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AP Physics 1 mock exam Free full practice test

Written by STEM-Tutor's tutors · Updated October 2026 · 48 questions and 4 free-response questions · about 180 minutes

A full-length practice test built from our exam-style questions on every unit of AP Physics 1. Each answer is marked as you go, and every question has a worked solution to open afterwards.

On this page
  1. Before you start
  2. Section I
  3. Section II
  4. After the mock

Before you start

Section I: 48 exam-style questions

Q1Exam-style
A stone is dropped from rest from a bridge and hits the water 2.5 s later. Ignoring air resistance, how high is the bridge above the water?
m

Hint

Dropped from rest means v0 = 0. Which equation links distance, time and acceleration?

Worked solution
  1. Take downward as positive, so a = g = 9.8 m/s² and v0 = 0.
  2. Δy = v0t + ½gt2 = 0 + ½(9.8)(2.5)2.
  3. Δy = 4.9 × 6.25 = 30.6 m.
Q2Exam-style
A ball rolls off a horizontal table 1.2 m high at 3.0 m/s. How far from the foot of the table does it land?
m

Hint

Treat the vertical and horizontal motions separately. The vertical motion tells you the time in the air.

Worked solution
  1. Vertically the ball starts with zero velocity and falls 1.2 m: 1.2 = ½(9.8)t2.
  2. t = √(2 × 1.2 / 9.8) = 0.495 s.
  3. Horizontally the velocity stays at 3.0 m/s, so Δx = 3.0 × 0.495.
  4. Δx = 1.48 m.
Q3Exam-style
A projectile is launched from level ground at 20 m/s, 30° above the horizontal. How long is it in the air?

Hint

Find the vertical component of the launch velocity first. The time up equals the time down.

Worked solution
  1. Vertical component: vy0 = 20 sin 30° = 10 m/s.
  2. Time to the top, where vy = 0: t = 10 / 9.8 = 1.02 s.
  3. The flight is symmetric on level ground, so the total time is 2 × 1.02 = 2.04 s, answer B.
  4. 1.02 s is only the time to the top, a common trap.
Q4Exam-style
The position-time graph of a cart is a straight line sloping downward. Which describes the cart's motion?

Hint

On a position-time graph, the slope is the velocity. What does a constant, negative slope mean?

Worked solution
  1. The slope of a position-time graph is the velocity.
  2. A straight line has a constant slope, so the velocity is constant and the acceleration is zero.
  3. The slope is negative, so the cart moves in the negative direction.
  4. The answer is B.
Q5Challenge
Car A starts from rest and accelerates at 2.0 m/s² at the moment car B passes it at a constant 12 m/s. How long does car A take to catch car B?
s

Hint

Write an expression for each car's position against time, then set them equal.

Worked solution
  1. Car A: xA = ½(2.0)t2 = t2.
  2. Car B: xB = 12t.
  3. Set them equal: t2 = 12t, so t(t − 12) = 0.
  4. t = 0 is the start, so A catches B at t = 12 s.
Q6Challenge
A ball is thrown straight up at 15 m/s from 2.0 m above the ground. How long is it in the air before it hits the ground?
s

Hint

Take up as positive and the launch point as y = 0. Where is the ground?

Worked solution
  1. Take up as positive, so a = −9.8 m/s², and the ground is at y = −2.0 m.
  2. −2.0 = 15t − 4.9t2, so 4.9t2 − 15t − 2.0 = 0.
  3. Quadratic formula: t = [15 + √(225 + 39.2)] / 9.8 = (15 + 16.25) / 9.8.
  4. The positive root is t = 3.19 s.
Q7Exam-style
A 2.0 kg block slides down a frictionless ramp inclined at 30° to the horizontal. What is its acceleration?
30°mgFNfriction
Free-body diagram for a block on a 30° slope. Weight mg acts straight down, the normal force FN acts at right angles to the slope, and friction acts up the slope if the block tends to slide down. Resolve weight into mg sinθ along the slope and mg cosθ into it.
m/s²

Hint

Resolve the weight into components along and perpendicular to the ramp.

