre-of-mass route
Afterthought. Make the counterweight heavier by a gram and the symmetry breaks, but only just: the two accelerations then differ by a term proportional to the mass difference. The classic puzzle is a knife-edge, and the knife-edge is the answer.
Up the rough incline, and back down
A block is launched up an incline of θ = 30° with coefficient of friction μ = 0.20. It rises, stops, and slides back down to its starting point.
Find the ratio of the time going up to the time coming back down, and the ratio of the two speeds at the bottom. Explain the direction of the inequality without doing any arithmetic.
Hint 1
Friction reverses when the motion reverses. It is the one force in the problem whose sign is not fixed by the geometry, and the whole problem is bookkeeping on that sign.
Hint 2
The distance up and the distance down are equal, which lets you compare the two phases through their accelerations alone: t ∝ 1/√a for a fixed distance from or to rest.
Answer, without the method
tup/tdown = 0.70 and vreturn/vlaunch = 0.70 as well — the same number, and no coincidence.
2 routes to it
Two accelerations
Energy route, and why the two ratios agree
Afterthought. As μ climbs toward tan θ the return trip slows without limit and the ratio falls to zero; past it the block never returns at all. The formula and the physics run out at exactly the same value.
The banked curve has two speed limits
A curve of radius r = 50 m is banked at θ = 25°. The coefficient of friction between tyre and road is μ = 0.30. Take g = 10 m s⁻².
Find the slowest and the fastest speed at which a car can hold the curve, and say which way friction points in each case.
Hint 1
There is nothing special about the "design speed" — it is just where friction happens to be zero. Above it the car tends to slide out, below it to slide down, and friction reverses between the two.
Hint 2
Resolve along the horizontal and the vertical rather than along the slope. The acceleration is horizontal and has no vertical part, which makes those the axes that keep the algebra honest.
Answer, without the method
From 8.5 m s⁻¹ to 21.1 m s⁻¹ (31 to 76 km h⁻¹). At the lower limit friction acts up the bank, stopping the car sliding down; at the upper limit it acts down the bank, helping to hold the car in.
2 routes to it
Both limits at once
Where the limits break down
Afterthought. The safe band widens with μ and with the bank angle, but not symmetrically: raising μ from 0.30 to 0.60 moves the floor down by 8.5 m s⁻¹ and the ceiling up by only 6. Grip buys more protection against sliding in than against sliding out.
The satellite that speeds up when you slow it down
A satellite is in a circular orbit 320 km above the Earth’s surface (r = 6.70 × 10⁶ m, GM = 3.986 × 10¹⁴ m³ s⁻²). Thin atmospheric drag slowly removes energy from it, and the orbit stays very nearly circular as it decays.
Find the orbital speed, then find it again 10 km lower — and explain how a force that opposes the motion the whole time leaves the satellite moving faster.
Hint 1
Write the total energy of a circular orbit in terms of r alone, and notice the factor relating it to the kinetic energy. That factor is the whole answer.
Hint 2
Ask what drag actually controls. It removes total energy — it does not directly set the speed, and in a bound orbit those two are related the wrong way round from intuition.
Answer, without the method
v = 7713 m s⁻¹ at 320 km and 7719 m s⁻¹ ten kilometres lower: drag has taken energy out and the satellite has sped up by 6 m s⁻¹. Nothing is wrong — the potential energy fell by twice as much as the kinetic energy rose.
2 routes to it
Energy route
What drag is really doing
Afterthought. The same factor of two says a satellite in a higher orbit is slower, so overtaking in orbit is done by dropping down, going round faster, and climbing back. Direction and thrust are almost unrelated up there.
The ball that comes back
A solid ball of radius R = 5.0 cm is rolled along a floor with forward speed v₀ = 4.0 m s⁻¹ and heavy backspin of ω₀ = 300 rad s⁻¹. Friction eventually brings it to rolling without slipping.
Find its final velocity, and find the general condition on the backspin for the ball to come back to the thrower. Do it without knowing the coefficient of friction or how long the sliding lasts.
Hint 1
Friction acts at the contact point, so it has no moment about the line on the floor through that point. Something is conserved about that line for the whole sliding phase.
Hint 2
Take one consistent sign convention for "forward" and let backspin be negative angular momentum. Half the failures here are a sign, not a concept.
Answer, without the method
It returns at 1.4 m s⁻¹. In general it comes back whenever ω₀R > 2.5 v₀ — here 15 m s⁻¹ against 10.
2 routes to it
Angular momentum about the contact line
Force-and-torque route, the long way
Afterthought. At exactly ω₀R = 2.5 v₀ the ball rolls to a dead stop and stays there, having converted all of its momentum and all of its spin into heat. A snooker player calls that a stun-back shot and judges it by eye.
The loop that is only just completed
A block of mass m = 0.50 kg is held against a spring of stiffness k = 800 N m⁻¹, compressed by x, and released. It crosses a rough horizontal stretch of length d = 1.5 m (μ = 0.20) and then runs onto a smooth vertical circular loop of radius R = 0.40 m. Take g = 10 m s⁻².
Find the least compression that carries the block right round the loop — then, with the spring compressed to only 90 % of that, find where on the loop it leaves the track and how fast it is going there.
Hint 1
At the very top the track can only push downwards, towards the centre. The slowest safe passage is the one where gravity alone supplies the whole centripetal force.
Hint 2
"Leaves the track" means N = 0. Away from the top, only the radial component of the weight is available to bend the path — not the whole of it.
Answer, without the method
xmin = 12.7 cm. At 0.90 xmin it lets go where cos α = 0.588 — α ≈ 54° short of the top, at a height of 0.64 m — travelling 1.53 m s⁻¹.
3 routes to it
Energy route, and the 5/2 rule
Where it lets go
Numeric route
Afterthought. The loop costs 52 m g R, not 2 m g R. The extra half is the speed the block must still have at the top: half the loop's height again, bought in kinetic energy. Notice too that it can never leave the track on the lower half — below the horizontal the track is pushing it inward whatever its speed, so N = 0 is unreachable there.
The angle that costs the least
A crate of mass m = 20 kg is dragged d = 8.0 m across a floor at a steady speed by a rope held at an angle α above the horizontal. The coefficient of friction is μ = 0.50. Take g = 10 m s⁻².
Find the angle that needs the smallest pull, that pull, and the work it does — then decide whether the same angle also minimises the work.
Hint 1
The normal force is not mg. Write it with the rope's vertical component in it before you differentiate anything.
Hint 2
Anything of the form a cos α + b sin α is √(a² + b²) times a single cosine. That turns the minimisation into reading off an angle.
Answer, without the method
α = arctan μ = 26.6°, Fmin = μ m g√(1 + μ²) = 89.4 N against 100 N pulled flat, doing 640 J against 800 J. It does not minimise the work: the work falls all the way to α = 90°.
3 routes to it
Calculus route
Friction-cone route (no calculus at all)
The work route — where the question changes its answer
Afterthought. tan α = μ is the angle of repose over again — pull along the edge of the friction cone and none of your effort is wasted pressing the crate into the floor. It is also why a suitcase handle sits at about 25–30° and not horizontal, and why the saving is so modest: at μ = 0.5 the best angle buys 11 % of the force, and at small μ almost nothing.
The tank that takes twice as long as you think
A cylindrical tank of cross-section A = 0.20 m² holds water to a depth H = 1.25 m. A hole of area a = 4.0 cm² is opened in its flat bottom — small enough that the water surface descends slowly.
Find the time to empty, the time for the first half of the water to leave, and how badly wrong you would be to divide the volume by the initial flow rate.
Hint 1
Torricelli's √(2gh) is set by the depth at that instant, and the depth is falling. Write a volume balance before you reach for a number.
Hint 2
The quantity that runs down at a steady rate is not h.
Answer, without the method
t = Aa √( 2Hg ) = 250 s. The first half is gone in 73 s, the second half takes 177 s. Volume ÷ initial flow gives 125 s — exactly half, always.
3 routes to it
Separate the variables
What actually falls linearly
Dimensions, then the bench
Afterthought. The same integral is why a clepsydra — a water clock — is never a cylinder. To make the level fall at a constant rate you need a vessel whose cross-section grows as √h, that is, a bowl whose depth goes as the fourth power of the radius. The ancients found that shape by trial; it is one line of this differential equation.
What the scale reads while it hangs there
A beaker of water stands on a scale; beaker and water together weigh 18.0 N. A steel sphere of mass 0.60 kg and volume 2.0 × 10⁻⁴ m³ hangs from a spring balance, fully submerged and touching nothing. Water's density is 1000 kg m⁻³ and g = 10 m s⁻².
Give both readings. Then the string is cut: give the scale reading just after the cut, while the sphere is descending at a steady speed, and once it has settled on the bottom.
Hint 1
Buoyancy is a force the water exerts on the sphere. Newton's third law then has something to say about a force on the water.
Hint 2
Take the beaker, the water and the sphere as ONE body, and buoyancy never has to be written down at all.
Answer, without the method
Spring balance 4.0 N, scale 20.0 N. After the cut: 20.0 N while it accelerates, 24.0 N once it descends steadily, and 24.0 N at rest on the bottom.
3 routes to it
Force by force
One-body route — all four answers in a line
Check it against the limits
Afterthought. The scale reads the total weight minus whatever the string is carrying — no buoyancy, no depth, no shape anywhere in it. And note the timing: the moment the sphere reaches steady speed the scale has already arrived at its final reading, though the sphere may still be ten centimetres from the bottom.
Two wires, one load, and a bar that must stay level
A rigid bar of negligible mass, L = 1.20 m long, hangs from two vertical wires at its ends, each of unstretched length 1.00 m. Left: steel, A₁ = 1.0 mm², Y₁ = 2.0 × 10¹¹ Pa. Right: copper, A₂ = 2.0 mm², Y₂ = 1.2 × 10¹¹ Pa. A load of W = 300 N hangs from the bar a distance x from the steel wire.
Find x for the bar to hang horizontal, with the tensions and the stretch. Then find the x that makes the two stresses equal, and how far from level the bar sits in that case.
Hint 1
"Horizontal" is a statement about the two extensions. It is not a statement about the two tensions, and it is not a statement about the two stresses.
Hint 2
A wire of area A, length ℓ and modulus Y is a spring of stiffness AY/ℓ. Two springs holding a rigid bar have a centre, and it is not the middle.
Answer, without the method
Horizontal at x = 611 L = 65.5 cm, with T₁ = 136 N, T₂ = 164 N and both wires 0.68 mm longer. Equal stress needs x = 80 cm, and the bar then tilts 0.33 mm across its length — 2.8 × 10⁻⁴ rad.
3 routes to it
Statics and compatibility
Centre of stiffness
Equal stress — and why the bar tilts
Afterthought. Engineers meet this as load sharing: put a stiff member beside a compliant one and the stiff one takes almost everything, whatever their strengths suggest. Strength lives in the stress, geometry lives in the strain, and only the strain knows whether the bar is level.
Weighing a pair of stars
Two stars of masses m₁ = 2.0 × 10³⁰ kg and m₂ = 6.0 × 10³⁰ kg move in circular orbits about their common centre of mass, a constant d = 4.0 × 10¹¹ m apart. Take G = 6.67 × 10⁻¹¹ SI.
Find the period and each star's speed, and the energy needed to pull the pair apart completely. Then say how wrong you would be to treat the heavy star as nailed down.
Hint 1
Neither star sits at the centre of the circle it describes. Find the point that does not move, and measure both radii from there.
Hint 2
Both stars come round together — one ω for the pair. That single fact collapses a two-body problem into a one-body one.
Answer, without the method
T = 2π √(d³ / G(m₁ + m₂)) = 6.9 × 10⁷ s = 2.2 years; v₁ = 27.4 km s⁻¹, v₂ = 9.1 km s⁻¹; unbinding costs 1.0 × 10³⁹ J. Pretending the heavy star is fixed gives 2.5 years — 15 % too long.
3 routes to it
One star at a time
Reduced mass, then Kepler
The energy, by the virial shortcut
Afterthought. ω² = G(m₁ + m₂)/d³ contains only the SUM of the masses, so a binary's period can never tell you how that mass is split. For the split you need the two speeds, whose ratio is the mass ratio upside down. That pair of measurements — one period, two Doppler curves — is how very nearly every stellar mass known has been weighed.
The raindrop that grows as it falls
A spherical drop falls from rest through still mist, sweeping up and keeping every droplet in its path, so it stays spherical and grows. It starts from a negligible size, and there is no air resistance. The mist holds 1.0 g of water in every cubic metre; the drop's own density is 1000 kg m⁻³. Take g = 10 m s⁻².
Show that the drop falls with a constant acceleration and find it — then find how long it takes to reach a radius of 1 mm, and how fast it is moving by then.
Hint 1
The mist it swallows is at rest, so the drop must spend momentum bringing it up to speed. The law to write is d(mv)/dt = mg, not m dv/dt = mg.
Hint 2
Starting from nothing, the problem contains no length and no time at all. Guess v ∝ t and let the equation fix the constant.
Answer, without the method
a = g7 = 1.43 m s⁻², whatever the mist's density or the drop's. The radius grows as t²; 1 mm takes ≈ 75 s, by which time the drop is doing 107 m s⁻¹ and has fallen 4 km.
3 routes to it
The equation, and a power law
Dimensions and self-similarity, before any solving
Why it is not simply a = g
Afterthought. Seven is a strange number to fall out of a cloud, and it is worth knowing where it lives: six parts of the weight accelerating swallowed mist, one part accelerating the drop. The model is also decisively wrong about the real sky — at 107 m s⁻¹ drag would be overwhelming, and actual raindrops arrive at about 8 m s⁻¹. What survives is the method: whenever mass is joining or leaving, d(mv)/dt is the law and ma is the trap.
How small must a pendulum's swing be to keep time to a second a day?
A seconds pendulum is only isochronous in the small-angle approximation. Its true period exceeds the ideal one by a fraction that depends on the amplitude θ₀.
Find the largest amplitude a clock may swing at if it is to keep time to 1 second a day. Then find its error at a 5° swing, and explain how a real pendulum clock manages to keep far better time than that number suggests.
Hint 1
The restoring torque is proportional to sin θ, not θ. Since sin θ = θ − θ³/6, the leading correction to the period must be second order in the amplitude.
