AP Physics C: Mechanics 2026 Practice Exam 2 | & 4
FRQs with Complete Answer Key, Rationales and
Step-by-Step Solution
1. A particle moves along the x-axis with acceleration a(t) = 6t m/s^2. At t=0, the particle is
at rest at the origin. What is its position at t=2 s?
A. 4 m
B. 6 m
C. 8 m
D. 10 m
Answer: C
Rationale: Integrate a(t) to get v(t)=3t^2, then x(t)=t^3. At t=2, x=8 m. Options A and B come
from misintegration, and D from using the wrong power.
2. A 2 kg block on a rough horizontal surface (k=0.4) is connected by a massless string over
a frictionless pulley to a 1 kg hanging block. The system is released from rest. What is the
acceleration of the blocks? (g=9.8 m/s^2)
A. 0.67 m/s^2
B. 1.33 m/s^2
C. 2.0 m/s^2
D. 3.2 m/s^2
Answer: A
Rationale: Using Newton's second law: m2g - T = m2a and T - ¼k m1g = m1a. Solving gives a =
(m2 - k m1)g/(m1+m2) = (1-0.8)*9.8/3 0.67 m/s^2. Other options omit friction or use incorrect
mass ratios.
3. A particle moves along the x-axis under the influence of a force F(x) = 3x^2 + 2 N. What
is the work done by this force as the particle moves from x=0 to x=2 m?
A. 10 J
B. 12 J
C. 14 J
D. 16 J
Answer: B
Rationale: Work = "+F dx = "+(3x^2+2) dx from 0 to 2 = [x^3+2x]_0^2 = 8+4=12 J. Option D
results from using average force (8 N) times distance (2 m), which is incorrect for a
non-constant force.
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,4. A 2 kg ball moving east at 4 m/s collides with a 1 kg ball moving north at 6 m/s on a
frictionless surface. If they stick together, what is the magnitude of their final velocity?
A. 2.0 m/s
B. 3.3 m/s
C. 4.0 m/s
D. 5.0 m/s
Answer: B
Rationale: Conservation of momentum: initial px=8, py=6, total mass=3, so vx=8/3, vy=2.
Magnitude = sqrt((8/3)^2+2^2)=10/33.33 m/s. Option A is average speed, C and D from
miscalculations.
5. A wheel starts from rest and rotates with constant angular acceleration. It makes 10
revolutions in the first 2 seconds. What is its angular acceleration?
A. 5 rad/s^2
B. 10 rad/s^2
C. 5 rad/s^2
D. 10 rad/s^2
Answer: B
Rationale: ¸ = ½ ± t^2, with ¸ = 10 rev = 20À rad, t=2 s !’ ± = 2¸/t^2 = 40À/4 = 10À rad/s^2.
Option A results from forgetting 2 per revolution; C and D omit .
6. A uniform rod of mass M and length L is pivoted at one end. A horizontal force F is
applied perpendicular to the rod at the free end. What is the angular acceleration?
A. 3F/(ML)
B. 2F/(ML)
C. F/(ML)
D. 4F/(ML)
Answer: A
Rationale: Torque = FL, moment of inertia about end I = (1/3)ML^2. Then ± = Ä/I = 3F/(ML).
Options B and C use incorrect moments of inertia (e.g., about center or point mass).
7. A merry-go-round (a solid disk of mass M and radius R) rotates at angular speed . A
child of mass m jumps radially onto the edge. What is the new angular speed?
A. M/(M+2m)
B. M/(M+m)
C. (M+2m)/M
D. (M+m)/M
Answer: A
Rationale: Conservation of angular momentum: I_i É = I_f É_f, with I_i=½ MR^2, I_f=½ MR^2 +
mR^2. Thus _f = (½ M)/(½ M + m) = M/(M+2m). Other options treat child at center or invert
the ratio.
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, 8. A damped harmonic oscillator has natural frequency 0. Which equation describes
critically damped motion?
A. x(t) = A e^{-t} cos('t)
B. x(t) = (A + Bt) e^{-0 t}
C. x(t) = A e^{-0 t} + B e^{-20 t}
D. x(t) = A cos(0 t) e^{-t}
Answer: B
Rationale: Critical damping occurs when ³ = É0, yielding the solution x(t) = (A + Bt) e^{-É0 t}.
Options A and D are underdamped (oscillatory), and C is overdamped (two exponential decays).
9. A satellite orbits Earth in a circular orbit of radius r with period T. If the orbit radius is
increased to 4r, what is the new period?
A. T/2
B. 2T
C. 4T
D. 8T
Answer: D
Rationale: Kepler's third law: T^2 " r^3, so T2/T1 = (r2/r1)^{3/2} = 4^{3/2} = 8. Option A would
correspond to T 1/r, B to T r^{1/2}, C to T r.
10. Three particles are located at (0,0) (1 kg), (2,0) (2 kg), and (0,2) (3 kg). Where is the
center of mass?
A. (1, 1)
B. (2/3, 1)
C. (1, 2/3)
D. (2/3, 2/3)
Answer: B
Rationale: x_cm = (1*0+2*2+3*0)/(1+2+3) = 4/6 = 2/3; y_cm = (1*0+2*0+3*2)/6 = 6/6 = 1.
Option A is the unweighted average, others are incorrect coordinate swaps.
11. A projectile is launched from ground level with initial speed v0 at angle above
horizontal. What is the radius of curvature of its trajectory at the highest point?
A. v0^2 cos^2 / g
B. v0^2 / g
C. v0^2 sin^2 / g
D. v0^2 / (2g)
Answer: A
Rationale: At the highest point, velocity is horizontal with magnitude v0 cos¸. The acceleration is
g downward, which provides centripetal acceleration: g = v^2/R. Thus R = v^2/g = v0^2 cos^2 /
g.