Worked solution
  1. Along the ramp the only force is the component of weight, mg sinθ.
  2. a = mg sinθ / m = g sinθ, so the mass cancels.
  3. a = 9.8 × sin 30° = 4.9 m/s², down the ramp.
Q8Exam-style
A 3.0 kg mass and a 5.0 kg mass hang on either side of a light string over a frictionless pulley. What is the size of their acceleration?
m/s²

Hint

Treat both masses and the string as one system. What is the net force that drives it?

Worked solution
  1. The net force on the system is the difference in weights: (5.0 − 3.0) × 9.8 = 19.6 N.
  2. The total mass being accelerated is 8.0 kg.
  3. a = 19.6 / 8.0 = 2.45 m/s².
Q9Exam-style
For the two masses in the previous question (3.0 kg and 5.0 kg over a pulley), what is the tension in the string?
N

Hint

Apply Newton's second law to the 3.0 kg mass alone. Which way does it accelerate?

Worked solution
  1. The 3.0 kg mass accelerates upward at 2.45 m/s².
  2. Newton's second law for it: T − mg = ma, so T = m(g + a).
  3. T = 3.0 × (9.8 + 2.45) = 36.8 N. Check: it lies between the two weights, 29.4 N and 49 N, as it must.
Q10Exam-style
A 1200 kg car drives round a flat bend of radius 50 m at a steady 15 m/s. What friction force must the road supply?
N

Hint

On a flat bend, friction is the only horizontal force, so it provides all of the centripetal force.

Worked solution
  1. The net force must point to the centre and equal mv2/r.
  2. F = 1200 × 152 / 50 = 1200 × 225 / 50.
  3. F = 5400 N, towards the centre of the bend.
Q11Challenge
A 0.50 kg ball on a 0.80 m string is swung in a vertical circle. At the lowest point its speed is 4.0 m/s. What is the tension in the string there?
N

Hint

At the lowest point, tension acts up and weight acts down. The net force must point up, towards the centre.

Worked solution
  1. Net force towards the centre (upward): T − mg = mv2/r.
  2. mv2/r = 0.50 × 16 / 0.80 = 10 N, and mg = 4.9 N.
  3. T = 4.9 + 10 = 14.9 N.
Q12Exam-style
The distance between two planets doubles. What happens to the gravitational force between them?

Hint

The force depends on 1/r2.

Worked solution
  1. Fg = Gm1m2/r2, an inverse-square law.
  2. Doubling r multiplies r2 by 4, so the force is divided by 4.
  3. The answer is B, one quarter.
Q13Exam-style
A roller-coaster car starts from rest at a height of 25 m. Ignoring friction, how fast is it moving when it is 10 m above the ground?
m/s

Hint

Only the drop in height matters: energy lost as potential energy becomes kinetic energy.

Worked solution
  1. Conservation of energy: mgΔh = ½mv2, and the mass cancels.
  2. The car has dropped Δh = 25 − 10 = 15 m.
  3. v = √(2 × 9.8 × 15) = √294.
  4. v = 17.1 m/s.
Q14Exam-style
A spring with k = 400 N/m is compressed 0.10 m and launches a 0.050 kg ball along a frictionless surface. What is the ball's speed when it leaves the spring?
m/s

Hint

The spring's elastic energy becomes the ball's kinetic energy.

Worked solution
  1. Elastic energy: ½kΔx2 = ½ × 400 × 0.102 = 2.0 J.
  2. Set 2.0 = ½ × 0.050 × v2, so v2 = 80.
  3. v = 8.94 m/s.
Q15Exam-style
A 4.0 kg box slides across a level floor. It starts at 5.0 m/s and stops after 3.0 m. What is the coefficient of kinetic friction?

Hint

The work done by friction removes all the box's kinetic energy. Does the mass matter?

Worked solution
  1. Work-energy theorem: μmg d = ½mv2, so the mass cancels.
  2. μ = v2 / (2gd) = 25 / (2 × 9.8 × 3.0) = 25 / 58.8.
  3. μ = 0.425, about 0.43 (no units).
Q16Exam-style
Two identical balls are thrown from the top of a cliff at the same speed, one straight up and one straight down. Ignoring air resistance, how do their speeds compare when they hit the ground?