Hint 2
One second in a day is a fractional error of 1/86 400 = 1.16 × 10⁻⁵. Work backwards from that to θ₀ and be prepared for a surprisingly small angle.
Answer, without the method
The amplitude must stay under 0.78° — a swing of just 14 mm of arc on a one-metre pendulum. At 5° the clock loses 41 s a day; at 30° it loses 25 minutes. Real clocks escape the problem not by swinging tiny arcs but by swinging a constant arc, typically 1.5°, whose fixed 3.7 s/day error is then trimmed away by shortening the rod. What matters to a clock is not the amplitude error but its variation.
3 routes to it
Where θ₀²/16 comes from
Why the error had to be quadratic, without the elliptic integral
How horology actually solves it
Afterthought. The habit is to ask not "how large is my error?" but "how much does my error vary?" — a large but stable systematic can be calibrated out, while a small drifting one cannot. That distinction runs through all of measurement: the offset of a thermocouple, the zero of a balance, the dark current of a detector, the fixed latency of a network. It is also why the eventual replacement for the pendulum was not a better pendulum but a quartz crystal, whose amplitude dependence is 10⁻¹⁰ instead of 10⁻⁵ — you win by changing the oscillator, not by regulating it harder.
The spring's own mass counts, and exactly a third of it counts
A 500 g block hangs from a spring of stiffness 20 N m⁻¹. The spring is not weightless: it is 60 g of steel.
Find the period ignoring the spring's mass, then find the correction and show where the factor of one third comes from. Then predict the period if you cut the spring in half and use one piece — and check the whole scheme against the exact answer.
Hint 1
The spring's far end is fixed and its near end moves with the block. Assume the speed of each element is proportional to its distance from the fixed end, and add up its kinetic energy.
Hint 2
For the cut spring, ask what each half of a stretched spring does. Two identical springs in series share the load but not the extension — or is it the other way around?
Answer, without the method
Naively 0.9935 s. Including the spring, the effective mass is m + ms/3 = 520 g and the period is 1.0132 s — 2 % longer, which as a clock would be half an hour a day. Cutting the spring in half doubles its stiffness to 40 N m⁻¹ and halves its mass, giving 0.7095 s — shorter than T/√2 = 0.7025 s would suggest, because only some of the spring's mass went away. The exact treatment gives 1.0137 s, so the one-third rule is good to 0.05 %.
3 routes to it
Add up the spring's kinetic energy
Why cutting a spring stiffens it
The exact answer: it is a wave problem
Afterthought. The general move — replace a distributed system by a lumped one with an effective parameter obtained from an energy integral — is one of the most reusable tricks in physics, and the coefficient is always some ∫(shape)²: a third for a uniform spring, 33/140 of a beam's mass for a cantilever, 2πρa³/3 of added fluid mass for a sphere accelerating through a liquid. It works because energies are quadratic, and it fails in a predictable way: it always gets the fundamental nearly right and knows nothing whatever about the higher modes.
Forty-two minutes to the other side of the world, whichever way you dig
Bore a straight, evacuated, frictionless tunnel through a planet of uniform density — Earth, taking R = 6370 km and g = 9.81 m s⁻² at the surface — and drop a car into it.
Find how long it takes to reach the far end through the centre, and its top speed. Then find how long a tunnel that misses the centre takes. Finally, say what the real Earth does to your answer.
Hint 1
Inside a uniform sphere, only the mass within radius r pulls on you, and that mass is proportional to r³. Work out how g depends on r before writing any equation of motion.
Hint 2
For the chord, resolve the (radial) gravitational force along the tunnel. The distance from the tunnel's midpoint is the only coordinate that appears.
Answer, without the method
42.2 minutes, one way, with a top speed of 7.9 km s⁻¹ — which is exactly the speed of a satellite skimming the surface. A chord tunnel that misses the centre takes the same 42.2 minutes, whatever chord you choose: London to Sydney and London to Paris are timetabled alike. The real Earth, whose density rises steeply inwards, does it in about 38 minutes.
3 routes to it
Gravity inside a uniform sphere is a spring
It is an orbit, seen edge on
The real planet, and the engineering that kills it
Afterthought. The lasting content is the equivalence of three things that look unrelated — a mass on a spring, a body oscillating inside a uniform sphere, and the projection of a circular orbit — because all three obey the same linear equation. Recognising that a system is linear buys you isochronism, superposition, and the whole machinery of normal modes at once; and the moment the force law departs from linearity, as it does in the real Earth, every one of those gifts is withdrawn together. That is why physics spends so much of its effort finding the regime in which a problem is linear, and is so wary of leaving it.
The pulse that starts from nothing and accelerates at exactly g/2
A uniform rope 12.0 m long and 4.00 kg in mass hangs freely from a ceiling. Someone flicks the bottom end sideways, sending a transverse pulse up it. Take g = 9.8 m s⁻².
Find how long the pulse takes to reach the ceiling. The wave speed at the free end is zero — say how the pulse gets going at all, and what its acceleration is. Then find the transit time if a 4.00 kg mass is hung from the bottom.
Hint 1
The tension at a height y above the free end is the weight of rope below it. Get v(y) first, and notice what the linear mass density does.
Hint 2
You have v as a function of position. Use a = v dv/dy rather than integrating for the time — the result is worth seeing.
Answer, without the method
2.21 s up, and 2.21 s back. It gets going because although the speed at the very end is zero, the pulse's acceleration there is not: it climbs with a constant acceleration of exactly g/2, all the way, whatever the rope. Hanging an equal 4.00 kg mass on the bottom raises the tension everywhere and cuts the climb to 0.92 s.
3 routes to it
Integrate dy/v
a = g/2, and why it must be a constant
Hang a mass on it, and the general case
Afterthought. The technical trick worth keeping is a = v dv/dx — converting a known speed-versus-position into an acceleration without ever finding the trajectory. It turned an integral into a one-line constant-acceleration problem here, and it is the standard route through any problem where you know the speed field rather than the history: rockets losing mass, charges in a static potential, fluid flowing through a varying pipe. The wider lesson is that a wave's speed is a property of the medium at each point, so a non-uniform medium refracts, stretches and steepens the wave — which is most of what oceanography, seismology and atmospheric acoustics consist of.
The organ goes flat in winter, and it is not the metal
An open organ pipe is tuned to A = 440.0 Hz in a church at 20 °C. In January the church is at 5 °C. Take the speed of sound as 331 m s⁻¹ at 0 °C, and the pipe metal's expansion coefficient as 2.5 × 10⁻⁵ K⁻¹.
Find the pitch it sounds in January, in hertz and in cents. Decide whether the contraction of the metal matters. Then explain how a resonance-tube measurement of the speed of sound avoids ever having to know the pipe's end correction.
Hint 1
The pipe's length is fixed, so the pitch follows the speed of sound. And the speed of sound in a gas depends on temperature — through the absolute temperature, in kelvin, under a square root.
Hint 2
Before comparing two effects, express both as fractional changes. One of them is a fraction of √T and the other is αΔT.
Answer, without the method
It sounds 428.6 Hz — 45 cents flat, nearly half a semitone, which is grossly out of tune. The metal's contraction pushes the pitch up by only 0.65 cents, seventy times smaller, and is negligible. The end correction cancels if you measure the difference between two successive resonance lengths, since consecutive resonances are exactly λ/2 apart however the open end behaves.
3 routes to it
Pitch follows the speed of sound
The competing effect, and why it loses by a factor of seventy
Measuring v without knowing where the pipe ends
Afterthought. The reusable habit is to reduce every candidate effect to a fractional change before comparing, because that is the only way competing mechanisms can be weighed against each other. Here it separated a 2.6 % effect from a 0.04 % one, and dismissed the plausible-sounding one on sight. The other transferable idea is the differential measurement: a nuisance parameter that appears identically in two readings disappears from their difference, which is why spectroscopists measure line spacings rather than positions, why a Vernier works, and why the two-slit fringe spacing is a better measurement than any single fringe's position.
Moving the source is not the same as moving the listener
A siren sounds 500.0 Hz in still air where sound travels at 340 m s⁻¹. In one experiment the siren approaches a stationary listener at 34 m s⁻¹; in the other the listener approaches a stationary siren at the same 34 m s⁻¹. The relative speed is identical.
Find the frequency heard in each case and account for the difference. Find the speed at which the two answers differ by 1 %. Then find the shift a 24.1 GHz radar speed gun sees from a car at 30 m s⁻¹, and say why light has no such asymmetry.
Hint 1
Do the two cases by different bookkeeping. In one the wavelength in the air is altered; in the other the wavelength is untouched and the rate of meeting crests changes.
Hint 2
For the radar, remember the wave makes two journeys. The car receives a shifted frequency and then re-radiates it as a moving source.
Answer, without the method
Moving source: 555.6 Hz. Moving listener: 550.0 Hz. They differ because the air is a preferred frame — the two experiments are not the same experiment viewed differently. Their ratio is exactly 1/(1 − β²), so the difference is β² and reaches 1 % at β = 0.1, i.e. 34 m s⁻¹. The radar gun sees a double shift, 2vf/c = 4.8 kHz. For light there is no asymmetry at all: only the relative velocity can appear, and that is the content of the relativistic Doppler formula.
3 routes to it
Count crests in the frame of the air
The exact ratio, and how large the asymmetry can get
The double shift a speed gun uses, and the case with no medium
Afterthought. The habit worth taking is to ask which frames a problem's answer is allowed to mention. A wave in a medium may refer to the medium's frame, so source motion and observer motion are physically distinct and the formula must be asymmetric; a wave with no medium may refer only to relative velocity, so the formula is forced to be symmetric before any derivation begins. That kind of reasoning — deciding what an answer can depend on, and then checking that it depends on nothing else — catches errors faster than checking algebra, and it is the same discipline that makes dimensional analysis so productive.
Two speakers at sixty decibels do not make a hundred and twenty
Two identical loudspeakers each produce 60.0 dB at your seat when run alone.
Find the level with both running, in three cases: fed the same signal and equidistant from you; fed independent noise; and fed the same signal from a point where the path difference is half a wavelength. Then say what ten speakers give, and why a real hall never behaves like the first case.
Hint 1
Decibels are logarithmic, so levels never add. Convert to intensity, add the right thing, and convert back.
Hint 2
When the two signals are the same, add amplitudes; when they are unrelated, add intensities. Ask what the cross-term 2√(I₁I₂)cos φ averages to in each case.
Answer, without the method
Same signal, in phase: 66 dB (amplitudes add, so intensity quadruples). Independent signals: 63 dB (intensities add). Same signal, antiphase: silence. Never 120 dB — that would be 10⁶ times the power. Ten speakers give 80 dB coherent and 70 dB incoherent. A real hall gives you something near the incoherent answer, because music is broadband: every frequency has its null in a different place, so instead of silence you get a comb.
3 routes to it
The decibel algebra, done once
Where the +6 and the +3 come from: one cross-term
Why a hall gives you neither answer, and what you can hear
Afterthought. The general point is that whether you add amplitudes or intensities is a question about coherence, not about waves, and the answer differs by the factor N. It decides why a laser beam of N atoms is N² times brighter in its own direction than N independent atoms, why radio telescope arrays gain resolution by combining phases rather than powers, and why the sum of many independent noise sources grows only as √N — which is the same statistics that makes a random walk grow as √N. The moment you can say whether the phases are locked, you know which power of N to write.
Where you pluck decides which harmonics are allowed to exist
A stretched string of length L is pulled aside at a single point a distance a from one end and released. The initial shape is a triangle with its apex at a, and the resulting sound is the sum of the string's harmonics with amplitudes An ∝ sin(nπa/L)/n².
Find which harmonics are absent, and give the physical reason without doing the Fourier integral. Compare the tone of a pluck at the middle with one at a fifth of the way along. Then say why a piano's hammers strike at about a seventh of the string, and what touching the midpoint of a plucked string does.
Hint 1
Ask what a force applied exactly at a node of some mode can do to that mode. Nothing else is needed for the first part.
Hint 2
For the timbre comparison, just evaluate sin(nπa/L)/n² for n = 1…5 at a/L = 1/2 and at a/L = 1/5, and convert the ratios to decibels.
Answer, without the method
Harmonic n is absent whenever the pluck point is a node of that harmonic — n a multiple of L/a. Plucked at the middle, every even harmonic vanishes and the tone is nearly pure (the 3rd is 19 dB down); plucked at a fifth, only the 5th and its multiples vanish and the 2nd is a mere 8 dB down, giving a far brighter tone. Piano hammers strike near 1/7 because the 7th harmonic is badly out of tune with the tempered scale, and striking at its node never excites it. Touching the midpoint kills every harmonic without a node there — all the odd ones — leaving the octave: the flageolet.
3 routes to it
A force at a node excites nothing
Two plucks, two spectra
What instrument makers do with it
Afterthought. The transferable principle is that you excite only what you overlap with: the amplitude of any mode is the projection of your driving pattern onto that mode's shape, so a drive applied at a node gives nothing, and a drive shaped like a mode gives that mode alone. It is the same statement as selection rules in spectroscopy, as the orthogonality that makes normal-mode analysis work, and as the reason a tuning fork struck on its stem sounds different from one struck on a tine. Whenever you want a particular mode — or want to be rid of one — the question is never how hard to push, but where.
Two ways to let a gas out, and only one of them pays
An insulated cylinder holds 2.00 mol of argon at 300 K in 10.0 L. It is allowed to reach 30.0 L in two different ways. (a) A tap is opened to an evacuated 20.0 L chamber and the gas rushes in. (b) A piston is drawn out so slowly that the gas is in equilibrium the whole way.
Find the final temperature, the work delivered and the entropy change in each case — and then say, with a number you can defend, how much work the sudden route threw away.
Hint 1
In one of the two cases, ask what the gas is pushing against. Nothing is receding, so nothing is being pushed.
Hint 2
Entropy is a state function, so you may compute it along a path you did not take. Pick the reversible path that shares both end states with the sudden one.
Answer, without the method
(a) T stays at exactly 300 K, W = 0, ΔS = +18.3 J K⁻¹. (b) T falls to 144 K, W = +3.89 kJ, ΔS = 0. The sudden route destroyed T₀ΔS = 5.5 kJ of work — precisely what a reversible isothermal expansion between the same two end states would have delivered.