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FRQs with Complete Answer Key, Rationales and
Step-by-Step Solution
1. A particle moves along the x-axis with acceleration a(t) = 6t m/s^2. At t=0, the particle is
at rest at the origin. What is its position at t=2 s?
A. 4 m
B. 6 m
C. 8 m
D. 10 m
Answer: C
Rationale: Integrate a(t) to get v(t)=3t^2, then x(t)=t^3. At t=2, x=8 m. Options A and B come
from misintegration, and D from using the wrong power.
2. A 2 kg block on a rough horizontal surface (k=0.4) is connected by a massless string over
a frictionless pulley to a 1 kg hanging block. The system is released from rest. What is the
acceleration of the blocks? (g=9.8 m/s^2)
A. 0.67 m/s^2
B. 1.33 m/s^2
C. 2.0 m/s^2
D. 3.2 m/s^2
Answer: A
Rationale: Using Newton's second law: m2g - T = m2a and T - ¼k m1g = m1a. Solving gives a =
(m2 - k m1)g/(m1+m2) = (1-0.8)*9.8/3 0.67 m/s^2. Other options omit friction or use incorrect
mass ratios.
3. A particle moves along the x-axis under the influence of a force F(x) = 3x^2 + 2 N. What
is the work done by this force as the particle moves from x=0 to x=2 m?
A. 10 J
B. 12 J
C. 14 J
D. 16 J
Answer: B
Rationale: Work = "+F dx = "+(3x^2+2) dx from 0 to 2 = [x^3+2x]_0^2 = 8+4=12 J. Option D
results from using average force (8 N) times distance (2 m), which is incorrect for a
non-constant force.
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,4. A 2 kg ball moving east at 4 m/s collides with a 1 kg ball moving north at 6 m/s on a
frictionless surface. If they stick together, what is the magnitude of their final velocity?
A. 2.0 m/s
B. 3.3 m/s
C. 4.0 m/s
D. 5.0 m/s
Answer: B
Rationale: Conservation of momentum: initial px=8, py=6, total mass=3, so vx=8/3, vy=2.
Magnitude = sqrt((8/3)^2+2^2)=10/33.33 m/s. Option A is average speed, C and D from
miscalculations.
5. A wheel starts from rest and rotates with constant angular acceleration. It makes 10
revolutions in the first 2 seconds. What is its angular acceleration?
A. 5 rad/s^2
B. 10 rad/s^2
C. 5 rad/s^2
D. 10 rad/s^2
Answer: B
Rationale: ¸ = ½ ± t^2, with ¸ = 10 rev = 20À rad, t=2 s !’ ± = 2¸/t^2 = 40À/4 = 10À rad/s^2.
Option A results from forgetting 2 per revolution; C and D omit .
6. A uniform rod of mass M and length L is pivoted at one end. A horizontal force F is
applied perpendicular to the rod at the free end. What is the angular acceleration?
A. 3F/(ML)
B. 2F/(ML)
C. F/(ML)
D. 4F/(ML)
Answer: A
Rationale: Torque = FL, moment of inertia about end I = (1/3)ML^2. Then ± = Ä/I = 3F/(ML).
Options B and C use incorrect moments of inertia (e.g., about center or point mass).
7. A merry-go-round (a solid disk of mass M and radius R) rotates at angular speed . A
child of mass m jumps radially onto the edge. What is the new angular speed?
A. M/(M+2m)
B. M/(M+m)
C. (M+2m)/M
D. (M+m)/M
Answer: A
Rationale: Conservation of angular momentum: I_i É = I_f É_f, with I_i=½ MR^2, I_f=½ MR^2 +
mR^2. Thus _f = (½ M)/(½ M + m) = M/(M+2m). Other options treat child at center or invert
the ratio.
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, 8. A damped harmonic oscillator has natural frequency 0. Which equation describes
critically damped motion?
A. x(t) = A e^{-t} cos('t)
B. x(t) = (A + Bt) e^{-0 t}
C. x(t) = A e^{-0 t} + B e^{-20 t}
D. x(t) = A cos(0 t) e^{-t}
Answer: B
Rationale: Critical damping occurs when ³ = É0, yielding the solution x(t) = (A + Bt) e^{-É0 t}.
Options A and D are underdamped (oscillatory), and C is overdamped (two exponential decays).
9. A satellite orbits Earth in a circular orbit of radius r with period T. If the orbit radius is
increased to 4r, what is the new period?
A. T/2
B. 2T
C. 4T
D. 8T
Answer: D
Rationale: Kepler's third law: T^2 " r^3, so T2/T1 = (r2/r1)^{3/2} = 4^{3/2} = 8. Option A would
correspond to T 1/r, B to T r^{1/2}, C to T r.
10. Three particles are located at (0,0) (1 kg), (2,0) (2 kg), and (0,2) (3 kg). Where is the
center of mass?
A. (1, 1)
B. (2/3, 1)
C. (1, 2/3)
D. (2/3, 2/3)
Answer: B
Rationale: x_cm = (1*0+2*2+3*0)/(1+2+3) = 4/6 = 2/3; y_cm = (1*0+2*0+3*2)/6 = 6/6 = 1.
Option A is the unweighted average, others are incorrect coordinate swaps.
11. A projectile is launched from ground level with initial speed v0 at angle above
horizontal. What is the radius of curvature of its trajectory at the highest point?
A. v0^2 cos^2 / g
B. v0^2 / g
C. v0^2 sin^2 / g
D. v0^2 / (2g)
Answer: A
Rationale: At the highest point, velocity is horizontal with magnitude v0 cos¸. The acceleration is
g downward, which provides centripetal acceleration: g = v^2/R. Thus R = v^2/g = v0^2 cos^2 /
g.
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