Hint

Compare their energies at the top of the cliff and at the ground.

Worked solution
  1. Both start with the same kinetic energy (same speed) and the same potential energy (same height).
  2. Both lose the same potential energy falling to the ground.
  3. So both arrive with the same kinetic energy, and the same speed. The answer is C. Only their times differ.
Q17Exam-style
A 60 kg skier starts from rest at the top of a slope 40 m high and reaches the bottom at 20 m/s. How much energy was transferred by friction and air resistance?
J

Hint

Compare the potential energy at the top with the kinetic energy at the bottom.

Worked solution
  1. Potential energy lost: mgh = 60 × 9.8 × 40 = 23,520 J.
  2. Kinetic energy gained: ½ × 60 × 202 = 12,000 J.
  3. Energy transferred by friction and air resistance: 23,520 − 12,000 = 11,520 J, about 11.5 kJ.
Q18Exam-style
The speed of a car triples. By what factor does its kinetic energy change?

Hint

Kinetic energy depends on the square of the speed.

Worked solution
  1. K = ½mv2, so K is proportional to v2.
  2. Tripling v multiplies K by 32 = 9.
  3. The answer is C. This is why stopping distances grow so quickly with speed.
Q19Exam-style
A 2.0 kg cart moving at 3.0 m/s collides with a 1.0 kg cart at rest, and they stick together. What is their speed afterwards?
m/s

Hint

Total momentum before equals total momentum after.

Worked solution
  1. Momentum before: 2.0 × 3.0 + 0 = 6.0 kg·m/s.
  2. Afterwards they move together with mass 3.0 kg: 6.0 = 3.0v.
  3. v = 2.0 m/s.
Q20Exam-style
In the previous collision (2.0 kg at 3.0 m/s sticking to 1.0 kg at rest), how much kinetic energy is lost?
J

Hint

Work out the total kinetic energy before and after.

Worked solution
  1. Before: ½ × 2.0 × 3.02 = 9.0 J.
  2. After: ½ × 3.0 × 2.02 = 6.0 J.
  3. Lost: 3.0 J, a third of the original kinetic energy.
Q21Exam-style
Two skaters stand at rest on ice. The 50 kg skater pushes the 70 kg skater, and moves off at 1.4 m/s. How fast does the 70 kg skater move?
m/s

Hint

The total momentum starts at zero, so it must stay zero.

Worked solution
  1. 0 = 50 × 1.4 + 70v, taking the 50 kg skater's direction as positive.
  2. v = −70 / 70 = −1.0 m/s.
  3. The 70 kg skater moves at 1.0 m/s in the opposite direction.
Q22Exam-style
An egg dropped onto a pillow survives, but one dropped from the same height onto a hard floor breaks. Why?

Hint

Both eggs stop from the same speed. What does that tell you about the impulse?

Worked solution
  1. Both eggs go from the same speed to rest, so the change in momentum, and the impulse, are the same.
  2. Impulse = FΔt. The pillow increases Δt, so the average force F is smaller.
  3. The answer is B, the same idea as crumple zones and airbags.
Q23Exam-style
A 0.40 kg ball at rest is kicked. The graph shows the force on it against time. What is the ball's speed as it leaves the foot?
t (s)F (N)60000.0200.040
Force on the ball against time during the kick.
m/s

Hint

The area under a force-time graph is the impulse.

Worked solution
  1. Impulse = area of the triangle = ½ × 0.040 × 600 = 12 N·s.
  2. Impulse = Δp = mv − 0, so 12 = 0.40v.
  3. v = 30 m/s.
Q24Challenge
A 1.0 kg cart moving at 4.0 m/s has an elastic, head-on collision with a 3.0 kg cart at rest. What is the 1.0 kg cart's velocity afterwards? Take its original direction as positive.
m/s

Hint

Write the momentum equation, and use the fact that in a head-on elastic collision the speed of approach equals the speed of separation.

