Final Practice Exam: AP Physics 1
A full practice exam covering all eight units, in the same shape as the real one. Sit it in one go if you can, with the timer running, then mark it and use the breakdown to see which units to review.
Before you start
| Section | What | Time | Weight |
|---|---|---|---|
| Section I | 42 multiple choice, one correct answer each (A to D), 5 from each unit plus a sixth from Units 2 and 3 | 85 minutes | 50% |
| Section II | 4 free response: Mathematical Routines, Translation Between Representations, Experimental Design and Analysis, Qualitative/Quantitative Translation | 95 minutes | 50% |
- A calculator is allowed. Keep the equation sheet open, just like in the real exam.
- Use g = 9.8 m/s² unless a question says otherwise. Ignore air resistance unless told.
- Section I: click a letter to choose it. You can change your mind until you press Submit Section I. There is no penalty for guessing, so answer every question.
- Section II: write on paper (or in the boxes), showing your reasoning. Then open each scoring guide and give yourself points honestly.
- Each question has a hidden worked solution. Try not to open them until you have finished the section.
Your Section I choices are saved in this browser only, so you can come back later.
Section I: Multiple choice (42 questions, 85 minutes)
Choose the one best answer for each question.
1. A spring with k = 200 N/m is squeezed 0.10 m and used to launch a 0.50 kg block along a frictionless floor. What is the block's speed after it leaves the spring?
Answer and worked solution
Answer: D. 2.0 m/s
Spring energy becomes kinetic energy: ½kx² = ½mv². ½(200)(0.10)² = 1.0 J. Then 1.0 J = ½(0.50)v², v² = 4.0, v = 2.0 m/s.
4.0 m/s forgets the square root.
Review: Topic 3.3 Potential Energy (Unit 3: Work, Energy, and Power).
2. A 4.0 kg box is pulled across a level floor by a 20 N horizontal rope. The coefficient of kinetic friction is 0.25. What is the box's acceleration?
Answer and worked solution
Answer: B. 2.6 m/s²
Normal force N = mg = (4.0)(9.8) = 39.2 N. Friction f = μN = 0.25 × 39.2 = 9.8 N.
Net force = 20 − 9.8 = 10.2 N, so a = 10.2 / 4.0 = 2.55 ≈ 2.6 m/s².
5.0 m/s² ignores friction. 7.5 m/s² adds friction instead of subtracting it.
Review: Topic 2.7 Kinetic and Static Friction (Unit 2: Force and Translational Dynamics).
3. A wheel starts from rest and has a constant angular acceleration of 3.0 rad/s². Through what angle does it turn in the first 4.0 s?
Answer and worked solution
Answer: A. 24 rad
Δθ = ω₀t + ½αt² = 0 + ½(3.0)(4.0)² = ½(3.0)(16) = 24 rad.
12 rad is ω = αt = 12 rad/s read as an angle. 48 rad forgets the ½.
Review: Topic 5.1 Rotational Kinematics (Unit 5: Torque and Rotational Dynamics).
4. A large open water tank has a small hole in its side 0.80 m below the water surface. About how fast does water leave the hole?
Answer and worked solution
Answer: A. 4.0 m/s
Bernoulli (Torricelli): both the surface and the jet are at air pressure, and the surface barely moves, so ρgh = ½ρv². v = √(2gh) = √(2 × 9.8 × 0.80) = √15.7 = 3.96 ≈ 4.0 m/s.
2.8 m/s forgets the 2. 16 m/s forgets the square root.
Review: Topic 8.4 Fluids and Conservation Laws (Unit 8: Fluids).
5. Water flows through a horizontal pipe. The cross-sectional area drops from 4.0 cm² to 1.0 cm². In the wide part the speed is 2.0 m/s. What is the speed in the narrow part?
Answer and worked solution
Answer: B. 8.0 m/s
Continuity: A₁v₁ = A₂v₂. (4.0)(2.0) = (1.0)v₂, so v₂ = 8.0 m/s. The same volume each second must squeeze through a smaller hole, so it goes faster. (Area units cancel, so cm² is fine.)
4.0 m/s uses the ratio of diameters instead of areas. 16 m/s squares the ratio.
Review: Topic 8.4 Fluids and Conservation Laws (Unit 8: Fluids).
6. A 2.0 kg box slides across a level floor at 6.0 m/s. The coefficient of kinetic friction is 0.30. How far does the box slide before it stops?