3 routes to it
The first law, applied twice
Entropy: the path you did not take
Why the molecules do not slow down
Afterthought. The habit worth keeping is to stop treating adiabatic and isentropic as the same word. Q = 0 only kills the ∫dQ/T term; irreversibility generates entropy from inside. Free expansion is the cleanest counterexample in the subject, and it is also the standard test of whether a gas is ideal — do it with a real gas and the temperature does shift slightly, because U depends a little on volume through the attractive tail. Joule looked for that shift in 1845 and could not resolve it; the effect is a fraction of a kelvin, and finding it was the beginning of taking intermolecular forces seriously.
The gas that cools while you heat it
One mole of a diatomic ideal gas is expanded along the path pV1.20 = constant, and its temperature is observed to fall from 400 K to 300 K.
Is heat entering the gas or leaving it? Find the work, the heat, and the molar heat capacity for this process — then find every exponent n for which a gas absorbs heat while getting colder.
Hint 1
Work out the work first: for pVn = constant it collapses to nR(T₁ − T₂)/(n − 1), with no integral left in sight.
Hint 2
Compare that work with the change in internal energy. If the gas does more work than it loses from U, the difference had to come in from somewhere.
Answer, without the method
Heat enters: Q = +2.08 kJ, while the gas does W = +4.16 kJ and its internal energy drops by 2.08 kJ. The molar heat capacity is C = −2.5R = −20.8 J mol⁻¹ K⁻¹ — negative. That happens for every polytropic exponent between 1 and γ, that is 1 < n < 1.4 for a diatomic gas.
3 routes to it
First law, with the polytropic work done once and for all
Build C(n) once, then read the sign off it
The picture: a path that leans between the two you know
Afterthought. The lasting lesson is that heat capacity is not a property of a substance at all — it is a property of a process. Cp and CV are simply the two members of this family that laboratories find convenient, and a table listing "the" specific heat of a gas is quietly naming one of them. Negative heat capacity is also not a curiosity confined to gas cylinders: a self-gravitating cloud of gas has one, which is why a star contracts, heats and radiates faster the more energy it loses, and why gravitating systems refuse to come to equilibrium with a heat bath.
The efficiency you get when you want power, not perfection
A steam plant draws heat from a boiler at 800 K and rejects it to a river at 300 K. Carnot promises 62.5 %; the plant delivers about 40 %. Suppose the only loss is that heat takes time to cross a temperature difference: the working fluid runs between Th′ and Tc′, absorbing heat at a rate proportional to (800 − Th′) and rejecting it at a rate proportional to (Tc′ − 300), and between those two temperatures the engine is perfectly reversible.
Find the efficiency at which this engine delivers the most power, and the two temperatures its fluid actually runs at. Then say what happened to the conductances.
Hint 1
A reversible engine between the fluid's own two temperatures has efficiency 1 − Tc′/Th′. Write the two transfer rates and impose that reversibility as a constraint linking Th′ and Tc′.
Hint 2
You are maximising power, not efficiency — so maximise the difference of the two heat rates, and expect the answer to be a geometric mean rather than an arithmetic one.
Answer, without the method
Maximum power comes at η* = 1 − √(Tc/Th) = 38.8 %, with the fluid running between 645 K and 395 K — it never sees either reservoir's temperature. The conductances cancel out of the efficiency entirely; they set how much power you get, not at what efficiency you get it. Carnot's 62.5 % is real, and it is available only at zero output.
3 routes to it
Maximise the power directly
Where the 500 K got spent
Tabulate it and see how flat the peak is
Afterthought. That 1 − √(Tc/Th) is worth carrying around, because it predicts real plants far better than Carnot does: 300 K/800 K gives 39 % against a measured ~40 %, and geothermal and ocean-thermal plants fall on it too. The reason it works is that it is a reversible engine embedded in an irreversible surrounding — all the entropy is generated in the two heat exchangers, which is where it really is generated. The deeper habit: whenever a limit is achieved only at zero rate, ask for the limit at finite rate, because that is the one engineering has to live with.
How the Earth lost its hydrogen and kept its nitrogen
At the top of the atmosphere, around 500 km up, the gas is at about 1000 K and the escape speed is 11.2 km s⁻¹. Molecules there rarely collide again on the way out, so any molecule moving upward faster than escape speed is gone.
Estimate the fraction of H₂ molecules and the fraction of N₂ molecules that are above escape speed, and account for the ratio. The two gases' typical speeds differ by only √14 ≈ 3.7.
Hint 1
Compare the escape speed with the most probable speed √(2kT/m), and call the ratio x. Everything after that is a function of x.
Hint 2
The Maxwell distribution's tail falls as e−x². You do not need the prefactor to answer the question — work out x² for each gas first and look at the two numbers.
Answer, without the method
For H₂ about 1.3 × 10⁻⁶ — one molecule in 800 000. For N₂ about 3 × 10⁻⁹¹. A speed ratio of 3.7 becomes a probability ratio of about 10⁸⁴, because the speed enters an exponent: mvesc²/2kT is 15 for hydrogen and 211 for nitrogen. The hydrogen leaks away in geological no-time; the nitrogen would need 10⁷⁰ times the age of the universe.
3 routes to it
The exponent is the whole answer
The Maxwell tail, with its prefactor
How fast is that, in metres per second?
Afterthought. Whenever a quantity sits in an exponent, precision in the prefactor is wasted effort and precision in the exponent is everything — a 10 % error in the temperature here moves the answer by a factor of four, while a factor-of-two error in the prefactor moves it by two. The same structure runs through reaction rates, thermionic and field emission, nucleation, and quantum tunnelling: the physics is in the argument of the exponential, and the rest is decoration. It is also why the argument works in reverse — measure how much deuterium is left relative to hydrogen and you can read off how much water a planet has lost.
How good must the vacuum in a vacuum flask be?
A vacuum flask has a 5.0 mm gap between its two silvered walls, at about 300 K. Air molecules have a collision cross-section of roughly 4.3 × 10⁻¹⁹ m² (a diameter of about 0.37 nm).
Find the pressure at which the mean free path first exceeds the gap. Say how the heat leak through the residual gas depends on pressure above and below that pressure — and then decide whether reaching it is good enough, given that the walls are silvered.
Hint 1
Mean free path is λ = kT/(√2 σp). Set it equal to the gap and solve for p — then, out of curiosity, work out how many molecules are still in a cubic metre.
Hint 2
Kinetic theory gives the conductivity as ⅓ n v̄ λ c. Substitute λ ∝ 1/n and watch what happens to n. What caps λ once it exceeds the gap?
Answer, without the method
The crossover is at p ≈ 1.4 Pa — about 10⁻⁵ atm, and still 3 × 10²⁰ molecules per cubic metre. Above it, conduction is almost independent of pressure; below it, conduction falls in proportion to p. And 1.4 Pa is not good enough: the residual gas would still carry roughly 100 W m⁻², against about 17 W m⁻² for radiation between two surfaces of emissivity 0.05. The flask must be pumped a further decade or two, until the gas leak drops below the radiation leak.
3 routes to it
The crossover pressure, and the crowd that is still there
Why pumping does nothing at first, and everything later
Is it good enough? Weigh gas against radiation
Afterthought. The general move is to identify the length that competes with your apparatus. A dimensionless ratio of mean free path to system size — the Knudsen number — decides whether a gas behaves as a fluid at all: above about 1 there is no viscosity worth the name, no conduction law, no pressure gradient driving flow, only molecules flying between walls. Aerogel and vacuum panels exploit the same trick at atmospheric pressure by making the pores smaller than the mean free path, and a re-entering spacecraft passes through the same crossover in the opposite direction, which is why its drag stops obeying the usual formulae above 100 km.
What it costs to mix hot water with cold
You have 1.00 kg of water at 80 °C and 1.00 kg at 20 °C (c = 4186 J kg⁻¹ K⁻¹). Tipped into one bucket they settle at 50 °C, and no energy has been lost — the first law is perfectly content.
Find how much entropy that mixing generated, and how much work you could have extracted instead by running an engine between the two buckets until they reached a common temperature. What is that common temperature, and is the work worth having?
Hint 1
Nothing is lost in the mixing, so the first law will not detect the damage. Compute the entropy of each body separately: mc ln(Tf/Ti), and add.
Hint 2
For the engine, the condition is that the total entropy does not change. Write that condition first; the final temperature drops out of it before energy conservation is used at all.
Answer, without the method
Mixing generates +36.3 J K⁻¹ of entropy and yields nothing. A reversible engine between them delivers 11.7 kJ of work and leaves both buckets at √(T₁T₂) = 321.6 K = 48.4 °C — cooler than 50 °C, because the missing energy left as work. It is worth having: 11.7 kJ would lift those two kilograms of water 600 m. But it is only 9 % of the 126 kJ of heat that flowed.
3 routes to it
Count the entropy the mixing made
Run an engine between the buckets instead
How much of the heat was worth anything?
Afterthought. The geometric mean is not a coincidence of water. Whenever two identical bodies with temperature-independent heat capacity are brought to equilibrium, mixing gives the arithmetic mean and a reversible engine gives the geometric one, and the difference is the work — so Tarith − Tgeom is a thermometer for irreversibility. The habit to build is to run both calculations on any process that merely lets a difference collapse: an entropy count says how bad it was, and a reversible version of the same process says how much you gave up in joules, which is the number anyone will pay attention to.
The ring that must be heated to fit, and the stress if it cannot shrink
A steel ring of inner diameter 39.95 mm, 4.0 mm thick and 10 mm wide, is to be shrunk onto a shaft of diameter 40.00 mm. Both start at 20 °C. For steel, α = 1.2 × 10⁻⁵ K⁻¹ and E = 200 GPa; it yields at about 300 MPa.
Find the temperature the ring must reach to slip on. Then find the hoop stress once it has cooled onto the shaft, and say whether the ring survives. Finally: what axial force would it take to push it off again, with μ = 0.15?
Hint 1
The gap you must close is 0.05 mm on a 39.95 mm diameter. Ask whether a hole in a heated ring gets bigger or smaller before you write anything down.
Hint 2
Once it is on and cooled, the ring wants to be 0.05 mm smaller in circumference than it is. That is a strain — and strain times E is a stress, with no length left in the expression.
Answer, without the method
Heat the ring to 124 °C — a 104 K rise. Cooled onto the shaft it carries a hoop stress of 250 MPa, uncomfortably close to yield, and independent of the ring's diameter. The grip is a contact pressure of 50 MPa, a normal force of 63 kN, so it takes about 9.4 kN — a tonne — to push it off, or 190 N m of torque to twist it.
3 routes to it
Close the gap, then refuse to let it close
Does a hole expand? Two ways to be sure
How hard does it actually grip?
Afterthought. Two generalisations are worth taking away. First, σ = EαΔT is dimension-free and startlingly large for metals: 2.4 MPa per kelvin for steel, so a fully restrained rail or pipe reaches yield with a 120 K swing — which is what expansion loops, bellows and the gaps in bridge decks exist to prevent. Second, whenever a problem mentions a hole, a cavity or a gap, expansion treats it as material: the standard error in this corner of the subject is not arithmetic but a picture of matter pushing inwards, and the cure is to think of heating as multiplying a map by a scale factor.
The planet is 33 degrees warmer than it has any right to be
Sunlight arrives at 1361 W m⁻² and the Earth reflects 30 % of it straight back. Treat the planet as a sphere in radiative equilibrium, then add a single atmospheric layer that is transparent to sunlight and completely opaque to infrared.
Find the bare planet's temperature and the surface temperature with the one opaque layer. Compare both with the observed 15 °C, and say what a second layer would add.
Hint 1
Sunlight is intercepted by a disc of area πR² and radiated away by a sphere of area 4πR². That factor of 4 is the most commonly dropped number in the whole calculation.
Hint 2
With the layer, write two balances: one for the top of the atmosphere (which must still emit exactly what the Sun delivers) and one for the surface (which now receives sunlight and the layer's downward glow).
Answer, without the method
Bare rock sits at 255 K (−18 °C). One opaque layer raises the surface by the factor 21/4 to 303 K (+30 °C). The truth, 288 K (+15 °C), lies between: the atmosphere is worth about 0.6 of an opaque layer. A second layer gives only 31/4, i.e. 335 K — each added layer buys steadily less, since Ts = (n + 1)1/4 Te.
3 routes to it
Two balances, one factor of four
n layers, and why the fourth root is merciful
What the model gets wrong, and why it still earns its keep
Afterthought. The habit here is to build the crudest model that contains the mechanism, and then to interrogate the exponent rather than the number: because T ∝ (flux)1/4, everything about planetary temperature is compressed — a factor of 16 in illumination is a factor of 2 in temperature. The same fourth root is why filament lamps are so sensitive to voltage, why a spacecraft radiator has to be enormous, and why a fractional change in emissivity is worth a quarter of itself in temperature. Whenever a quantity appears under a root, expect the world to be more stable than you feared and harder to change than you hoped.
You cannot raise the energy of the air in a room
A room 4.0 m × 3.0 m × 2.5 m (30 m³) is heated from 10 °C to 20 °C. The room is not airtight — it leaks, so the pressure inside stays at atmospheric, 1.013 × 10⁵ Pa, throughout. Treat air as a diatomic ideal gas.
By how much has the internal energy of the air in the room increased? Then find the heat required, how much air left, and where the energy went.
Hint 1
Write U for n moles of a diatomic gas, then use pV = nRT to eliminate n and T together. Look at what is left.
Hint 2
The room is an open system. Gas leaving carries its internal energy and the work it did pushing the outside air aside — that is, it carries CpT per mole, not CVT.
Answer, without the method
The internal energy of the air in the room has increased by exactly nothing — U = 5⁄2pV = 7.6 MJ, and both p and V are fixed. The heat needed is 370 kJ, 44 mol (1.3 kg) of air leaves the room, and every one of those 370 kJ walked out with it as enthalpy.
3 routes to it
The temperature cancels
Where 370 kJ went, if not into U
Why the room still takes half an hour
Afterthought. The general lesson is to check whether your system is closed before reaching for ΔU = Q − W, because an open system's bookkeeping runs on enthalpy, not internal energy — that is precisely what H is for, and why Cp appears wherever matter flows through a boundary at fixed pressure. The same U = 5⁄2pV identity, run in reverse, says something equally odd about the atmosphere: its total internal energy is fixed by the surface pressure and the volume of the sky, so warming it must mean redistributing it, not adding to it.