Worked solution
  1. Momentum: 1.0(4.0) = 1.0v1 + 3.0v2.
  2. Elastic, head-on: v2 − v1 = 4.0, so v2 = v1 + 4.0.
  3. Substitute: 4.0 = v1 + 3v1 + 12, so 4v1 = −8.0.
  4. v1 = −2.0 m/s: the light cart bounces back. Check: v2 = 2.0 m/s, and kinetic energy 8 J before = 2 J + 6 J after.
Q25Exam-style
A 30 kg child sits 2.0 m from the pivot of a seesaw. How far from the pivot, on the other side, must a 40 kg child sit to balance it?
m

Hint

For balance, the clockwise and anticlockwise torques about the pivot are equal.

Worked solution
  1. Torque balance: 30g × 2.0 = 40g × d, so g cancels.
  2. d = 60 / 40 = 1.5 m.
Q26Exam-style
A uniform 4.0 m beam of mass 20 kg rests on supports at each end. A 50 kg person stands 1.0 m from the left end. What force does the left support exert?
N

Hint

Take torques about the right-hand support, so its force has no torque.

Worked solution
  1. About the right support: the beam's weight acts 2.0 m away and the person 3.0 m away.
  2. Balance: FL × 4.0 = (20 × 9.8)(2.0) + (50 × 9.8)(3.0) = 392 + 1470.
  3. FL = 1862 / 4.0 = 466 N. Check: the right support gives 686 − 466 = 220 N, and the person is nearer the left, so the left force is larger, as expected.
Q27Exam-style
A net torque of 6.0 N·m acts on a disc with rotational inertia 0.50 kg·m². What is its angular acceleration?
rad/s²

Hint

Use Newton's second law for rotation.

Worked solution
  1. α = Στ / I = 6.0 / 0.50.
  2. α = 12 rad/s².
Q28Exam-style
A string is wrapped round a pulley that is a uniform disc of mass 2.0 kg and radius 0.10 m (I = ½MR2). The string is pulled with a steady 4.0 N. What is the pulley's angular acceleration?
rad/s²

Hint

Find the rotational inertia and the torque separately.

Worked solution
  1. I = ½ × 2.0 × 0.102 = 0.010 kg·m².
  2. The string pulls tangentially at the rim: τ = 0.10 × 4.0 = 0.40 N·m.
  3. α = 0.40 / 0.010 = 40 rad/s².
Q29Exam-style
A wheel spins at a constant rate. How does the linear speed of a point on the rim compare with that of a point halfway between the axle and the rim?

Hint

Every point on the wheel has the same angular velocity ω.

Worked solution
  1. v = rω, and ω is the same for every point on the rigid wheel.
  2. The rim point is twice as far from the axle, so its linear speed is twice as large.
  3. The answer is B.
Q30Challenge
A 1.0 kg block hangs from a light string wrapped round a pulley: a uniform disc of mass 2.0 kg and radius 0.10 m on a frictionless axle. The block is released. What is its acceleration?
m/s²

Hint

Write Newton's second law for the block and for the pulley, and link them with a = Rα.

Worked solution
  1. Block: mg − T = ma.
  2. Pulley: TR = Iα = (½MR2)(a/R), so T = ½Ma = 1.0a.
  3. Add the equations: mg = (m + ½M)a, so 9.8 = 2.0a.
  4. a = 4.9 m/s², less than g because some of the energy goes into turning the pulley.
Q31Exam-style
A skater spins at 2.0 revolutions per second with rotational inertia 3.0 kg·m². She pulls her arms in, reducing it to 1.2 kg·m². What is her new rate of spin, in revolutions per second?
rev/s

Hint

Angular momentum is conserved. You can keep the rates in revolutions per second, since the units cancel.

Worked solution
  1. I1ω1 = I2ω2.
  2. 3.0 × 2.0 = 1.2 × ω2.
  3. ω2 = 5.0 rev/s.
Q32Exam-style
A solid cylinder (I = ½MR2) rolls without slipping from rest down a slope, dropping a height of 1.5 m. What is its speed at the bottom?
m/s

Hint

Share the potential energy between translational and rotational kinetic energy, and use v = Rω.