Answer and worked solution
Answer: B. 6.1 m
Work-energy theorem: the friction work removes all the kinetic energy. K = ½(2.0)(6.0)² = 36 J. Friction f = μmg = (0.30)(2.0)(9.8) = 5.88 N.
fd = K, so d = 36/5.88 = 6.1 m. (The mass cancels: d = v²/(2μg).)
12 m forgets the ½ in K. 1.8 m leaves out μ. 60 m leaves out g.
Review: Topic 3.2 Work (Unit 3: Work, Energy, and Power).
7. A 6.0 kg block and a 4.0 kg block sit side by side, touching, on a frictionless floor. A horizontal 20 N push is applied to the 6.0 kg block, so both blocks speed up together. What force does the 6.0 kg block exert on the 4.0 kg block?
Answer and worked solution
Answer: B. 8.0 N
Treat both blocks as one system: a = F/(m₁ + m₂) = 20/(6.0 + 4.0) = 2.0 m/s².
Now the 4.0 kg block alone: the only horizontal force on it is the push from the 6.0 kg block, so F = (4.0)(2.0) = 8.0 N.
20 N assumes the whole push is passed on. 12 N is the net force on the 6.0 kg block (20 − 8.0). 2.0 N is the acceleration, not a force.
Review: Topic 2.5 Newton's Second Law (Unit 2: Force and Translational Dynamics).
8. A block slides down a rough ramp at a constant speed. Consider the block-Earth system. Which statement is correct?
Answer and worked solution
Answer: A. The kinetic energy stays the same, and the mechanical energy (K + U) decreases.
Constant speed means K is constant and the net work is zero. The block loses height, so Ug drops. K + U drops, and the lost energy becomes thermal energy from friction. Gravity does positive work; friction does an equal amount of negative work.
Review: Topic 3.4 Conservation of Energy (Unit 3: Work, Energy, and Power).
9. A skateboarder starts from rest at the top of a smooth (frictionless) ramp 5.0 m high. What is her speed at the bottom?
Answer and worked solution
Answer: B. 9.9 m/s
Energy is conserved: mgh = ½mv², so v = √(2gh) = √(2 × 9.8 × 5.0) = √98 = 9.9 m/s. The mass cancels.
7.0 m/s = √(gh) forgets the 2. 98 and 49 forget the square root.
Review: Topic 3.4 Conservation of Energy (Unit 3: Work, Energy, and Power).
10. A 2.0 kg block on a spring (k = 200 N/m) oscillates with amplitude 0.10 m on a frictionless surface. What is its maximum speed?
Answer and worked solution
Answer: A. 1.0 m/s
All the energy is spring energy at the ends and kinetic energy at the middle: ½kA² = ½mvmax². ½(200)(0.10)² = 1.0 J = ½(2.0)v², so v² = 1.0 and vmax = 1.0 m/s.
Same answer from vmax = A√(k/m) = 0.10 × 10 = 1.0 m/s.
Review: Topic 7.4 Energy of Simple Harmonic Oscillators (Unit 7: Oscillations).
11. What is the absolute pressure 2.0 m below the surface of a swimming pool? (Water density 1000 kg/m³, air pressure at the surface 1.01 × 10⁵ Pa.)
Answer and worked solution
Answer: A. 1.2 × 10⁵ Pa
The water adds ρgh = (1000)(9.8)(2.0) = 19,600 Pa ≈ 2.0 × 10⁴ Pa (gauge pressure). Absolute pressure = 1.01 × 10⁵ + 0.196 × 10⁵ = 1.21 × 10⁵ ≈ 1.2 × 10⁵ Pa.
2.0 × 10⁴ Pa is only the gauge pressure. 8.1 × 10⁴ Pa subtracts instead of adds.
Review: Topic 8.2 Pressure (Unit 8: Fluids).
12. A uniform seesaw is balanced at its centre.
A 30 kg child sits 2.0 m to the left of the pivot. Where must a 40 kg child sit so the seesaw stays balanced and level?
Answer and worked solution
Answer: A. 1.5 m to the right of the pivot
For balance the clockwise and counterclockwise torques are equal: (30 kg)(g)(2.0 m) = (40 kg)(g)d. g cancels: d = 60/40 = 1.5 m, on the right side.
The heavier child sits closer to the pivot. 2.7 m puts the masses the wrong way round.
Review: Topic 5.5 Rotational Equilibrium and Newton's First Law in Rotational Form (Unit 5: Torque and Rotational Dynamics).