Two ways to empty a gas cylinder, and the gauge that lies afterwards
A 20 L cylinder holds nitrogen at 150 bar and 300 K. It is vented down to 50 bar in two ways: (a) through a needle valve over an hour, the steel walls keeping the gas at 300 K; (b) wide open, in a few seconds, too fast for any heat to reach the gas. Treat the gas as ideal, γ = 1.4.
For each route, find the mass of gas released and the temperature of the gas left inside. Find the heat the slow route draws from the walls. Then say what the pressure gauge reads an hour after the fast vent has been shut off.
Hint 1
Count moles, not pressures: n = pV/RT at each stage, and the mass released is the difference. Only one of the two routes has the same T at both ends.
Hint 2
In the fast vent, follow a parcel of gas that stays in the cylinder. Nothing heats it, and it expands smoothly while pushing the gas ahead of it out of the valve. What kind of process is that?
Answer, without the method
Slow: the gas stays at 300 K, 2.25 kg escapes, and the walls hand over 200 kJ of heat — which is exactly (Δn)RT. Fast: the remaining gas cools to 219 K (−54 °C) and only 1.83 kg escapes, 18 % less for the same pressure drop. An hour later the gauge has climbed by itself from 50 bar to 68 bar, with nothing added.
3 routes to it
The slow vent: isothermal, and the heat is (Δn)RT
The fast vent: the gas that stays behind expands reversibly
The gauge that rises after you close the valve
Afterthought. The habit worth taking is to ask what a parcel of matter that stays in the control volume experiences, separately from what happens to the matter that leaves. Here that single distinction separates a reversible adiabat from a throttling process happening centimetres apart, and it recurs everywhere — in rocket tank pressurisation, in the blowdown of a steam drum, in a pneumatic cylinder, in the puff of mist when a fizzy drink is opened. And the deeper reason our two answers differ at all is that pressure is not a measure of how much gas you have: it is a measure of gas and temperature, and only mass is honest.
The pool of water that is always 170 metres ahead
On a summer day the tarmac is at 60 °C while the air at a driver's eye, 1.2 m up, is at 30 °C. Air's refractivity is proportional to its density, so n − 1 = 2.9 × 10⁻⁴ × (293/T): the hot air just above the road is optically thinner than the air above it.
Find the steepest downward line of sight that is still turned back by the hot layer, and hence how far ahead the shimmering "water" begins. Then show that a mirage can never appear close to you, whatever the temperature of the road.
Hint 1
In a medium stratified in horizontal layers, Snell's law says n sin i is the same in every layer. Measure angles from the horizontal instead, call the angle θ, and see what conserved quantity you get.
Hint 2
The angles are milliradians, so expand: 1 − θ²/2 is accurate to a part in a million here. The answer will come out as a square root of Δn, not as Δn.
Answer, without the method
Only lines of sight within 0.41° (7.1 mrad) of horizontal are turned back, so the mirage's near edge sits about 170 m ahead — and it retreats as you drive, always staying 170 m off. It can never be close: since Δn cannot exceed n − 1 = 2.9 × 10⁻⁴ even with vacuum at the road, the turning angle can never exceed 1.4°, so the nearest possible mirage is about 40 eye-heights away, or 50 m for a driver.
3 routes to it
The conserved quantity in a stack of layers
Rays as projectiles
The hard bound, and the mirage that works the other way
Afterthought. The move that makes this problem tractable is to stop treating refraction as something that happens at surfaces. There is no surface here at all — no boundary, no reflecting layer, nothing to draw — only n varying continuously, and the language for that is a conserved quantity along the ray (n cos θ) or a differential equation with a projectile's shape. The same machinery, unchanged, gives sound bending back down over cold ground at night (which is why distant trains are audible then), radio ducting along the sea surface, and the bending of starlight by the atmosphere that puts the setting Sun half a degree above where it really is — enough that when you see it touch the horizon, it has already set.
The whole sky through a hole 4.5 metres wide
A diver floats 2.0 m below a flat, glassy water surface (n = 1.333) and looks up.
Find the size of the circular window through which the entire above-water world is visible, and how the sky is distributed across it. Where does a bird 10 m up appear to be? And what does the diver see outside the window?
Hint 1
Reverse the light: a ray arriving from just above the horizon refracts at the critical angle. Everything above water must therefore arrive inside that cone.
Hint 2
For the bird, use two rays from it and follow them in as a paraxial pencil — or notice that the apparent-depth factor works in both directions, one way giving 1/n and the other n.
Answer, without the method
The window is a cone of half-angle 48.6°, so a disc of diameter 4.5 m at that depth. The whole 180° hemisphere above is squeezed into it, and squeezed very unevenly: everything within 10° of the horizon lands in the outer 7 % of the disc's area. A bird 10 m up appears at 13.3 m — n times higher. Outside the window the underside of the surface is a perfect mirror, showing the diver the lake bed.
3 routes to it
The cone, and how badly the sky is compressed
The bird at 13.3 m, and the fish at 1.5 m
Outside the window: the silvered ceiling, and why the diver can still see you
Afterthought. The habit worth carrying is to reverse the light whenever a boundary looks confusing. Every statement about what can get in becomes a statement about what can get out, and the critical angle — which is a fact about escaping — instantly becomes a fact about the field of view. The same argument governs why an optical fibre accepts light only inside a cone, why a light-emitting diode wastes most of its output inside its own high-index chip, and why the brilliance of a cut diamond is a deliberate exercise in giving light no legal way out except through the top.
The lens that loses three quarters of its power in water
A thin glass lens, n = 1.52, has a focal length of +20.0 cm in air. It is now submerged in water, n = 1.33.
Find its focal length in water. Then find what an identically shaped lens made of ice (n = 1.31) would do in water. Finally, explain why the human eye is hopelessly out of focus underwater and why a flat glass mask fixes it.
Hint 1
In the lensmaker's equation the index that matters is the ratio of lens to surroundings. Write the shape factor once and never touch the radii again.
Hint 2
For the ice lens, ask what the sign of (nlens − nmedium) is. A shape does not decide whether a lens converges.
Answer, without the method
In water f rises to +72.8 cm: the power drops by a factor of 3.64. The ice lens, convex and thick in the middle, is diverging in water, f ≈ −6.9 m. The eye fails underwater because about 42 of its 60 dioptres come from the air-to-cornea step, which water removes; the mask restores that step and gives the whole 42 D back.
3 routes to it
Lensmaker's equation with the medium put back in
A lens is a phase plate, so the power lives in the index step
Why your eye fails and a flat mask cures it
Afterthought. The generalisable point is that no optical component has properties of its own — only differences from its surroundings do. Power, reflectivity and even visibility all scale with an index step, which is why immersion oil recovers a microscope's resolution, why a chipped windscreen repaired with matched resin becomes invisible, and why cladding an optical fibre with glass of index only 1 % lower is enough to guide light for kilometres. Whenever a number is quoted for a lens, a mirror, a coating or a fibre, the honest question is: against what?
Measure a focal length without ever measuring from the lens
An object and a screen are clamped 100.0 cm apart. A converging lens slid between them gives a sharp image at two distinct positions, and those two positions are 20.0 cm apart. The two images are not the same size.
Find the focal length. Show that the two magnifications multiply to exactly one, and use that to get the object's size from the two image sizes alone. Then find the smallest separation for which any sharp image is possible at all.
Hint 1
Do not solve for u. Add the two conjugate distances and multiply them, using 1/u + 1/v = 1/f — you will have the sum and the product of two numbers, which is all a quadratic needs.
Hint 2
For the magnifications, reverse the light. What does the second lens position do to the object and image that the first one did not?
Answer, without the method
f = 24.0 cm, with the lens 40.0 cm and 60.0 cm from the object. The magnifications are 1.5 and 2/3 — their product is exactly 1, so the object's height is the geometric mean of the two image heights, √(I₁I₂). No sharp image exists at all unless the separation is at least 4f = 96 cm, and at exactly 4f the two positions merge into one, with the lens in the middle and unit magnification.
3 routes to it
Sum and product: the quadratic that hides in the lens equation
Reversibility, and why the product of the magnifications is one
Why this is the method a laboratory actually uses
Afterthought. The transferable habit is to look for the symmetric combinations of your unknowns before solving for them. The conjugate equation is symmetric in u and v, so their sum and product are the natural coordinates, and once you are in those coordinates the two-position structure, the reversibility, the unit product of magnifications and the D ≥ 4f condition are all one fact rather than four. The same manoeuvre reappears wherever a relation is symmetric — in the two launch angles that hit the same target, in the two resistances that dissipate the same power, in the pair of frequencies at which a resonant circuit gives half power.
A 60° prism deviates 11 degrees more than the formula says
A prism of apex angle 60.0° is made of glass with n = 1.62. The formula most people reach for gives a deviation of (n − 1)A = 37.2°.
Find the true minimum deviation and the incidence at which it occurs. Then find the range of incidence angles for which any light emerges from the second face at all, and the largest apex angle a prism of this glass can usefully have.
Hint 1
Do the exact chain: i₁ → r₁ → r₂ = A − r₁ → i₂, with δ = i₁ + i₂ − A. Nothing is approximated anywhere in it.
Hint 2
For the minimum, you do not need calculus: reversibility says δ(i₁) and δ(i₂) describe the same curve, so the extremum must sit where i₁ = i₂.
Answer, without the method
The minimum deviation is 48.2°, at an incidence of 54.1° with the ray passing symmetrically — the thin-prism formula is 11° low, an error of nearly a quarter. Light emerges only for incidence above 37.1°; below that it is totally internally reflected at the exit face. And no prism of this glass with an apex angle above 2 × 38.1° = 76.2° will transmit at any incidence whatever.
3 routes to it
The exact prism, and the minimum by symmetry
When nothing comes out at all
How fast the thin-prism formula rots
Afterthought. The habit is to ask what small quantity an approximation is expanding in, and then to check its size against 1 before using the result. Here the small quantity is A itself, and 60° is 1.05 radians — not small by any reading, so the surprise should be that the formula is ever quoted for such prisms rather than that it fails. The same discipline saves you with the small-angle pendulum (θ²/16), the paraxial lens (aberrations in ρ²), the thin-film approximation, and the binomial expansions that litter relativity: the expansion parameter is usually written down in the first line of the derivation and then never looked at again.
The best pinhole camera you can build is a 0.07 megapixel camera
A pinhole camera has its film 100 mm behind the hole and works in green light, λ = 550 nm. Make the hole large and every point of the scene paints a patch the size of the hole; make it small and diffraction spreads that patch out again.
Find the hole diameter that gives the sharpest picture, and the blur that remains. Convert it to an angular resolution, decide how many resolvable spots fit across the Moon, and work out the exposure penalty against an ordinary lens.
Hint 1
Write the two blurs as functions of d — one growing, one shrinking — and set them equal rather than differentiating a sum. For a pair like this that lands within a few per cent of the true optimum.
Hint 2
Diffraction from a circular hole gives an Airy disc of angular diameter 2.44λ/d, so its size on the film is 2.44λL/d.
Answer, without the method
The optimum is d = √(2.44λL) = 0.37 mm, and the blur is that same 0.37 mm — you cannot do better at this camera length. That is an angular resolution of 0.21°, so the Moon's half-degree disc is about 2.4 blur widths across: two or three pixels. The camera works at f/270, needing roughly 300 × the exposure of f/16, and the whole 100 mm plate holds only about 70 000 resolvable spots.
3 routes to it
Two blurs, one crossing point
The scaling, and why you cannot fix it by building a bigger one
What a lens actually buys you
Afterthought. The general pattern is worth more than the pinhole: when two competing errors depend oppositely on one parameter, the optimum sits where they are equal, and the achievable performance is then the geometric mean of the two mechanisms. It reappears in the optimum slit width of a spectrometer, the optimum integration time against drift and noise, the optimum aperture of a camera lens (diffraction against aberration, which is why f/8 is the sweet spot), and the optimum thickness of a shielding layer. And the pinhole's fate — an optimum that improves only as a square root — is the signature of a mechanism that cannot be engineered around, only replaced.
The third fringe that refuses to appear
Two parallel slits, each 0.10 mm wide, have their centres 0.30 mm apart. They are lit by λ = 600 nm and the pattern is caught on a screen 2.0 m away.
Find the fringe spacing. Then find which bright fringes are missing from the pattern, and how many survive inside the central diffraction envelope. What happens to the centre of the pattern if you cover one slit?
Hint 1
The pattern is a product, not a sum: what each slit sends in a direction, times how the two contributions interfere. Write both factors before you evaluate either.
Hint 2
A fringe is missing when the interference maximum lands exactly on a diffraction zero. Set the two conditions equal and see what the ratio d/a has to do with it.
Answer, without the method
Fringes every 4.0 mm, with the envelope's first zero at 12.0 mm — so every third order is missing: the 3rd, 6th, 9th and so on, because d/a = 3. Five bright fringes fit inside the central envelope (orders 0, ±1, ±2). Covering one slit drops the central intensity to a quarter, not a half, and leaves a single-slit pattern 24 mm wide.
3 routes to it
The product, and where a maximum meets a zero
You cannot build a fringe out of two zeros
What the pattern tells you, and the factor of four
Afterthought. The transferable idea is the factorisation itself: whenever a system is a regular array of identical elements, the far field is (what one element does) × (how the array interferes) — the element factor and the array factor. Radio engineers design antenna arrays with exactly this decomposition, X-ray crystallographers call the two pieces the atomic form factor and the structure factor, and the missing orders here are the same phenomenon as a systematically absent Bragg reflection, which is how a crystallographer detects a centring or a glide plane without ever seeing an atom.
Where two glass plates touch, the fringe is dark
Two optical flats are pressed together at one edge, leaving a thin wedge of air. Lit from above with sodium light (λ = 589 nm) at normal incidence, they show straight fringes ten to the centimetre — and the fringe along the contact line is dark, where the path difference is zero.
Account for the dark contact fringe, find the wedge angle, and then design an anti-reflection coating for glass of n = 1.52 — the ideal index and thickness, and what magnesium fluoride (n = 1.38) actually achieves. Finally, name an experiment that decides the phase question outright.
Hint 1
Count the phase changes at the two surfaces separately. Reflection at a boundary where the index goes up flips the sign of the wave; reflection where it goes down does not.