Worked solution
  1. Energy: Mgh = ½Mv2 + ½Iω2.
  2. With I = ½MR2 and ω = v/R, the rotational term is ¼Mv2.
  3. So gh = ¾v2, giving v = √(4gh/3) = √(4 × 9.8 × 1.5 / 3) = √19.6.
  4. v = 4.43 m/s.
Q33Exam-style
A 3.0 kg hoop (I = MR2) rolls without slipping at 2.0 m/s. What is its total kinetic energy?
J

Hint

Add the translational and rotational kinetic energies.

Worked solution
  1. Translational: ½Mv2 = ½ × 3.0 × 4.0 = 6.0 J.
  2. Rotational: ½(MR2)(v/R)2 = ½Mv2 = 6.0 J.
  3. Total: 12 J. For a hoop, the two parts are always equal.
Q34Challenge
A 0.10 kg lump of clay moving at 5.0 m/s hits the rim of a disc at rest, tangentially, and sticks. The disc has radius 0.50 m and rotational inertia 0.40 kg·m², and turns freely about its centre. What is the angular velocity afterwards?
rad/s

Hint

Angular momentum about the axle is conserved. Find the clay's angular momentum before it hits.

Worked solution
  1. Clay's angular momentum about the centre: L = mvr = 0.10 × 5.0 × 0.50 = 0.25 kg·m²/s.
  2. Afterwards the clay is part of the system: I = 0.40 + 0.10 × 0.502 = 0.425 kg·m².
  3. ω = L / I = 0.25 / 0.425 = 0.59 rad/s.
Q35Exam-style
A constant torque of 5.0 N·m turns a wheel (I = 0.25 kg·m²) from rest through 10 rad. What is its final angular velocity?
rad/s

Hint

The work done by a torque is τΔθ.

Worked solution
  1. Work done: W = τΔθ = 5.0 × 10 = 50 J.
  2. This becomes rotational kinetic energy: 50 = ½ × 0.25 × ω2.
  3. ω2 = 400, so ω = 20 rad/s.
Q36Exam-style
Sand is sprinkled onto a turntable spinning freely with no external torque, and sticks to it. What happens to the turntable's angular momentum and rotational kinetic energy?

Hint

Write K in terms of L: K = L2 / 2I.

Worked solution
  1. With no external torque, L stays constant.
  2. The sand increases I, so ω falls.
  3. K = L2/2I: with L fixed and I larger, K decreases. The lost energy becomes internal energy as the sand sticks.
  4. The answer is B.
Q37Exam-style
The mass on a spring oscillator is doubled. What happens to the period?

Hint

T depends on the square root of m.

Worked solution
  1. T = 2π√(m/k), so T is proportional to √m.
  2. Doubling m multiplies T by √2, about 1.41.
  3. The answer is B.
Q38Exam-style
A 0.50 kg mass on a spring (k = 50 N/m) oscillates with amplitude 0.20 m. What is its maximum speed?
m/s

Hint

At equilibrium, all of the spring energy ½kA2 has become kinetic energy.

Worked solution
  1. ½kA2 = ½mvmax2, so vmax = A√(k/m).
  2. vmax = 0.20 × √(50 / 0.50) = 0.20 × 10.
  3. vmax = 2.0 m/s.
Q39Exam-style
For the oscillator in the previous question (k = 50 N/m, amplitude 0.20 m), what is its total mechanical energy?
J

Hint

At the amplitude the mass is momentarily at rest, so all the energy is stored in the spring.

Worked solution
  1. E = ½kA2 = ½ × 50 × 0.202.
  2. E = 1.0 J. Check: ½ × 0.50 × 2.02 = 1.0 J too.
Q40Exam-style
A pendulum clock is taken to the Moon, where g is about one sixth of its value on Earth. How does its period change?

Hint

T depends on 1/√g.