13. A 1200 kg car drives around a flat curve of radius 50 m at a steady 15 m/s. How big is the friction force from the road that keeps it on the curve?
Answer and worked solution
Answer: B. 5400 N
Friction supplies the centripetal force: F = mv²/r = (1200)(15²)/50 = (1200)(225)/50 = 5400 N, pointing to the centre of the curve.
11760 N is the car's weight. 18000 N is m × v, which is not a force.
Review: Topic 2.9 Circular Motion (Unit 2: Force and Translational Dynamics).
14. A car starts from rest and speeds up with a constant acceleration of 2.5 m/s² for 6.0 s. How far does it travel in that time?
Answer and worked solution
Answer: D. 45 m
Use d = v₀t + ½at² with v₀ = 0: d = ½(2.5 m/s²)(6.0 s)² = ½(2.5)(36) = 45 m.
90 m forgets the ½. 15 m is v = at = 15 m/s read as a distance. 30 m mixes up the formula.
Review: Topic 1.2 Displacement, Velocity, and Acceleration (Unit 1: Kinematics).
15. Two clay balls collide and stick together. No outside forces act on the two-ball system. Which is true for the system?
Answer and worked solution
Answer: C. Momentum is conserved, but kinetic energy is not.
With no net outside force, total momentum is always conserved. Sticking together is a perfectly inelastic collision: some kinetic energy turns into thermal energy and the work of changing the balls' shape.
Review: Topic 4.4 Elastic and Inelastic Collisions (Unit 4: Linear Momentum).
16. The graph shows the position of an object in simple harmonic motion. Point P is at t = 0.50 s.
At point P, which describes the object's velocity?
Answer and worked solution
Answer: A. Largest speed, moving in the negative direction
Velocity is the slope of the x-t graph. At P the object passes through x = 0 (equilibrium), where the graph is steepest, so the speed is largest. The graph is going down there, so the velocity is negative.
Its value: vmax = 2πA/T = 2π(0.20 m)/(2.0 s) = 0.63 m/s, so v = −0.63 m/s.
Review: Topic 7.3 Representing and Analyzing SHM (Unit 7: Oscillations).
17. A child pulls a sled 10 m along level snow with a 30 N force on a rope that makes a 60° angle above the horizontal. How much work does the rope do on the sled?
Answer and worked solution
Answer: C. 150 J
W = Fd cos θ = (30 N)(10 m)cos 60° = (30)(10)(0.50) = 150 J. Only the part of the force along the motion does work.
300 J uses the full force. 260 J uses sin 60° instead of cos 60°.
Review: Topic 3.2 Work (Unit 3: Work, Energy, and Power).
18. A 60 kg skater and a 40 kg skater stand still on smooth ice and push off each other. The 60 kg skater moves away at 2.0 m/s. How fast does the 40 kg skater move?
Answer and worked solution
Answer: D. 3.0 m/s
Total momentum starts at zero and stays zero: (60)(2.0) = (40)v, so v = 120/40 = 3.0 m/s in the opposite direction.
1.3 m/s puts the masses the wrong way round.
Review: Topic 4.3 Conservation of Linear Momentum (Unit 4: Linear Momentum).
19. A block of wood with density 600 kg/m³ floats at rest in fresh water (1000 kg/m³). What fraction of the block's volume is under water?
Answer and worked solution
Answer: A. 60%
Floating: buoyant force = weight. ρwaterVunderg = ρwoodVtotalg, so Vunder/Vtotal = 600/1000 = 0.60 = 60%.
40% is the part above the water.
Review: Topic 8.1 Internal Structure and Density (Unit 8: Fluids).
20. A simple pendulum has a period T. Its length is made 4 times longer. (Small swings.) What is the new period?
Answer and worked solution
Answer: C. 2T
T = 2π√(L/g). Multiply L by 4: the square root of 4 is 2, so the period doubles to 2T.
4T forgets the square root. A longer pendulum always swings more slowly, never faster.
Review: Topic 7.2 Frequency and Period of SHM (Unit 7: Oscillations).
21. A 0.20 kg ball at rest is kicked. The graph shows the net force on the ball during the kick.
What is the ball's speed after the kick?
Answer and worked solution
Answer: C. 20 m/s
Impulse = area under the F-t graph = ½ × base × height = ½(0.020 s)(400 N) = 4.0 N·s.
Impulse = change in momentum: 4.0 = (0.20)v, so v = 20 m/s.
40 m/s forgets the ½ for the triangle. Watch the units: 20 ms = 0.020 s.