Hint 2
For the wedge, the extra path is twice the local gap, so consecutive fringes are half a wavelength of gap apart. Turn that into a horizontal spacing.
Answer, without the method
Dark because only one of the two reflections suffers a π phase shift: air→glass does, glass→air does not, so at zero path difference the amplitudes cancel. The wedge angle is λ/2Δx = 2.9 × 10⁻⁴ rad = 61 arcsec. The ideal coating has n = √1.52 = 1.23 and thickness λ/4n = 112 nm; real MgF₂ takes the reflectance from 4.3 % to 1.2 %, not to zero. The decisive experiment: fill the wedge with a liquid of index between the two glasses — the contact fringe turns bright.
3 routes to it
The phase bookkeeping, then the geometry
The experiment that settles the sign
Designing the coating, and why one layer is not enough
Afterthought. The recurring point is that a phase is physical and must be tracked, not absorbed into a convention. The π at a hard boundary is not an optical quirk: a pulse on a light string reflects inverted from a junction with a heavy one, a voltage wave inverts on reflection from a short circuit, and a quantum wavefunction picks up the same sign against a rising potential step — one statement, three subjects. Whenever a calculation gives the right magnitude and the wrong contrast, the missing ingredient is almost always a sign at a boundary rather than an error in the path lengths.
Six hundred times, on a telescope that cannot use a hundred and twenty
At night the eye's pupil opens to about 5 mm; take λ = 550 nm. A car's headlights are 1.5 m apart. A shop sells a 60 mm telescope advertised as "600 ×".
Find the distance at which diffraction alone would make the two headlights merge. Then find the smallest magnification at which a 60 mm telescope hands all of its detail to the eye, the largest that is worth using, and say plainly what 600 × delivers.
Hint 1
Rayleigh's criterion, θ = 1.22λ/D, applies to the eye's pupil just as it does to a telescope's objective — the eye is a 5 mm telescope.
Hint 2
Magnification does not resolve anything; it only rescales what the aperture has already decided. Ask what angular size the telescope's smallest detail must be given to be visible to the eye.
Answer, without the method
Diffraction alone merges the headlights at 11 km (real eyes manage 1.5–2 km, for other reasons). The 60 mm telescope resolves 2.3 arcsec and needs about 26 × to deliver that to a 1-arcmin eye; beyond about 120 × the exit pupil falls under 0.5 mm and the view is dim and swimming. At 600 × you see exactly the detail 120 × shows, five times larger, five times dimmer per unit area, and shaking five times as much. The extra magnification is empty.
3 routes to it
Rayleigh, twice
Magnification only rescales; and the eye is beautifully matched to itself
What actually limits a telescope on the ground
Afterthought. The reusable idea is to identify which stage of a chain sets the information content, and to be sceptical of any later stage that claims to add to it. Aperture sets angular resolution; the eyepiece, the sensor, the display and the zoom can only fail to throw it away. The same audit applies to sampling rate against filter bandwidth in electronics, to bit depth against noise floor in audio, and to interpolation in images: past the point where a stage matches the information available, extra capability is not merely useless but actively misleading, because it looks like performance.
Add a third polariser and light gets through
Two ideal polarising sheets are crossed, and unpolarised light of intensity I₀ falls on them. Nothing emerges. A third sheet is now slid in between the two, with its axis at 45°.
Find what emerges, and whether 45° is the best angle. Then find how much can be pushed through if you use N sheets spread evenly across the 90°, and say what that limit means. Absorbers only — nothing here rotates anything.
Hint 1
Malus's law applies to each sheet in turn, and each sheet defines a fresh reference direction for the next. Only the angle between consecutive axes ever matters.
Hint 2
A polariser does not filter the light it receives — it projects it. Ask what direction the light is polarised along after the middle sheet, not what it was polarised along before.
Answer, without the method
I₀/8 emerges, and 45° is indeed optimal, since the output is (I₀/8)sin²2θ. With N sheets evenly spread the output is (I₀/2)cos2N(90°/N): 3 sheets give 0.21 I₀, 10 give 0.40 I₀, 45 give 0.47 I₀, and as N → ∞ it tends to I₀/2 — the whole of the polarised light. A stack of pure absorbers becomes a lossless rotator of the plane of polarisation in the limit.
3 routes to it
Malus, applied one sheet at a time
Projection, not filtration
The staircase: absorbers that add up to a rotator
Afterthought. The habit worth taking away is to treat a filter as an operator rather than a sieve: it maps an input state to an output state, and the output may have components the input lacked. Once that is in place, the ordering effects, the third-sheet surprise and the lossless-rotator limit stop being paradoxes and become consequences. The same reframing pays off in every measurement chain where sequential projections occur — polarisation optics, spin resonance, filter cascades in signal processing — and the general moral is that composition of operations is not the same as intersection of restrictions.
The cavity that hides no charge
A solid sphere of radius R = 0.10 m carries uniform charge density ρ = 3.0 µC m⁻³. A spherical cavity of radius b = 0.030 m is then hollowed out, its centre a = 0.040 m from the sphere's centre. No charge is placed in the cavity.
Find the field at the cavity's near wall, at its far wall, and at its centre — then the potential difference across it.
Hint 1
You cannot use Gauss's law on the cavity directly — no surface through it has enough symmetry. Add something that is not there instead.
Hint 2
Inside a uniformly charged sphere the field is ρr/3ε₀ — linear in the position vector from its own centre. Two such fields, subtracted, lose the r.
Answer, without the method
The field is the same everywhere inside the cavity: E = ρ a3 ε₀ = 4.5 kV m⁻¹, pointing from the sphere's centre towards the cavity's. Across the cavity that gives E × 2b = 271 V.
3 routes to it
Superposition route
Why it is allowed — and where it stops working
Superposition route
Afterthought. The field in the cavity depends on where the cavity is, but not on how big it is — and not at all on where you stand inside it. That is a strong statement, and it is entirely a consequence of E ∝ r for a uniform ball. Whenever a problem offers you an off-centre hole, a chipped corner, or a missing chunk, the move is the same: add the missing piece back with the opposite sign.
The energy the second capacitor never receives
A capacitor C₁ = 4.0 µF is charged to 100 V and disconnected from its supply. It is then connected, through a switch and a resistor R, across an uncharged C₂ = 6.0 µF.
Find the final common voltage and the energy missing at the end. Then find how the answer depends on R — and what happens instead if R is replaced by an inductor.
Hint 1
Charge has nowhere to go, and at the end the two plates are at one potential. That is two equations and the final state, with no mention of R.
Hint 2
For the R-independence, you do not need to solve the circuit: write the heat as ∫i²R dt and notice what the time constant does to it.
Answer, without the method
40 V, and 12 mJ of the original 20 mJ is gone — 60 % of it. The loss is C₁C₂2(C₁+C₂) ΔV², completely independent of R. With an inductor in place of R nothing is lost: the charge oscillates, and C₂ overshoots to 80 V.
3 routes to it
Final state, then the books
Where it went, and why R does not matter
Put an inductor there instead
Afterthought. 60 % is not a small leak, and it is set entirely by the ratio C₂/C₁: with C₂ ≫ C₁ almost everything is lost, and with C₂ ≪ C₁ almost nothing. Real switched-capacitor converters live on this equation and defeat it the same way this problem does — by putting an inductor in the path so the charge is carried rather than dumped.
The uncharged sphere that is pulled in anyway
A point charge q = 1.0 µC sits r = 0.10 m from the centre of a small uncharged conducting sphere of radius a = 2.0 mm, free to move.
Find the force on the sphere, say how it depends on r and on the sign of q, and use it to explain why a thin stream of tap water bends towards a charged comb whichever way the comb was charged.
Hint 1
An uncharged conductor feels no force in a uniform field. This field is not uniform, and that is the whole problem.
Hint 2
Induced dipole moment p = 4πε₀a³E. The energy of an induced dipole is −½αE², not −pE — the missing half is the work of polarising it.
Answer, without the method
F = 2 a³ q²4πε₀ r⁵ = 14 µN, always attractive, and falling off as 1/r⁵. Reversing q leaves it unchanged, because the induced dipole reverses with it.
3 routes to it
Induced dipole and the gradient of E²
Build the dipole out of two charges, and see where r⁻⁵ comes from
Dimensions, the water stream, and the honest caveat
Afterthought. "Uncharged means no force" is safe only in a uniform field, and no real field is uniform. Every dielectric is drawn towards the strong-field region, which is why toner sticks to a latent image, why smoke precipitators work, and why a neutral grain of dust finds the one charged spot on a screen.
One step across an endless grid
An infinite square grid is soldered up from identical resistors, each 12 Ω, one to every edge.
Find the resistance between two adjacent nodes. Then do the same for an infinite cubic lattice of the same resistors, and say why the argument cannot give you the resistance between two nodes a diagonal apart.
Hint 1
You cannot reduce this by series and parallel; nothing in it is either. Use the one thing an infinite lattice gives you for free.
Hint 2
Feed current in at A and take it out at B as two separate problems, and add the answers.
Answer, without the method
6.0 Ω — exactly half of one resistor. On a cubic lattice it is 4.0 Ω: in general 2R/z, where z is the number of edges meeting at a node.
3 routes to it
Superposition of two symmetric problems
The same argument, counting edges at a node
Why the diagonal is a different kind of problem
Afterthought. Half a resistor, whatever the lattice looks like beyond the two nodes — and it is exactly half because each of the two symmetric sub-problems contributes a quarter. That one split-and-add manoeuvre is the whole of lattice network theory at this level, and it works at any node of any lattice; it is the non-adjacent pairs that need real mathematics.
Two loads, the same power
A battery of emf ε = 12 V and internal resistance r = 2.0 Ω drives a variable load R.
Find the R that draws the most power and that power. Then find the two values of R that both deliver exactly 16 W, note what their product is, and compare the efficiency of the two choices.
Hint 1
Write P(R), then divide top and bottom by R. What is left in the denominator is a sum of two terms whose product is fixed.
Hint 2
"Which R gives 16 W" is a quadratic in R. You do not have to solve it to know what its two roots multiply to.
Answer, without the method
R = r = 2.0 Ω gives Pmax = ε²4r = 18 W at 50 % efficiency. 16 W is delivered by R = 1.0 Ω or 4.0 Ω — their product is r², always — at efficiencies of 33 % and 67 %.
3 routes to it
AM–GM route, no calculus
The quadratic, and the product of its roots
Which of the two you should choose
Afterthought. Maximum power and maximum efficiency are different goals with different answers, and the peak in P is so flat that you can have 89 % of the power at twice the efficiency by simply doubling the load. Whenever a question says "best", ask best at what.
The meter that lies, and by how much
A resistor of about 100 Ω is measured by dividing a voltmeter reading by an ammeter reading. The ammeter has resistance rA = 1.0 Ω, the voltmeter RV = 2.0 kΩ. Two wirings are possible: A — voltmeter across the resistor alone, ammeter outside it; B — ammeter in series with the resistor, voltmeter across the pair.
Find the apparent resistance in each wiring, say which is better here, recover the true value from either reading, and find the resistance at which the two wirings are equally wrong.
Hint 1
Neither meter is ideal, so in each wiring one of them is measuring something that includes a piece it should not. Say what, in words, before writing anything.
Hint 2
Compare the two fractional errors, R/RV and rA/R, and ask when they are equal.
Answer, without the method
A gives 95.2 Ω (4.8 % low), B gives 101 Ω (1.0 % high), so B wins for this resistor. They are equally wrong at R = √(rARV) = 45 Ω: below that use A, above it use B.
3 routes to it
Analyse both wirings
The crossover is a geometric mean
Do not correct the error — invert it
Afterthought. Every measurement disturbs what it measures, and here the disturbance is a resistor you can name. The Wheatstone bridge and the potentiometer exist to sidestep the problem entirely: at balance no current flows through the detector, so its resistance stops mattering — which is how they beat a good voltmeter and a good ammeter used together.
Released from rest into crossed fields
A proton (m = 1.67 × 10⁻²⁷ kg, q = 1.6 × 10⁻¹⁹ C) is released from rest at the origin in an electric field E = 500 V m⁻¹ along +y and a magnetic field B = 0.020 T along +z.
Find how far it ever gets along y, its greatest speed, and its average velocity in the long run.
Hint 1
The magnetic force does no work, so the speed at any point is fixed by the y-coordinate alone. That is one equation for free.
Hint 2
Is there a frame in which the electric field disappears? If there is, the motion in that frame is something you already know completely.
Answer, without the method
It never passes y = 2.6 cm, its speed oscillates between 0 and 2E/B = 50 km s⁻¹, and it drifts steadily along +x at E/B = 25 km s⁻¹ — a cycloid, retracing the same arch every 3.3 µs.
3 routes to it
Change frames until the field vanishes
Stay in the lab frame and find two constants
The bench, and where the picture breaks
Afterthought. A constant force on a magnetised particle does not accelerate it — it makes it drift sideways, at right angles to the force, at E/B. That single fact governs the confinement of laboratory plasmas, the sorting of ions in a mass filter, and the Hall voltage that lets you measure a magnetic field with a slab of semiconductor.
How much magnetism can you spin into a disc?
A charge of Q = 2.0 µC is spread uniformly over one face of an insulating disc of radius R = 0.10 m, which is then spun about its axis at ω = 3000 rad s⁻¹.
Find the field at the centre and the disc's magnetic moment — then show that for any body whose charge and mass are distributed alike, the moment and the angular momentum are locked in a fixed ratio, and say what that ratio is.
Hint 1
A ring of charge going round is a current loop. Which ring, and carrying what current?
Hint 2
Every ring contributes the same amount to the field at the centre — check that before integrating, and the integral becomes a multiplication.
Answer, without the method
B = µ₀ σ ω R2 = 12 nT, and m = σ ω π R⁴4 = 1.5 × 10⁻⁵ A m². For any such body m/L = Q/2M — independent of shape, size and spin rate.
3 routes to it
Ring by ring
The gyromagnetic ratio, without integrating anything
How small the answer is, and why that matters
Afterthought. m/L = Q/2M is the most useful line in this problem and it contains no geometry at all. It is why the classical model of atomic magnetism gets the magnitude of the Bohr magneton right on the first attempt, and why the factor-of-two discrepancy for electron spin was such a serious clue.