Worked solution
  1. T = 2π√(L/g), so T is proportional to 1/√g.
  2. Dividing g by 6 multiplies T by √6, about 2.45.
  3. The answer is B: the clock runs slow.
Q41Exam-style
What length of simple pendulum has a period of 1.5 s?
m

Hint

Rearrange T = 2π√(L/g) for L.

Worked solution
  1. Square both sides: T2 = 4π2L/g, so L = gT2 / 4π2.
  2. L = 9.8 × 1.52 / 39.48 = 22.05 / 39.48.
  3. L = 0.559 m, about 56 cm.
Q42Challenge
Hanging a mass on a vertical spring stretches it by 4.9 cm. The mass is then pulled down and released. What is the period of its oscillation?
s

Hint

You do not need the mass or k separately: the stretch tells you m/k.

Worked solution
  1. At equilibrium kx = mg, so m/k = x/g = 0.049 / 9.8 = 0.0050 s².
  2. T = 2π√(m/k) = 2π√0.0050.
  3. T = 0.444 s.
Q43Exam-style
An object of volume 0.0020 m³ is held completely under water (density 1000 kg/m³). What is the buoyant force on it?
N

Hint

The buoyant force equals the weight of the water displaced.

Worked solution
  1. Fb = ρwaterVg = 1000 × 0.0020 × 9.8.
  2. Fb = 19.6 N, whatever the object is made of.
Q44Exam-style
A block of wood with density 600 kg/m³ floats in water (1000 kg/m³). What fraction of its volume is under the water? Give your answer as a decimal.

Hint

Set the buoyant force equal to the weight, and write both in terms of densities and volumes.

Worked solution
  1. Floating: ρwaterVunderg = ρwoodVtotalg.
  2. Vunder / Vtotal = ρwood / ρwater = 600 / 1000.
  3. Fraction under water = 0.60.
Q45Exam-style
Water flows through a horizontal pipe. Its speed is 2.0 m/s at point 1 and 6.0 m/s at point 2. By how much is the pressure at point 1 greater than at point 2?
Pa

Hint

In a horizontal pipe the ρgy terms cancel in Bernoulli's equation.

Worked solution
  1. P1 + ½ρv12 = P2 + ½ρv22.
  2. P1 − P2 = ½ × 1000 × (6.02 − 2.02) = 500 × 32.
  3. P1 − P2 = 16,000 Pa. Faster flow, lower pressure.
Q46Exam-style
Water leaks from a small hole in the side of a large open tank, 1.8 m below the water surface. How fast does the water leave the hole?
m/s

Hint

Apply Bernoulli's equation between the surface and the hole. Both are at atmospheric pressure, and the surface barely moves.

Worked solution
  1. The pressure terms cancel and the surface speed is about zero, so ρgh = ½ρv2.
  2. v = √(2gh) = √(2 × 9.8 × 1.8) = √35.28.
  3. v = 5.94 m/s, the same speed as an object falling 1.8 m.
Q47Exam-style
A ship sails from a river (fresh water) into the sea (denser salt water). What happens?

Hint

The buoyant force must still equal the ship's weight. How much denser water must it displace for that?

Worked solution
  1. Floating, the buoyant force equals the weight in both cases, so the buoyant force does not change.
  2. Fb = ρVg: with larger ρ, a smaller volume V is needed.
  3. The ship displaces less water, so it rises: B.
Q48Challenge
A rock weighs 30 N in air and 20 N when hung fully under water (1000 kg/m³). What is the density of the rock?
kg/m³

Hint

The loss in weight is the buoyant force, which tells you the rock's volume.

Worked solution
  1. Buoyant force = 30 − 20 = 10 N = ρwaterVg.
  2. The rock's weight is ρrockVg = 30 N.
  3. Divide: ρrock / ρwater = 30 / 10 = 3.0.
  4. ρrock = 3000 kg/m³.