Review: Topic 4.2 Change in Momentum and Impulse (Unit 4: Linear Momentum).
22. The skater in a spin pulls her arms in and spins faster. What happens to her rotational kinetic energy, and why?
Answer and worked solution
Answer: C. It increases, because her muscles do positive work pulling her arms in.
K = L²/(2I). L stays the same while I drops, so K goes up. The extra energy comes from the work her muscles do pulling her arms inward (internal energy turns into kinetic energy). Conserving angular momentum does not mean conserving kinetic energy.
Review: Topic 6.4 Conservation of Angular Momentum (Unit 6: Energy and Momentum of Rotating Systems).
23. A solid disk (I = ½MR²) of mass 2.0 kg and radius 0.30 m spins at 10 rad/s about its centre. What is its rotational kinetic energy?
Answer and worked solution
Answer: B. 4.5 J
I = ½(2.0)(0.30)² = 0.090 kg·m². K = ½Iω² = ½(0.090)(10)² = 4.5 J.
9.0 J uses I = MR².
Review: Topic 6.1 Rotational Kinetic Energy (Unit 6: Energy and Momentum of Rotating Systems).
24. A 2.0 kg cart moving at 3.0 m/s hits a 1.0 kg cart at rest. They stick together. What is their speed just after the collision?
Answer and worked solution
Answer: C. 2.0 m/s
Momentum is conserved: (2.0)(3.0) = (2.0 + 1.0)v, so 6.0 = 3.0v and v = 2.0 m/s.
Review: Topic 4.4 Elastic and Inelastic Collisions (Unit 4: Linear Momentum).
25. A net torque of 2.0 N·m acts on a disk with rotational inertia 0.50 kg·m². The disk starts from rest. What is its angular speed after 3.0 s?
Answer and worked solution
Answer: B. 12 rad/s
α = τ/I = 2.0/0.50 = 4.0 rad/s². Then ω = αt = (4.0)(3.0) = 12 rad/s.
4.0 rad/s is α, not ω.
Review: Topic 5.6 Newton's Second Law in Rotational Form (Unit 5: Torque and Rotational Dynamics).
26. A 60 kg person stands on a bathroom scale in an elevator. The elevator speeds up while moving upward with an acceleration of 2.0 m/s². What does the scale read?
Answer and worked solution
Answer: D. 708 N
Up is positive. Forces on the person: normal force N up, weight mg down. N − mg = ma, so N = m(g + a) = 60(9.8 + 2.0) = 708 N. The scale reads N.
588 N is the weight (no acceleration). 468 N would be the reading when accelerating downward.
Review: Topic 2.5 Newton's Second Law (Unit 2: Force and Translational Dynamics).
27. A solid sphere, a solid cylinder and a thin hoop are released from rest at the same height on the same ramp. Each rolls without slipping. Which reaches the bottom first?
Answer and worked solution
Answer: B. The solid sphere
Each starts with the same Mgh per kilogram. Rolling splits that energy into translational and rotational kinetic energy. The less rotational inertia per MR², the more energy goes into moving forward. Sphere: I = (2/5)MR² (smallest), cylinder ½MR², hoop MR² (largest). So the sphere is fastest at every point, and the hoop is last. Mass and radius do not matter.
Review: Topic 6.5 Rolling (Unit 6: Energy and Momentum of Rotating Systems).
28. A spinning ice skater has rotational inertia 4.0 kg·m² and spins at 2.0 rev/s. She pulls her arms in, and her rotational inertia drops to 1.6 kg·m². Ignore friction. What is her new spin rate?
Answer and worked solution
Answer: D. 5.0 rev/s
No outside torque, so angular momentum is conserved: I₁ω₁ = I₂ω₂, so (4.0)(2.0) = (1.6)ω₂ and ω₂ = 8.0/1.6 = 5.0 rev/s. (Rev/s works here because the same units are on both sides.)
0.80 rev/s puts the inertias the wrong way round: a smaller I must mean faster spin.
Review: Topic 6.4 Conservation of Angular Momentum (Unit 6: Energy and Momentum of Rotating Systems).
29. A constant torque of 5.0 N·m turns a wheel (I = 0.20 kg·m²) from rest through an angle of 8.0 rad. What is the wheel's final angular speed?
Answer and worked solution
Answer: C. 20 rad/s
Work done by the torque: W = τΔθ = (5.0)(8.0) = 40 J. This becomes rotational KE: 40 = ½(0.20)ω², so ω² = 400 and ω = 20 rad/s.