The rod that never reaches a top speed
A frictionless rod of mass m = 50 g slides on horizontal rails L = 0.50 m apart in a vertical field B = 2.0 T. The rails are joined not by a resistor but by an uncharged capacitor C = 20 mF, and everything is resistanceless. A constant force F = 0.35 N pushes the rod from rest.
Find the rod's acceleration, and account for all the energy at the moment it reaches 10 m s⁻¹. Then say how the whole story changes if the capacitor is swapped for a resistor.
Hint 1
The capacitor's charge is set by the emf at that instant, so the current is set by the rate at which the emf changes.
Hint 2
Write the retarding magnetic force in terms of the acceleration, then collect it with ma. Something recognisable appears.
Answer, without the method
a = Fm + B²L²C = 5.0 m s⁻², constant for ever — the circuit adds B²L²C = 20 g of inertia to a 50 g rod. At 10 m s⁻¹ the rod holds 2.5 J and the capacitor 1.0 J, together the 3.5 J the force has done. With a resistor there is a terminal speed instead, and half the input ends as heat.
3 routes to it
Circuit, then Newton
Effective mass, from the energy
Put a resistor there instead
Afterthought. A capacitor in an induction loop is not a brake, it is a mass — it resists change of speed rather than speed itself, and stores what it takes instead of burning it. That is the electrical face of inertia, and the same algebra with L and C exchanged makes an inductor behave like a spring. Whether a circuit element feels like mass, damping or springiness is a question about which derivative of the position it responds to.
The charge that does not care how fast you pull
A square loop of side ℓ = 0.10 m and total resistance 0.50 Ω lies in a uniform field B = 0.40 T perpendicular to its plane. It is pulled completely out of the field region at a steady v = 2.0 m s⁻¹.
Find the total charge that flows, the heat produced and the work you had to do. Then say which of the three change, and by how much, if you pull it out ten times faster.
Hint 1
Integrate the current rather than evaluating it: ∫i dt takes a form that never mentions time.
Hint 2
The work you do goes somewhere. The loop's kinetic energy is unchanged, so there is only one candidate.
Answer, without the method
q = ΔΦR = 8.0 mC, and it is completely independent of the speed. The heat and the work are both 0.64 mJ and both are proportional to v — pull ten times faster and they are ten times larger, while q does not move.
3 routes to it
Integrate the current
Work it out from the force you had to fight
What the independence is good for — and when it fails
Afterthought. Charge cares only about the total flux change; heat cares about how quickly you forced it. That split — one quantity a pure difference between states, the other a path integral — is the same distinction as potential energy versus friction work, and knowing which of the two a question is asking about usually decides how much work it will take.
The small sphere that ruins the big one
A conducting sphere of radius R = 0.10 m is joined by a long thin wire to a second one of radius r = 0.020 m, far enough away that neither disturbs the other's field. Air breaks down at 3.0 × 10⁶ V m⁻¹.
Share a total charge of 6.0 µC between them and find both surface fields. Then find the highest potential the pair can be raised to before something leaks — and compare it with what the large sphere could hold on its own.
Hint 1
The wire makes them one conductor, so there is one thing they must share. It is not the charge.
Hint 2
Write both the potential and the surface field in terms of the same radius and see which way each of them scales.
Answer, without the method
Charge divides as R : r, so 5.0 µC and 1.0 µC — but the surface fields go as 1/R, so the small sphere's field is five times the large one's. The pair tops out at 60 kV; the large sphere alone would reach 300 kV.
3 routes to it
One potential, two fields
Capacitances add, and that is the trap
Why lightning rods are pointed, and corona
Afterthought. Charge collects where the surface is curved, and field is what breaks things — so a system is limited by its worst point and not by its size. Doubling the radius of a terminal doubles the voltage it can hold; polishing off one burr can be worth more than either.
The slab that is pulled in
A parallel-plate capacitor has square plates of side 0.10 m separated by d = 1.0 mm, held at V = 200 V by a battery. A slab of dielectric constant κ = 4.0, of the same thickness as the gap, is inserted a distance x from one edge.
Find the force on the slab and its direction, then account for the energy exactly. Then disconnect the battery and say what changes.
Hint 1
The capacitance is two capacitors in parallel: slab-filled width x and empty width (L − x). Write C(x) and you have written most of the problem.
Hint 2
With the battery attached, energy is not conserved between the field and the slab alone — the battery is a third party, and it delivers V dQ.
Answer, without the method
F = ½V² dCdx = 53 µN, drawing the slab inwards, and constant all the way in. The battery supplies V²dC of energy per unit length — half stored, half spent pulling. At constant charge the pull is still inwards but weakens as the slab enters.
3 routes to it
Constant voltage, with the battery in the books
Constant charge, and the sign that never flips
Where the force actually acts
Afterthought. Systems move to increase their capacitance. That single sentence predicts the direction of every dielectric and every plate motion in the chapter, and it is safer than any energy expression, because "which energy" depends on what the battery is doing.
The lamp that draws thirteen times too much
A tungsten filament lamp is rated 100 W at 230 V. Cold, on a bench meter, its resistance measures 40 Ω. Tungsten's temperature coefficient of resistance is α = 4.5 × 10⁻³ K⁻¹.
Find its working resistance, the current at the instant of switch-on, and the temperature the filament runs at. Then say why the lamp does not blow every time you switch it on.
Hint 1
The rating describes the lamp when it is hot. The meter describes it when it is cold. They are not the same resistor.
Hint 2
R = R₀(1 + αΔT) has only one unknown once you have both resistances.
Answer, without the method
Working resistance V²/P = 529 Ω, so switch-on current is 230/40 = 5.75 A against 0.435 A running — 13 times as much. That ratio gives ΔT ≈ 2700 K, so the filament sits near 2700 °C. It survives because the surge lasts a few tens of milliseconds, less than the filament's thermal time.
3 routes to it
Two resistances, and the temperature between them
Why it survives — the thermal time constant
The V–I curve is not a straight line
Afterthought. "Ohmic" is a claim about a range, not about a material. The filament obeys Ohm's law perfectly at every fixed temperature and disobeys it wildly across a switch-on, and the same is true of every resistor you will ever overload — which is why a power resistor's rating is a temperature, expressed as watts.
The wire whose shape does not matter
A wire carrying I = 3.0 A is bent into a semicircle of radius R = 0.20 m and placed in a uniform field B = 0.50 T perpendicular to its plane. The straight leads run along the line joining its two ends.
Find the force on the semicircular part. Then find it for the same current and endpoints bent into three quarters of a circle, and for a closed loop — and say what the three answers have in common.
Hint 1
Write the force as an integral of I dl × B and take the constant B outside. What is left of the integral?
Hint 2
∫dl along any path is a quantity you already know how to evaluate without doing the integral.
Answer, without the method
F = B I (2R) = 0.60 N, at right angles to the line joining the ends. For the three-quarter arc the endpoints are R√2 apart, so F = 0.42 N. For a closed loop, zero. In a uniform field only the end-to-end vector matters, never the path.
3 routes to it
Take B out of the integral
Do the integral the hard way, once, to believe it
What it means for loops and motors
Afterthought. The whole result is the one line ∫Idl × B = I(∫dl) × B, and it holds only because B is constant enough to come out of the integral. That is the same manoeuvre, and the same caveat, as pulling g out of a centre-of-mass calculation — the reason both tricks work is uniformity, and the reason both fail is a gradient.
How deep does it get into the field?
A proton (m = 1.67 × 10⁻²⁷ kg, q = 1.6 × 10⁻¹⁹ C) enters a slab-shaped region of field B = 0.010 T, out of the page, at right angles to its face and moving at v = 2.0 × 10⁵ m s⁻¹. The slab is d = 0.10 m thick.
Find the angle at which it emerges and the time it spends inside. Then find the field that would just prevent it emerging at all, and the deepest it gets in that case.
Hint 1
Inside the slab the path is a circular arc whose centre is on the entry face. Draw that circle and the question becomes trigonometry.
Hint 2
The time inside depends on the angle turned through, not on the speed. Write it that way and one of the numbers you were given becomes unnecessary.
Answer, without the method
r = mv/qB = 0.209 m, so sin θ = d/r = 0.479 and it leaves the far face deflected 28.6°, after 0.52 µs inside. It fails to emerge once B ≥ 0.021 T (r ≤ d), and then turns back after reaching a depth of exactly r.
3 routes to it
The chord, drawn properly
Time, without the speed
Turn it into an instrument, then break it
Afterthought. Everything here came from one construction: the centre of the arc is perpendicular to the velocity, a distance r away. Draw that centre before writing anything and most magnetic-deflection problems become geometry; skip it and they become simultaneous equations.
A millimetre of air undoes half a metre of iron
An iron ring of mean circumference 0.50 m and relative permeability µr = 2000 is wound with N = 500 turns carrying I = 1.0 A. A gap of 1.0 mm is then cut across it.
Find the field in the gap, and compare it with the field in the unbroken ring. Then find how wide the gap has to be before it dominates the ring, and say what that means for an electromagnet's design.
Hint 1
Ampère's law around the ring still holds, but H is not the same in the iron and in the gap. What is the same?
Hint 2
Write the loop integral of H as a sum of two terms, and each term as B/µ times a length.
Answer, without the method
Bgap = 0.50 T, against 2.5 T in the unbroken ring — a 1 mm gap in a 500 mm ring has removed 80 % of the field. The gap takes over as soon as it exceeds ℓiron/µr = 0.25 mm.
3 routes to it
Ampère's law, with B continuous
The magnetic circuit — Ohm's law in disguise
The honest caveats, and the bench
Afterthought. A magnetic circuit is a series circuit, and air is the big resistor — so the field is set by the worst part of the path, not the best. That is the magnetic twin of the earlier lesson that a charged system is limited by its sharpest point: in both, a long stretch of excellent material is undone by a short stretch of poor.
Why a paramagnet barely bothers
A solid holds 5 × 10²⁸ atoms per cubic metre, each with one unpaired electron spin of moment µ = 9.3 × 10⁻²⁴ J T⁻¹. It sits at 300 K in a field of 1.0 T — a strong laboratory magnet.
Find what fraction of the moments are effectively aligned, and hence the magnetisation and the susceptibility. Then find the temperature, or the field, that would come close to aligning them all.
Hint 1
Compare two energies before you compute anything: what the field offers a moment, and what the temperature already gives it.
Hint 2
A two-state moment in thermal equilibrium has ⟨µz⟩ = µ tanh(µB/kBT), and for small argument tanh x ≈ x.
Answer, without the method
µB/kBT = 2.3 × 10⁻³, so the alignment is about two parts in a thousand: M ≈ 1.0 × 10³ A m⁻¹ and χ ≈ 1.3 × 10⁻³. Full alignment needs T below about 1 K, or a field of order 400 T at room temperature.
3 routes to it
The two energies, then the two-level average
The classical average, and an instructive discrepancy
So where do real magnets come from?
Afterthought. One dimensionless ratio, µB/kBT, decides the whole of paramagnetism — and at room temperature in the strongest field you are likely to meet, it is a thousandth. Write that ratio first in every thermal-alignment problem, whether the aligning agent is a magnetic field, an electric field on a polar molecule, or gravity on a colloid; it tells you the answer's size before you know its formula.
The disc dynamo that Faraday's flux rule cannot explain
A copper disc of radius R = 0.15 m spins at ω = 200 rad s⁻¹ about its axis in a field B = 0.30 T parallel to that axis. One brush touches the axle and another the rim; they are joined through a load of 0.10 Ω (take the disc and brushes as resistanceless).
Find the emf, the current and the retarding torque. Then say what goes wrong if you try to get the emf from "rate of change of flux", and what to use instead.
Hint 1
Ask what force pushes an electron in the disc, and integrate its work per unit charge from the axle to the rim.
Hint 2
The retarding torque is not a new calculation. Energy has to come from somewhere, at a known rate.
Answer, without the method
ε = ½BωR² = 0.675 V, i = 6.75 A, and the torque is 0.023 N m — exactly P/ω, as it must be. The flux through any fixed circuit here never changes, so dΦ/dt gives zero; the emf comes from the Lorentz force on the carriers, and only the moving-conductor form of the law survives.
3 routes to it
Integrate the Lorentz force along a radius
The torque, from energy alone
Why the flux rule fails here
Afterthought. Two different mechanisms wear one name. A changing field makes an electric field that pushes stationary charges; a moving conductor makes its own charges feel v × B. Any problem where the circuit's shape or membership changes should be done with the second, and the flux rule should be treated as a convenience, not a law.
Three thousand volts out of a twelve-volt supply
A coil of inductance L = 2.0 H carries 1.5 A from a 12 V supply. A mechanical switch in series with it is opened, and the current falls to zero in about 1.0 ms as the contacts part.
Find the average voltage that appears across the coil, and the energy that has to go somewhere. Then design the cheapest fix that keeps the peak below 100 V, and say where the energy ends up then.
Hint 1
An inductor's current cannot change instantly, and a switch insists that it does. One of them has to give, and it is not the inductor.
Hint 2
The energy stored has one number and does not care how you open the switch. Only the voltage does.
Answer, without the method
ε = L Δi/Δt ≈ 3.0 kV from a 12 V supply, carrying 2.25 J that has nowhere else to go — it appears as an arc across the contacts. A diode or a resistor of 67 Ω or less across the coil caps the spike at 100 V; the same 2.25 J then decays into that resistor with time constant L/R = 30 ms.
3 routes to it
The spike and the stored energy
Give it somewhere to go
The same effect, sold as a feature
Afterthought. An inductor with its current interrupted behaves like an incompressible fluid with its pipe pinched: something has to yield, and the pressure spike goes as high as it must. Wherever a circuit switches an inductive load, look for the path the current takes in the instant after — if you cannot name it, it is going through something that was not designed to carry it.
Four hundred volts across a component, from a twenty-volt supply
A series circuit has L = 0.10 H, C = 10 µF and R = 5.0 Ω, driven by 20 V rms at its resonant frequency.
Find the resonant frequency, the current, and the voltage across the inductor. Explain how that voltage can exceed the supply without violating Kirchhoff's loop rule — and find the bandwidth over which the circuit stays near resonance.
Hint 1
At resonance the two reactances are equal in size. What are they individually, and what does the source then see?