Section II: 4 free-response questions

Free-response question 1: Kinematics

Mathematical routines style · 6 points · about 20 minutes

A student throws a ball horizontally at 8.0 m/s from the top of a building of height H = 20 m. Air resistance is negligible.

u (horizontal)a = g, downhorizontal: constant velocitylands
A projectile launched horizontally: constant horizontal velocity, constant downward acceleration g. Time links the two directions.
  1. (a) Derive an expression for the time t the ball takes to reach the ground, in terms of H and physical constants.
  2. (b) Calculate the horizontal distance from the base of the building to where the ball lands.
  3. (c) A second ball is thrown horizontally from the same point at 16 m/s. Compare its time in the air and its landing distance with the first ball's. Justify your answers.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

Vertically the ball starts from rest: H = ½gt2, so t = √(2H/g).

Part (b)

t = √(2 × 20 / 9.8) = 2.02 s. Horizontal velocity is constant, so d = 8.0 × 2.02 = 16.2 m.

Part (c)

The time is the same, because the vertical motion (start at rest vertically, fall H with acceleration g) does not depend on the horizontal velocity. The distance doubles, because d = vxt and vx doubles while t stays the same.

Your mark: 0 of 6

Free-response question 2: Work, energy and power

Qualitative/quantitative translation style · 6 points · about 22 minutes

A block of mass m is released from rest at height h on a frictionless curved ramp. At the bottom it slides onto a rough horizontal surface with coefficient of kinetic friction μ, and comes to rest after sliding a distance d along it.

  1. (a) Derive an expression for d in terms of h and μ.
  2. (b) A student claims that a heavier block released from the same height would slide further, because it has more energy at the bottom. Explain whether the student is correct.
  3. (c) The block is released from height 2h instead. By what factor does d change? Justify your answer using part (a).

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

The potential energy mgh is all removed by the work done by friction, μmg d. So mgh = μmg d, which gives d = h / μ.

Part (b)

The student is wrong. A heavier block does have more kinetic energy at the bottom, but the friction force μmg is larger by the same factor. The mass cancels, so d = h/μ is the same for any mass.

Part (c)

d = h/μ is directly proportional to h, so doubling the height doubles the stopping distance.

Your mark: 0 of 6

Free-response question 3: Energy and momentum of rotating systems

Qualitative/quantitative translation style · 6 points · about 20 minutes

Disc A spins at angular velocity ω0 on a frictionless axle. An identical disc B, not spinning, is dropped onto it on the same axle. Friction between the discs makes them spin together.

  1. (a) Derive an expression for the final angular velocity of the two discs, in terms of ω0.
  2. (b) Find the fraction of the original kinetic energy that remains after the discs spin together.
  3. (c) Explain why angular momentum is conserved in this process but kinetic energy is not.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

No external torque about the axle, so angular momentum is conserved: Iω0 = (2I)ω, so ω = ω0/2.

Part (b)

Before: ½Iω02. After: ½(2I)(ω0/2)2 = ¼Iω02. So half the kinetic energy remains.

Part (c)

The friction torques between the discs are internal: they are equal and opposite, so the total angular momentum is unchanged. But the discs slide over each other while friction acts, and that converts kinetic energy into internal (thermal) energy.

Your mark: 0 of 6

Free-response question 4: Fluids

Translation between representations style · 5 points · about 20 minutes

A block hangs from a spring scale. The scale reads W in air. The block is then lowered slowly into a beaker of water until it is completely under water, and then lowered further without touching the bottom.

depth of bottom facescale readingWW′h (fully under)
Scale reading against depth: it falls steadily while the block goes under, then stays constant.
  1. (a) Sketch, or describe, a graph of the scale reading against the depth of the block's bottom face below the water surface.
  2. (b) Derive an expression for the density of the block, in terms of W, the fully submerged reading W′, and the density of water ρw.
  3. (c) W = 12 N and W′ = 7.5 N. Calculate the density of the block.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

The reading starts at W. As the block goes under, the buoyant force grows in proportion to the volume under water, so the reading falls in a straight line. Once the block is fully under water, the displaced volume stops changing, so the reading stays constant at a lower value W′.

Part (b)

Buoyant force when fully submerged: W − W′ = ρwVg. Weight: W = ρVg. Dividing gives ρ = ρwW / (W − W′).

Part (c)

ρ = 1000 × 12 / 4.5 = 2670 kg/m³.

Your mark: 0 of 5

After the mock

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