40 rad/s confuses work (J) with angular speed. 14 rad/s forgets the ½ in ½Iω². 200 rad/s forgets both the ½ and the square root (W/I).
Review: Topic 6.2 Torque and Work (Unit 6: Energy and Momentum of Rotating Systems).
30. A ball rolls horizontally off the edge of a cliff 19.6 m high with a speed of 6.0 m/s. Ignore air resistance. How far from the base of the cliff does it land?
Answer and worked solution
Answer: C. 12 m
Vertical: starts with vy = 0, so 19.6 m = ½(9.8 m/s²)t², t² = 4.0 s², t = 2.0 s.
Horizontal: constant speed, x = (6.0 m/s)(2.0 s) = 12 m.
The horizontal speed does not change the fall time. 24 m would need 4 s in the air.
Review: Topic 1.5 Vectors and Motion in Two Dimensions (Unit 1: Kinematics).
31. The graph shows the velocity of a cyclist moving along a straight road (positive = east).
What is the cyclist's displacement from t = 0 to t = 6 s?
Answer and worked solution
Answer: C. 32 m east
Displacement = area under the v-t graph. Triangle from 0 to 4 s: ½(4 s)(8 m/s) = 16 m. Rectangle from 4 to 6 s: (2 s)(8 m/s) = 16 m. Total 32 m east.
48 m treats the whole 6 s as 8 m/s. 8 m is the final velocity, not a displacement.
Review: Topic 1.3 Representing Motion (Unit 1: Kinematics).
32. An electric motor lifts a 50 kg crate straight up at a constant speed of 2.0 m/s. What power does the motor deliver to the crate?
Answer and worked solution
Answer: B. 980 W
Constant speed, so the motor's force equals the weight: F = mg = 50 × 9.8 = 490 N. Power P = Fv = 490 × 2.0 = 980 W.
490 W forgets to multiply by the speed. 100 W is just m × v.
Review: Topic 3.5 Power (Unit 3: Work, Energy, and Power).
33. A 0.50 kg block on a spring with k = 50 N/m oscillates on a frictionless surface. What is the period?
Answer and worked solution
Answer: A. 0.63 s
T = 2π√(m/k) = 2π√(0.50/50) = 2π√(0.010) = 2π(0.10) = 0.63 s.
0.10 s forgets the 2π. 1.6 s is the frequency in Hz (1/0.63), not the period.
Review: Topic 7.2 Frequency and Period of SHM (Unit 7: Oscillations).
34. A metal block with weight 50 N and volume 0.0020 m³ hangs from a spring scale, fully under water (1000 kg/m³). What does the scale read?
Answer and worked solution
Answer: D. 30 N
Buoyant force = weight of water pushed aside = ρVg = (1000)(0.0020)(9.8) = 19.6 N up.
The block is at rest: T + 19.6 = 50, so T = 30.4 ≈ 30 N.
20 N is the buoyant force itself. 70 N adds the buoyant force instead of subtracting it.
Review: Topic 8.3 Fluids and Newton's Laws (Unit 8: Fluids).
35. A 50 N force is applied at the end of a 0.30 m wrench. The force makes a 30° angle with the wrench handle. What torque does it produce about the bolt?
Answer and worked solution
Answer: D. 7.5 N·m
τ = rF sin θ, where θ is the angle between the wrench (r) and the force: τ = (0.30)(50)sin 30° = (0.30)(50)(0.50) = 7.5 N·m.
15 N·m assumes the force is perpendicular. 13 N·m uses cos 30°.
Review: Topic 5.3 Torque (Unit 5: Torque and Rotational Dynamics).
36. A heavy truck hits a small parked car. During the collision, how does the force of the truck on the car compare with the force of the car on the truck?
Answer and worked solution
Answer: B. They are equal in size and opposite in direction.
Newton's third law: the two forces are a pair. They always have equal size and opposite direction, whatever the masses or speeds. The car is damaged more and speeds up more because it has less mass (a = F/m), not because it feels a bigger force.
Review: Topic 2.3 Newton's Third Law (Unit 2: Force and Translational Dynamics).
37. A block on a horizontal spring oscillates on a frictionless surface. At the moment it reaches its maximum displacement from equilibrium, which is true?
Answer and worked solution
Answer: D. Its speed is zero and its acceleration has its largest size, pointing toward equilibrium.