Hint 2
Kirchhoff's rule is about instantaneous voltages. Add the two big ones as phasors before declaring a contradiction.
Answer, without the method
ω₀ = 1/√(LC) = 1000 rad s⁻¹ (159 Hz), i = V/R = 4.0 A, and VL = VC = 400 V — twenty times the supply. They are in antiphase and cancel exactly, so the loop rule is untroubled. The bandwidth is R/L = 50 rad s⁻¹, a Q of 20.
3 routes to it
Phasors, and the magnification
Q from the energy, with no phasors at all
What it is for, and how it goes wrong
Afterthought. Q is one number wearing three hats: the voltage magnification across L or C, the ratio of stored to dissipated energy per radian, and the reciprocal of the fractional bandwidth. Compute it first, and you know the circuit's sharpness, its danger and its efficiency in one line.
The capacitor that pays for itself
A workshop draws 5.0 kW of real power from a 230 V, 50 Hz supply at a power factor of 0.60 lagging. The cable feeding it has resistance 0.40 Ω each way.
Find the line current and the power wasted in the cable. Then find the capacitor that brings the power factor to unity, and say what the utility gains and what the workshop gains.
Hint 1
Real power is VI cos φ, so a poor power factor means the same kilowatts arrive on a larger current. The cable does not care about phase, only about I².
Hint 2
The capacitor is not there to supply power. Work out the reactive power the load demands and give it exactly that.
Answer, without the method
I = P/(V cos φ) = 36.2 A, wasting 1.05 kW in the cable — a fifth of what the workshop uses. Correcting to unity drops the current to 21.7 A and the cable loss to 376 W, a saving of 64 %, and needs 400 µF across the supply.
3 routes to it
The power triangle
What reactive power actually is
Who pays, and the caveats
Afterthought. Cables and transformers are rated in amperes; work is done in watts; and the power factor is the exchange rate between them. Improving it moves no energy and saves real money, which is a rare combination and worth recognising when a system's cost is dominated by a peak rather than a total.
Dim a lamp for nothing
A lamp needs 60 V across it at 0.50 A — treat it as a 120 Ω resistor — and the only supply is 230 V, 50 Hz. It can be run in series with a resistor, or in series with a choke.
Find the resistor and the choke that each do the job, and the power wasted in each. Then explain why the two answers differ so wildly, and why the choke's voltage plus the lamp's does not add up to 230 V.
Hint 1
In each case the series element has to drop the difference between 230 V and 60 V. In one case that difference is a subtraction, and in the other it is not.
Hint 2
Average power is VI cos φ. What is φ for an ideal inductor?
Answer, without the method
Resistor: 340 Ω, burning 85 W to deliver 30 W — 26 % efficient. Choke: 1.41 H (X = 444 Ω), wasting nothing. Its 222 V is in quadrature with the lamp's 60 V, and √(60² + 222²) = 230 V exactly.
3 routes to it
Both designs, side by side
Why an inductor drops voltage without taking energy
So why is anything ever dimmed with a resistor?
Afterthought. Reactance drops voltage; resistance drops voltage and takes payment. Nearly every efficiency trick in power electronics is some version of replacing the second with the first, and the price is always a component that stores energy — bigger, heavier, or switched faster.
The sail that light can lift
Sunlight at the Earth's orbit delivers 1.36 kW m⁻². The Sun radiates 3.83 × 10²⁶ W and has mass 1.99 × 10³⁰ kg.
Find the radiation pressure on a perfectly reflecting sail at that distance, and the sail area needed to hold up 1.0 kg against the Sun's gravity there. Then show that the answer is the same at every distance, and find the areal density a sail must beat.
Hint 1
An electromagnetic wave carries momentum U/c with its energy U. What happens to that momentum when the light is reflected rather than absorbed?
Hint 2
Write both the light force and the gravitational force on the sail at a distance r and look at what the r's do.
Answer, without the method
P = 2I/c = 9.1 µPa. Balancing gravity on 1.0 kg needs 654 m² — a sail 26 m square. Both forces fall as 1/r², so the balance is distance-independent: a sail wins if its total mass per area is under 1.5 g m⁻², which is thinner than kitchen foil.
3 routes to it
Momentum flux, then the sum
The two 1/r² laws, and a universal number
Where the model bends
Afterthought. Two inverse-square laws in competition produce a criterion with no distance in it. That structure recurs everywhere — a dust grain's radiation-to-gravity ratio decides whether it stays in a solar system at all, and the same cancellation sets the Eddington luminosity of a star. When two forces share a distance dependence, look for the dimensionless number they leave behind.
The field inside a milliwatt
A 1.0 mW helium-neon laser at 633 nm is focused to a spot of diameter 10 µm.
Find the intensity at the focus and the peak electric and magnetic fields there. Compare the electric field with the field that binds an electron in an atom, and find how many photons cross the focus each second.
Hint 1
Intensity is power per area, and for a wave it is also ½ε₀cE₀². Equate them.
Hint 2
The atomic field is the Coulomb field of one proton at one Bohr radius. That is a number you can put together in a line.
Answer, without the method
I = 1.3 × 10⁷ W m⁻², E₀ = √(2I/ε₀c) = 98 kV m⁻¹, B₀ = E₀/c = 0.33 mT. The atomic field is ~5 × 10¹¹ V m⁻¹, so the laser is a ten-millionth of it — which is why light passing through glass behaves linearly. And 3.2 × 10¹⁵ photons pass every second.
3 routes to it
From power to fields
Energy density, and the equal split
Two comparisons that explain modern optics
Afterthought. A milliwatt is a small power and 10⁵ V m⁻¹ is a large field, and the reconciliation is the spot size: intensity is what matters, and focusing buys it for nothing. The same arithmetic run at higher power is why a few joules in a few femtoseconds, focused hard, can exceed the field inside an atom — the frontier of laser physics is just this calculation pushed until the ratio in route three reaches one.
The muon that should never arrive
Cosmic-ray muons are created about 15 km up and travel down at v = 0.995 c. A muon at rest decays with a mean lifetime of 2.2 µs.
Find what fraction of them reach the ground, and what fraction a physicist who had never heard of relativity would predict. Then repeat the calculation as the muon itself would tell it.
Hint 1
Work out how far a muon travels in one lifetime at that speed, and compare it with 15 km, before you invoke anything relativistic.
Hint 2
In the muon's frame the muon is at rest and its lifetime is the ordinary 2.2 µs. Something else has to be different.
Answer, without the method
About 10 % arrive. The non-relativistic prediction is 1.2 × 10⁻¹⁰ — too small by nearly a factor of 10⁹. In the muon's own frame nothing is dilated at all: the atmosphere is only 1.5 km thick, and it crosses that in 2.28 lifetimes, giving the same 10 %.
3 routes to it
The ground's account: a dilated lifetime
The muon's account: a contracted atmosphere
How it was actually measured
Afterthought. The muon flux at sea level is a standing measurement of time dilation that runs day and night, and about one muon crosses your palm every second because of it. Two frames, two mechanisms, one count — and if you ever find yourself applying both dilation and contraction to the same calculation, you have used one frame and a half.
The pole that fits and does not fit
A pole 20 m long is carried at v = 0.866 c (so γ = 2) straight through a barn 10 m long with a door at each end. The farmer plans to shut both doors at the same instant, with the pole entirely inside.
Decide whether the farmer succeeds, and describe the same events from the runner's frame. Find the time between the two door-closings as the runner measures them — and say what happens if the doors are then bolted shut.
Hint 1
The paradox is entirely about the word "same instant". Simultaneity is not a property of two events, it is a property of two events and a frame.
Hint 2
Transform the two door-closing events. In the barn frame they have the same t and differ in x by 10 m — which of those facts survives the transformation?
Answer, without the method
In the barn's frame the pole is contracted to 10 m and fits exactly: both doors can shut at once. In the runner's frame the barn is only 5 m long and the pole cannot fit — but the two closings are not simultaneous there. They are 57.7 ns apart, the far door shutting first and reopening before the near one closes. Bolt them and the pole must buckle: no rod is rigid in relativity.
3 routes to it
Transform the two events
The invariant that settles it
Now bolt the doors
Afterthought. "At the same time" is not a fact about the world; it is a fact about a frame. Nearly every relativistic paradox is assembled by borrowing simultaneity from one frame and lengths from another — so when one appears, list the events, transform them honestly, and check whether the pair you are worrying about is spacelike or timelike.
The last one per cent of c
A payload of mass 1.0 kg is to be accelerated, ideally, with no losses at all.
Find the energy needed to reach 0.50 c, 0.99 c and 0.999 c. Find the speed at which the kinetic energy first equals the rest energy. Then find the speed below which the schoolbook ½mv² is accurate to 1 %.
Hint 1
Kinetic energy is (γ − 1)mc², not ½mv² and not ½γmv². Get γ first and the rest is arithmetic.
Hint 2
For the last part, expand γ in powers of v/c and keep one term more than you think you need.
Answer, without the method
1.4 × 10¹⁶ J, 5.5 × 10¹⁷ J and 1.9 × 10¹⁸ J — the last figure is a few days of the world's entire energy supply, for one kilogram. KE = mc² at γ = 2, i.e. v = 0.866 c. And ½mv² is good to 1 % only below 0.115 c, about 3.5 × 10⁷ m s⁻¹.
3 routes to it
Three speeds, three factors
Where ½mv² goes wrong, and by how much
Momentum, and the ultrarelativistic shortcut
Afterthought. The cost of speed is not quadratic, it is unbounded — and the practical dividing line sits at about a tenth of c, which electrons cross in a few kilovolts and macroscopic objects never approach. Compute β² before choosing a formula, and the whole chapter becomes a question of which regime you are in.
Two photons that make matter
Two identical photons collide and create an electron–positron pair (mec² = 0.511 MeV each).
Find the least energy each photon can have if they meet head-on, and if they meet at right angles. Then explain why a single photon, however energetic, can never do this in empty space — but can beside a heavy nucleus.
Hint 1
Energy conservation alone is not enough — momentum must balance too. Find the one quantity that combines both and is the same in every frame.
Hint 2
For two photons of energies E₁, E₂ meeting at angle θ, work out Etot² − ptot²c². The angle will not go away.
Answer, without the method
Head-on: 0.511 MeV each. At right angles: 0.723 MeV each — a factor √2 more. A lone photon can never do it because its invariant mass is zero in every frame, and the pair's is at least 1.022 MeV/c²; a nucleus fixes that by absorbing momentum, making the threshold 1.022 MeV.
3 routes to it
The invariant mass of two photons
Why one photon cannot, and a nucleus can
Where the two-photon process actually happens
Afterthought. Energy is not the currency of particle creation — invariant mass is. E² − p²c² is the same for everyone, cannot be changed by any interaction, and answers "is this reaction possible?" in one line where energy bookkeeping alone gives the wrong answer with confidence.
Why colliders point beams at each other
Two identical particles of rest mass m each move at 0.60 c, collide head-on and stick together.
Find the mass of the object they make. Then work out the same collision with one particle at rest and the other at 0.60 c, and compare. Finally say how each arrangement scales when you spend more energy on the beam.
Hint 1
Kinetic energy is not conserved when things stick — but two other quantities are, and together they fix the product completely.
Hint 2
Mass is not additive. Compute the total energy and the total momentum first, and get the mass from them at the end.
Answer, without the method
Head-on: the product has mass 2.5 m — 25 % more than the two that made it, and it sits at rest. Fixed target: only 2.12 m, moving off at 0.33 c. Worse, a collider's reach grows as E while a fixed target's grows only as √E.
3 routes to it
Invariant mass, both arrangements
Do it from the centre of momentum
The scaling that decided how accelerators are built
Afterthought. Mass is not additive, and kinetic energy is not lost when things stick — it is re-labelled as mass. Both statements are the same statement, and √s is the quantity that says how much of your expensive beam energy is actually available to make something new.
Two wavelengths are enough to weigh a photon's worth
Light of 400 nm falling on a metal gives a stopping potential of 0.60 V; at 300 nm the stopping potential is 1.63 V.
From these two readings alone find Planck's constant, the work function and the threshold wavelength. Then say what happens to each of the three measured quantities if the lamp is made twice as bright.
Hint 1
Subtract the two Einstein equations. One unknown vanishes and the other falls straight out.
Hint 2
Work in electronvolts and nanometres and use hc = 1240 eV nm; the arithmetic then needs no powers of ten at all.
Answer, without the method
h = 6.6 × 10⁻³⁴ J s, φ = 2.50 eV, threshold 496 nm. Doubling the intensity doubles the current and changes nothing else — same stopping potential, same threshold, same instant response.
3 routes to it
Two equations, one subtraction
Why the slope is the real measurement
The three things intensity does not change
Afterthought. The photoelectric effect does not show that light carries energy — a wave does that. It shows that light delivers energy in indivisible lumps fixed by frequency, and every one of the three "no change with intensity" results is a separate proof of it. Einstein's Nobel citation was for this equation, not for relativity.
Why Compton needed X-rays
X-rays of wavelength 70.9 pm scatter from loosely bound electrons in graphite. The Compton wavelength of the electron is h/mec = 2.43 pm.
Find the scattered wavelength at 90° and at 180°, and the energy the recoiling electron carries away in each case. Then show why the same experiment with green light would have proved nothing.
Hint 1
The shift Δλ is an absolute amount, not a percentage. That single fact answers the last part before you compute anything.
Hint 2
The electron's energy is not a new calculation: it is the photon's loss.
Answer, without the method
At 90°: 73.3 pm, a 3.4 % shift, with 0.57 keV to the electron. At 180°: 75.8 pm, 6.8 %, with 1.1 keV. With 500 nm light the shift would be 5 parts in a million — utterly unmeasurable, which is why the effect went unseen for twenty years.
3 routes to it
The shift, and the electron's share
The Compton wavelength is a fixed length, and that is everything
What it proved that the photoelectric effect had not
Afterthought. Every quantum effect has a natural scale, and it is invisible to any probe much coarser than that scale. Compton's shift is 2.43 pm whatever you shine, so the experiment is a story about ruler choice as much as about photons — and the same reasoning tells you in advance which wavelength, energy or temperature a new quantum effect will need.
Why you do not diffract through a doorway
Compare three de Broglie wavelengths: an electron accelerated through 100 V, a neutron in thermal equilibrium at 300 K, and a 160 g cricket ball at 30 m s⁻¹.
Find all three, say which of the three can be diffracted and by what, and find how slowly the ball would have to move for its wavelength to reach 1 mm.
Hint 1
λ = h/p, so everything turns on momentum. Get p from the accelerating voltage in one case and from kBT in another.
Hint 2
Diffraction needs a wavelength comparable with the spacing of whatever you send it through. Name the spacing before judging the wavelength.
Answer, without the method
Electron 0.123 nm, neutron 0.145 nm — both the size of an atomic spacing, so both diffract off crystals. Ball 1.4 × 10⁻³⁴ m, twenty powers of ten below a nucleus. To reach 1 mm it would have to crawl at 4 × 10⁻³⁰ m s⁻¹ — one atom's width per ten million times the age of the universe.
3 routes to it
Three momenta, three wavelengths
The scaling that explains all three at once
How far up the mass scale has it actually been pushed?
Afterthought. λ = h/p is not a statement about small things; it is a statement about small momenta. Everyday objects have huge momenta in units of h, which is the entire reason the world looks classical — and the way to make anything quantum is always the same, make it light, make it slow, and leave it alone.
Ninety thousand g, from a beam of light
A sodium atom (23 u) absorbs photons of wavelength 589 nm head-on and re-emits them in random directions. It can cycle at most 3 × 10⁷ times a second.
Find the velocity change per photon, the greatest deceleration a laser can give the atom, and the distance needed to stop an atom leaving a 600 K oven at 600 m s⁻¹. Then explain why the re-emitted photons do not undo the work.
Hint 1
A photon's momentum is h/λ. Set that against the atom's mass to get one kick.
Hint 2
The two halves of the cycle are not symmetric. Ask what direction each photon comes from, and what direction each one leaves in.
Answer, without the method
One photon changes its speed by 2.9 cm s⁻¹; at 3 × 10⁷ cycles per second that is 9 × 10⁵ m s⁻² — about 90 000 g. The atom stops in 20 cm, after some 20 000 photons. Absorption is always from one direction, emission is in random ones, so the emissions average to zero and only the absorptions accumulate.
3 routes to it
One kick, then a great many
Why the emissions cancel — and what limit they set
The trick that makes it work at all
Afterthought. Momentum h/λ is a laughably small number and 10⁷ per second is a large one, and their product is 90 000 g. Most of quantum technology has that structure — a negligible single-quantum effect, repeated fast enough to dominate — so when an estimate gives an absurdly tiny per-event answer, ask the rate before dismissing it.
Four hundred kilograms against a million tonnes
Fission of ²³⁵U releases about 200 MeV per nucleus. The fusion ²H + ³H → ⁴He + n releases 17.6 MeV. Burning coal releases about 3 × 10⁷ J per kilogram.
Find the energy per kilogram of each fuel, and the mass of each needed to run a 1 GW station for a year. Then, given that fusion wins per kilogram, say why it is the harder of the two.
Hint 1
Energy per kilogram is energy per reaction divided by the mass that reacts. Count the nucleons carefully — for fusion it is five, not two.
Hint 2
For the last part, work out the electrostatic energy of two nuclei just touching, and compare it with kBT at a fusion plasma's temperature.
Answer, without the method
Fission 8.2 × 10¹³ J kg⁻¹ (384 kg a year), fusion 3.4 × 10¹⁴ J kg⁻¹ (92 kg), coal 3 × 10⁷ J kg⁻¹ (a million tonnes). Fusion is harder because it must first pay a ~290 keV Coulomb barrier, against a fuel temperature worth only about 10 keV — it proceeds by tunnelling.
3 routes to it
Per kilogram, and per year
Why both directions release energy
The barrier, and why tunnelling is the whole story
Afterthought. Energy per kilogram is a poor guide to what gets built. Fusion wins that comparison and loses on the barrier; coal loses it by seven orders of magnitude and powered a century. What decides is usually the activation cost, not the yield — and the barrier is where the engineering lives.
When is the daughter most active?
A pure sample of a parent nuclide of half-life 66 h decays into a daughter of half-life 6.0 h, which decays in turn. (These are the numbers for 99Mo → 99mTc, the workhorse of nuclear medicine.)
Find the time at which the daughter's activity is greatest, and the ratio of the two activities from then on. Then find the one-line condition that holds at the maximum — it makes the calculation almost trivial.
Hint 1
Write down what makes the daughter's population grow and what makes it shrink. The maximum is where those two rates are equal.
Hint 2
"Rate of formation" is the parent's activity and "rate of loss" is the daughter's. Setting them equal is one line.
Answer, without the method
The daughter peaks at t = 22.8 h, and thereafter the two activities settle into a fixed ratio Ad/Ap = λd/(λd − λp) = 1.10. At the maximum the two activities are exactly equal — which gives the answer with no differentiation at all.
3 routes to it
The one-line condition
Transient equilibrium — what happens afterwards
Why a hospital cares
Afterthought. A maximum is where a rate changes sign, and here that rate has a physical name on each side — formation and decay. Setting "in" equal to "out" is quicker than any differentiation, and the same move finds the peak of a charging capacitor's current, a reactor's xenon transient, and the top of any two-stage pipeline.
The alpha that could not classically get out
²²⁶Ra alpha-decays to ²²²Rn with a total released energy of Q = 4.87 MeV. Nuclear radii follow R = 1.2 A1/3 fm, and e²/4πε₀ = 1.44 MeV fm.
Find how the energy is shared between the alpha and the recoiling radon nucleus. Then find the height of the Coulomb barrier the alpha must escape, and say what the comparison implies.
Hint 1
The nucleus was at rest, so the two fragments carry equal and opposite momentum. Write the kinetic energies in terms of that one momentum.
Hint 2
The barrier is just the electrostatic energy of the alpha and the daughter when they are touching. Add their radii.
Answer, without the method
The alpha takes 4.78 MeV and the recoil 0.086 MeV — a share of 222 : 4. The barrier is about 27 MeV high, more than five times the alpha's energy, so the alpha cannot climb out at all: it tunnels, and that is why alpha half-lives span twenty-five orders of magnitude.
3 routes to it
Sharing the energy
The barrier it has no right to cross
Check the numbers
Afterthought. Two lines of ordinary momentum conservation give the energy sharing, and one line of electrostatics shows the decay is classically impossible. Both are worth doing on any decay before reaching for a table — the first because the light fragment always takes nearly everything, and the second because a barrier five times the escape energy is what makes half-lives span the age of the universe.
How many collisions to tame a neutron?
A fission neutron leaves with about 2.0 MeV and must be slowed to thermal energy, 0.025 eV, by elastic collisions with nuclei of mass number A.
Find the largest fraction of its energy a neutron can lose in one collision with hydrogen, deuterium and carbon-12. Then find how many head-on collisions with carbon are needed, and say why heavy water is a better moderator than either ordinary water or graphite.
Hint 1
A head-on elastic collision between masses 1 and A is the one-dimensional problem you already know. Get the neutron's outgoing speed first, then square it.
Hint 2
Slowing from 2 MeV to 0.025 eV is a factor of 10⁸. Multiplying by 0.716 repeatedly means taking a logarithm.
Answer, without the method
Head-on, the neutron keeps a fraction ((A−1)/(A+1))²: hydrogen 0 % kept (all of it lost in one hit), deuterium 1/9, carbon 0.716. Carbon therefore needs about 55 head-on collisions — and about 115 real ones. Heavy water wins because deuterium is nearly as light as hydrogen but, unlike hydrogen, almost never absorbs the neutron it has slowed.
3 routes to it
One collision, then a logarithm
Do it in the centre-of-mass frame instead
Why the best moderator is not the lightest
Afterthought. Equal masses transfer everything and unequal masses transfer almost nothing, which is why moderators are made of the lightest available nuclei — and why the best one is the second lightest. A figure of merit almost always beats a single quantity: fast slowing is useless if the moderator eats what it slows.
The same atom, with a heavier electron
Ordinary hydrogen has a ground state at −13.6 eV and a Bohr radius of 52.9 pm. Consider two relatives: positronium, an electron bound to a positron of equal mass, and muonic hydrogen, a proton with a muon (207 me) in place of its electron.
Find the ground-state energy and the orbit size of each, and the wavelength of each one's first Lyman line. Then say what single quantity in the Bohr formula accounts for all of it.
Hint 1
The nucleus is not nailed down. Both particles orbit their common centre of mass, and the two-body problem becomes a one-body problem with a corrected mass.
Hint 2
Once you have µ/me for each system, you need no new physics — only two proportionalities.
Answer, without the method
Positronium: −6.80 eV, radius 106 pm, Lyman-α at 243 nm. Muonic hydrogen: −2.53 keV, radius 256 fm, Lyman-α at 0.65 nm — an X-ray. Everything follows from the reduced mass: E ∝ µ and r ∝ 1/µ.
3 routes to it
Reduced mass, then two proportionalities
The scaling, and what else it predicts
Why anyone builds these
Afterthought. One symbol in the Bohr formula does all this work. Wherever a two-body bound state appears — an exciton in a semiconductor, a quarkonium state, a deuteron — the first move is to write down the reduced mass, and the second is to ask how the answer scales with it. Most "new" systems are the old one with µ and Z changed.
The atom that broadcasts on radio
In the Bohr model the electron in level n circles at a definite frequency, and a jump from n to n − 1 emits a photon of its own frequency. There is no reason for the two to be equal — except in one limit.
Find both frequencies for n = 2 and for n = 100, and the wavelength emitted at n = 100. Then show that the two frequencies must agree as n grows, and say why that agreement matters.
Hint 1
The orbital frequency is v/2πr, and Bohr gives you v ∝ 1/n and r ∝ n². Combine them before putting numbers in.
Hint 2
For the emitted frequency, expand 1/(n−1)² − 1/n² for large n rather than subtracting two nearly equal numbers.
Answer, without the method
At n = 2 the orbital frequency is 8.2 × 10¹⁴ Hz and the emitted one 2.5 × 10¹⁵ Hz — a factor of 3 apart. At n = 100 both are 6.6 × 10⁹ Hz, agreeing to 1.5 %: a 4.6 cm radio wave from a single atom. Both go as 1/n³, which is Bohr's correspondence principle.
3 routes to it
Both frequencies, as powers of n
Why the agreement is compulsory
These atoms are real, and they are enormous
Afterthought. A new theory must contain the old one as a limit, and here you can watch it happen in one algebraic step — 1/n³ from two entirely different calculations. Whenever a quantum result looks arbitrary, take its large-quantum-number limit; if it does not become the classical answer, the result is wrong.
Two spectra from one X-ray tube
An X-ray tube with a molybdenum target (Z = 42, K-shell binding energy 20.0 keV) is run at 30 kV, and then at 15 kV.
Find the short-wavelength cutoff at each voltage and the wavelength of the Kα line. Then say which features are present at each voltage, and which of them would change if the target were replaced by tungsten.
Hint 1
The shortest wavelength corresponds to an electron giving up all its energy in one go. Nothing about the target enters that statement.
Hint 2
A characteristic line needs a vacancy first. Ask what it costs to make one, not what the line's energy is.
Answer, without the method
Cutoff 41.3 pm at 30 kV and 82.7 pm at 15 kV — set by the voltage alone and the same for any target. Kα is at 72 pm (17.2 keV). At 30 kV both the continuum and the sharp lines appear; at 15 kV there are no lines at all, because 15 keV cannot eject a 20 keV K electron. Change the target and the lines move; the cutoff does not.
3 routes to it
The cutoff, and why it is target-blind
The lines need a vacancy first
Change the target, and what Moseley did with it
Afterthought. One spectrum, two independent physics stories: a continuum that knows only the accelerating voltage, and lines that know only the atom. Whenever a measured spectrum has both smooth and sharp features, ask which apparatus parameter moves each — the ones that move with the source belong to the source, and the ones that do not are telling you about the sample.
Sixty millivolts a decade
A silicon diode obeys I = I₀(eeV/kBT − 1). At 300 K, kBT/e = 25.9 mV.
Find the extra forward voltage needed to multiply the current by ten, and the current ratio for a 100 mV increase. Then find how the forward drop changes with temperature at constant current, and say why "a diode drops 0.7 V" is a useful lie.
Hint 1
Take logarithms. The exponential means every equal step in voltage multiplies the current by the same factor.
Hint 2
For the temperature part, do not differentiate the diode equation naively — I₀ itself depends on temperature, and far more strongly than the exponential does.
Answer, without the method
A decade of current costs 59.6 mV — the famous 60 mV per decade. A 100 mV increase multiplies the current by 48. At fixed current the drop falls by about 2 mV for every kelvin, so "0.7 V" is true only near room temperature at one particular current.
3 routes to it
The log slope
Why the drop falls as it warms
What the equation cannot do
Afterthought. An exponential device has no characteristic voltage — only a characteristic voltage per decade, and here it is kBT/e ln 10, a pure thermal quantity. Whenever a component is described by a fixed drop, ask over what range of current, and expect the true answer to be a slope on a log plot.
One carrier in five million million
Pure silicon at 300 K has ni = 1.0 × 10¹⁶ m⁻³ free carriers of each sign, from an atom density of 5 × 10²⁸ m⁻³. Its band gap is 1.1 eV.
Find how many silicon atoms share one conduction electron. Then dope it with one donor per million atoms and find the new electron and hole concentrations. Finally find the temperature rise that doubles ni, and say why germanium lost.
Hint 1
Doping does not change the product np. That single conservation-like statement gives you the minority carriers in one step.
Hint 2
For the doubling temperature, differentiate ln ni with respect to T rather than evaluating ni twice.
Answer, without the method
One carrier per 5 × 10¹² atoms — purer than any chemistry can express. One donor per million gives n = 5 × 10²² m⁻³, five million times more electrons, while holes fall to 2 × 10⁹ m⁻³. ni doubles every 9 K; germanium's 0.67 eV gap doubles every 5.5 K and leaks a thousand times more, which is why silicon won.
3 routes to it
Mass action: doping one carrier pushes the other down
The exponential in temperature
Where the doping stops working
Afterthought. np = ni² is the most useful line in semiconductor physics: doping does not create carriers so much as trade one kind for the other, at a fixed product. And every temperature limit in electronics — leakage, thermal runaway, the death of a junction — is the same exponential in Eg/2kBT seen from a different side.