At the turning point the block stops for an instant (v = 0). The spring is stretched (or squeezed) the most, so the restoring force F = −kx is largest, and so is a = −kx/m, pointing back toward equilibrium.
Review: Topic 7.1 Defining Simple Harmonic Motion (Unit 7: Oscillations).
38. A boat points straight across an 80 m wide river and moves at 4.0 m/s relative to the water. The river flows at 3.0 m/s. How long does the boat take to cross?
Answer and worked solution
Answer: A. 20 s
Only the across-the-river part of the velocity gets the boat across. That part is 4.0 m/s (the current is along the river, perpendicular). t = 80 m / 4.0 m/s = 20 s.
16 s uses the 5.0 m/s speed relative to the shore, but that velocity is at an angle, so the boat travels farther than 80 m. The current only carries the boat downstream (3.0 m/s × 20 s = 60 m).
Review: Topic 1.4 Reference Frames and Relative Motion (Unit 1: Kinematics).
39. A ball is thrown straight up. Ignore air resistance. At the very top of its flight, which is true?
Answer and worked solution
Answer: A. Its velocity is zero and its acceleration is 9.8 m/s² downward.
At the top the ball stops for an instant, so v = 0. Gravity still acts, so the acceleration is still g = 9.8 m/s² downward. If the acceleration were zero there, the ball would stay at the top forever.
Review: Topic 1.2 Displacement, Velocity, and Acceleration (Unit 1: Kinematics).
40. A solid cylinder and a hollow cylinder (thin-walled tube) have the same mass and the same radius. The same torque is applied to each about its central axis. Which has the greater angular acceleration?
Answer and worked solution
Answer: D. The solid cylinder, because its rotational inertia is smaller.
α = τ/I. The solid cylinder has mass spread from the axis outward, I = ½MR². The hollow one has all its mass at the rim, I = MR². Smaller I means larger α for the same torque, so the solid cylinder wins.
Review: Topic 5.4 Rotational Inertia (Unit 5: Torque and Rotational Dynamics).
41. A planet has twice the mass of Earth and twice Earth's radius. What is the gravitational field strength (g) at its surface?
Answer and worked solution
Answer: C. 4.9 N/kg
g = GM/R². Doubling M doubles g; doubling R divides g by 2² = 4. Overall g changes by 2/4 = ½, so g = ½(9.8) = 4.9 N/kg.
Review: Topic 2.6 Gravitational Force (Unit 2: Force and Translational Dynamics).
42. A 0.50 kg ball moving right at 4.0 m/s bounces off a wall and moves left at 3.0 m/s. What is the size of its change in momentum?
Answer and worked solution
Answer: D. 3.5 kg·m/s
Right is positive. pbefore = (0.50)(+4.0) = +2.0 kg·m/s. pafter = (0.50)(−3.0) = −1.5 kg·m/s. Δp = −1.5 − 2.0 = −3.5 kg·m/s, so the size is 3.5 kg·m/s.
0.50 kg·m/s forgets that velocity changed direction (sign).
Review: Topic 4.2 Change in Momentum and Impulse (Unit 4: Linear Momentum).
Section I results
Each question now shows ✓ or ✗. Open its worked solution: it links to the topic page to review.
Section II: Free response (4 questions, 95 minutes)
Show your work. Where a question asks you to justify, use physics principles in words, not just equations. Start each answer with the principle you are using. Aim for about 24 minutes per question.
Question 1: Mathematical Routines (10 points)
A block A of mass 0.40 kg slides to the right at 3.0 m/s across a horizontal frictionless surface. It hits block B, of mass 0.60 kg, and the two blocks stick together. Block B is at rest at the free end of a horizontal spring (spring constant 250 N/m) whose other end is fixed to a wall on the right. Before the collision, the spring is at its natural length. Take right as positive.
- Calculate the speed of the two blocks just after the collision. (2 points)
- Calculate the maximum compression of the spring. (2 points)
- Calculate the time from the collision until the spring first reaches maximum compression. (2 points)
- Calculate what fraction of block A's original kinetic energy is changed into other forms (such as thermal energy) during the collision. (2 points)
- Block B is replaced with a block of larger mass. Block A again hits it at 3.0 m/s and sticks. Will the maximum compression of the spring be greater than, less than, or the same as in part (b)? Justify your answer. (2 points)
Scoring guide and worked answer
1 point for using conservation of momentum: mAvA = (mA + mB)v.
1 point for the answer: (0.40)(3.0) = (1.0)v, so v = 1.2 m/s.1 point for using energy conservation after the collision: ½(mA+mB)v² = ½kx²max (energy is not conserved during the collision, but it is afterwards, since the surface is frictionless).
1 point for the answer: ½(1.0)(1.2)² = 0.72 J = ½(250)x², x² = 0.00576 m², x = 0.076 m (7.6 cm).1 point for recognising the time is a quarter of a period (from equilibrium to the turning point) and using T = 2π√(m/k) with the combined mass 1.0 kg.
1 point for the answer: T = 2π√(1.0/250) = 2π(0.0632) = 0.397 s, so t = T/4 = 0.099 s.1 point for both kinetic energies: before ½(0.40)(3.0)² = 1.8 J; after ½(1.0)(1.2)² = 0.72 J.
1 point for the fraction: (1.8 − 0.72)/1.8 = 1.08/1.8 = 0.60 (60%).1 point for "less than".
1 point for a correct justification: the momentum mAvA is the same, so the combined speed v = mAvA/(mA+M) is smaller. The kinetic energy right after the collision is p²/(2(mA+M)), which is smaller for larger M. Less kinetic energy to store means a smaller ½kx², so a smaller compression.
Question 2: Translation Between Representations (12 points)
A light string is wrapped around a pulley that is a uniform solid disk of mass M and radius R (rotational inertia ½MR²). The pulley turns on a frictionless horizontal axle. A block of mass m hangs from the free end of the string. The block is released from rest. The string does not slip and its mass is negligible.
- Describe the free-body diagram of the block: list each force, its direction, and say which force is larger. Then list the forces on the pulley that exert a torque about the axle. (3 points)
- Starting with Newton's second law for the block and the rotational form of Newton's second law for the pulley, derive an expression for the acceleration a of the block in terms of m, M and g. (3 points)
- For m = 2.0 kg, M = 4.0 kg and R = 0.10 m, calculate the acceleration of the block and the tension in the string. (2 points)
- Which statements best describe the graphs, for the time the block is falling? Choose one for each: the block's speed vs. time is (i) a horizontal line, (ii) a straight line through the origin with positive slope, (iii) a curve that gets steeper; the pulley's angular momentum vs. time is (i), (ii) or (iii). Explain using your answer to (b). (2 points)
- After the block has fallen 1.0 m, fill in the energy table for the block-pulley-Earth system, using the values in (c). (2 points)
Quantity Value after 1.0 m (J) Decrease in gravitational potential energy ? Translational kinetic energy of the block ? Rotational kinetic energy of the pulley ?
Scoring guide and worked answer
1 point: block has weight mg straight down and tension T straight up, and no other forces.
1 point: mg is larger than T (the block speeds up downward), so the weight arrow is longer.
1 point: on the pulley, only the string tension T (pulling down at the rim, distance R from the axle) makes a torque. The pulley's weight and the force from the axle act at the axle, so their torque is zero.1 point for the block (down positive): mg − T = ma.
1 point for the pulley: TR = Iα = (½MR²)(a/R), so T = ½Ma (using a = αR because the string does not slip).
1 point for combining: mg − ½Ma = ma, so a = mg / (m + M/2) = 2mg/(2m + M).Check: if M → 0, a → g (free fall), and a very heavy pulley gives a → 0. That makes sense.
1 point: a = (2.0)(9.8)/(2.0 + 2.0) = 19.6/4.0 = 4.9 m/s².
1 point: T = ½Ma = ½(4.0)(4.9) = 9.8 N (check: mg − T = 19.6 − 9.8 = 9.8 N = ma = 2.0 × 4.9 ✓).1 point: both are (ii), straight lines through the origin with positive slope.
1 point: from (b), a is constant, so v = at grows at a steady rate. The pulley's angular speed ω = v/R also grows steadily, and L = Iω, so L grows steadily too (constant net torque TR means constant rate of change of L).Speed after 1.0 m: v² = 2ad = 2(4.9)(1.0) = 9.8 m²/s², v = 3.13 m/s.
Decrease in Ug = mgh = (2.0)(9.8)(1.0) = 19.6 J. Block K = ½(2.0)(9.8) = 9.8 J. Pulley: I = ½(4.0)(0.10)² = 0.020 kg·m², ω = v/R = 31.3 rad/s, K = ½(0.020)(980) = 9.8 J.
1 point for the Ug and block values; 1 point for the pulley value, consistent with 19.6 J = 9.8 J + 9.8 J (energy conserved).
Question 3: Experimental Design and Analysis (10 points)
A student wants to find the spring constant k of a spring by timing oscillations. She has the spring, a stand to hang it from, a set of slotted masses (0.10 kg each), a stopwatch and a metre stick.
- Describe a procedure she could use to collect data to find k from oscillation times. Say what she measures, how she measures it so the results are accurate, and what she changes. (3 points)
- Her data are below. Which quantities should she plot on the axes to get a straight line whose slope lets her find k? Explain why it is straight. (2 points)
Hanging mass m (kg) 0.10 0.20 0.30 0.40 0.50 Period T (s) 0.45 0.62 0.78 0.88 1.00 - Use the data to calculate k. Show the values you plot and how you get the slope. (3 points)
- A classmate says, "If you pull the mass down farther before letting go, the period will be longer, because it has farther to go." Is the classmate correct? Explain, and describe a quick test of the claim using the same equipment. (2 points)
Scoring guide and worked answer
1 point: hang a known mass, pull it down a small distance and release it; time the oscillations with the stopwatch.
1 point: time many oscillations (for example 10) and divide by 10, and/or repeat and average, to reduce reaction-time error. Start timing as the mass passes a marked point (equilibrium is best, where it moves fastest).
1 point: repeat for several different masses (at least 5), keeping the amplitude small.1 point: plot T² on the vertical axis and m on the horizontal axis.
1 point: since T = 2π√(m/k), squaring gives T² = (4π²/k)m. That is y = (slope)x with slope 4π²/k, a straight line through the origin.T² values: 0.20, 0.38, 0.61, 0.77, 1.00 s².
1 point for the T² values (or a correct linearised graph).
1 point for the slope from a best-fit line (not just two data points): roughly (1.00 − 0.20)/(0.50 − 0.10) = 0.80/0.40 = 2.0 s²/kg; a least-squares fit gives 1.99 s²/kg.
1 point for k = 4π²/slope = 39.5/2.0 ≈ 20 N/m (anything from about 19 to 21 N/m is fine).1 point: not correct. For a mass on a spring the period T = 2π√(m/k) does not depend on amplitude. A bigger pull means a bigger restoring force, so the mass also moves faster; the extra distance and extra speed cancel.
1 point: test it by keeping the mass the same (for example 0.30 kg), timing 10 oscillations at two or three different amplitudes (say 2 cm, 4 cm, 6 cm), and comparing: the periods should agree within the timing uncertainty.
Question 4: Qualitative/Quantitative Translation (8 points)
A solid wooden block floats at rest in a tank of fresh water (density 1000 kg/m³). The same block is then moved to a tank of oil with density 800 kg/m³, where it also floats at rest. The wood has density 600 kg/m³.
- Without using equations, explain whether the block floats with more, less, or the same volume under the surface in the oil compared with the water. (3 points)
- Derive an expression for the fraction of the block's volume that is under the surface, f = Vunder/Vblock, in terms of the density of the block ρb and the density of the liquid ρL. Start from a fundamental principle. (2 points)
- Explain how your expression in (b) agrees with your reasoning in (a). (1 point)
- Calculate the fraction under the surface in each liquid. (2 points)
Scoring guide and worked answer
1 point: in both liquids the block floats at rest, so the buoyant force equals the block's weight, which does not change. The buoyant force is the same in both.
1 point: the buoyant force equals the weight of liquid pushed aside. Oil is less dense, so each cubic metre of oil weighs less.
1 point: to push aside the same weight of a lighter liquid, the block must push aside more volume, so more of the block is under the surface in oil (it floats lower).1 point for starting from equilibrium (Newton's first law): FB = mg, with FB = ρLVunderg and mg = ρbVblockg.
1 point for f = Vunder/Vblock = ρb/ρL (g cancels).1 point: ρL is in the denominator, so a smaller liquid density (oil) gives a larger fraction under the surface, matching (a).
1 point water: f = 600/1000 = 0.60.
1 point oil: f = 600/800 = 0.75.
Your total
Section I is worth 50% and Section II 50%, like the real exam. Submit Section I and enter your free response points above to see your combined percentage.
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This is a practice percentage, not an official AP score. Use it to track progress from one attempt to the next.
Review by unit
Every unit has topic pages with sims, a games page and a practice set: Unit 1 · Unit 2 · Unit 3 · Unit 4 · Unit 5 · Unit 6 · Unit 7 · Unit 8. Also see the equation sheet explained.