AP Physics C: Mechanics 2026 Practice Exam 1 - 40
Multiple-Choice Questions with Verified Answer Key, Detailed
Solutions & Calculations. - 48 Questions and Answers Already
Graded A+ Premium Exam Tested And Verified
Subject Area AP Physics C: Mechanics 2026 Practice Exam 1 - 40 Multiple-Choice
Questions with Verified Answer Key, Detailed Solutions & Calculations.
Description Comprehensive examination on AP Physics C: Mechanics 2026 Practice Exam 1 -
40 Multiple-Choice Questions with Verified Answer Key, Detailed Solutions &
Calculations..
Expected Grade A+
Total Questions 48
Duration 3 hours
Learning Outcomes 1. Demonstrate mastery of core concepts
Accreditation Aligned with US university standards.
Page 1
,1. A particle moves along the x-axis with acceleration a(t) = 6t - 4, where t is in
seconds. At t = 0, its velocity is 2 m/s and position is 1 m. What is its average velocity
from t = 1 s to t = 3 s?
Answer: 10 m/s
Integrate a(t) to get v(t) = 3t^2 - 4t + 2. Integrate to get x(t) = t^3 - 2t^2 + 2t + 1. Then
x(1)=2 m, x(3)=22 m. Average velocity = (22-2)/(3-1)=10 m/s.
2. A block of mass m is on an incline with angle and coefficient of kinetic friction k.
A horizontal force F is applied to the block. The block moves up the incline at
constant speed. Which expression gives F?
Answer: mg(sin + k cos) / (cos + k sin)
For constant speed, net force = 0. Components: parallel to incline: Fcos - mg sin - f = 0,
perpendicular: Fsin + mg cos = N, and f = k N. Solve for F yields option A.
3. A force F(x) = 4x + 3 (in N) acts on a 2 kg particle as it moves along the x-axis
from x=1 m to x=4 m. If the particle starts from rest at x=1, what is its speed at x=4?
Answer: (39) m/s
Work done = F dx from 1 to 4 = (4x+3) dx = 2x^2+3x | from 1 to 4 = (32+12)-(2+3)=39
J. Work = KE = (1/2)mv^2, so v = (2*39/2) = 39 m/s.
4. A 3 kg mass moving at 4 m/s collides head-on with a 2 kg mass moving at 6 m/s in
the opposite direction. The collision is perfectly inelastic. What is the loss of kinetic
energy?
Answer: 30 J
Momentum conservation: (3)(4) + (2)(-6) = 12-12=0, so final velocity = 0. Initial KE =
0.5*3*16 + 0.5*2*36 = 24+36=60 J. Final KE = 0. Loss = 60 J.
5. A uniform rod of mass M and length L is pivoted at one end. It is released from
rest in the horizontal position. What is the angular acceleration at the instant of
release?
Answer: 3g / (2L)
Torque due to gravity about pivot: = Mg(L/2). Moment of inertia about end: I =
(1/3)ML^2. = /I = (MgL/2)/(ML^2/3) = (3g)/(2L).
Page 2
, 6. A physical pendulum consists of a uniform disk of mass M and radius R pivoted at
a point on its edge. Find the period of small oscillations.
Answer: 2 (3R/(2g))
Distance from pivot to CM = R. I about pivot = I_CM + MR^2 = (1/2)MR^2 + MR^2 =
(3/2)MR^2. T = 2 (I/(mgd)) = 2 ((3/2)MR^2/(MgR)) = 2 (3R/(2g)).
7. A satellite orbits Earth in a circular orbit of radius r. The gravitational force is
proportional to 1/r^2. If the orbit radius is increased to 4r, what is the ratio of the
new orbital period to the original?
Answer: 8
Kepler's third law: T^2 r^3. So T r^(3/2). For r -> 4r, T_new = (4^(3/2)) T_old = 8
T_old. Ratio = 8.
8. In an Atwood machine with two masses m1 and m2 (m1 > m2) hanging over a
pulley of moment of inertia I and radius R, the string does not slip. What is the
acceleration of mass m1?
Answer: g (m1 - m2) / (m1 + m2 + I/R^2)
Equations: m1g - T1 = m1a, T2 - m2g = m2a, (T1 - T2)R = I = I a/R. Solve for a: a =
g(m1 - m2) / (m1 + m2 + I/R^2).
9. A particle's position on the x-axis is given by x(t) = A t^4 + B t^3, where A and B
are constants. At t = 0, the particle is at rest. At what non-zero time t does the
particle again come to rest?
Answer: t = -3B/(4A)
Velocity v(t)=dx/dt=4A t^3+3B t^2 = t^2(4A t+3B). v=0 at t=0 and when 4A t+3B=0 =>
t=-3B/(4A). Thus correct is A.
10. Two blocks of masses m1 and m2 (m1 > m2) are connected by a light string over
a frictionless pulley. The system is released from rest on a frictionless incline with
angles 1 and 2 for m1 and m2 respectively. For the system to accelerate with m1
moving downward, which condition must hold?
Answer: m1 sin 1 > m2 sin 2
The net force along the incline is m1g sin1 - m2g sin2 (assuming m1 moves downward).
For acceleration downward, net force positive, so m1 sin1 > m2 sin2. Option A.
Page 3
Multiple-Choice Questions with Verified Answer Key, Detailed
Solutions & Calculations. - 48 Questions and Answers Already
Graded A+ Premium Exam Tested And Verified
Subject Area AP Physics C: Mechanics 2026 Practice Exam 1 - 40 Multiple-Choice
Questions with Verified Answer Key, Detailed Solutions & Calculations.
Description Comprehensive examination on AP Physics C: Mechanics 2026 Practice Exam 1 -
40 Multiple-Choice Questions with Verified Answer Key, Detailed Solutions &
Calculations..
Expected Grade A+
Total Questions 48
Duration 3 hours
Learning Outcomes 1. Demonstrate mastery of core concepts
Accreditation Aligned with US university standards.
Page 1
,1. A particle moves along the x-axis with acceleration a(t) = 6t - 4, where t is in
seconds. At t = 0, its velocity is 2 m/s and position is 1 m. What is its average velocity
from t = 1 s to t = 3 s?
Answer: 10 m/s
Integrate a(t) to get v(t) = 3t^2 - 4t + 2. Integrate to get x(t) = t^3 - 2t^2 + 2t + 1. Then
x(1)=2 m, x(3)=22 m. Average velocity = (22-2)/(3-1)=10 m/s.
2. A block of mass m is on an incline with angle and coefficient of kinetic friction k.
A horizontal force F is applied to the block. The block moves up the incline at
constant speed. Which expression gives F?
Answer: mg(sin + k cos) / (cos + k sin)
For constant speed, net force = 0. Components: parallel to incline: Fcos - mg sin - f = 0,
perpendicular: Fsin + mg cos = N, and f = k N. Solve for F yields option A.
3. A force F(x) = 4x + 3 (in N) acts on a 2 kg particle as it moves along the x-axis
from x=1 m to x=4 m. If the particle starts from rest at x=1, what is its speed at x=4?
Answer: (39) m/s
Work done = F dx from 1 to 4 = (4x+3) dx = 2x^2+3x | from 1 to 4 = (32+12)-(2+3)=39
J. Work = KE = (1/2)mv^2, so v = (2*39/2) = 39 m/s.
4. A 3 kg mass moving at 4 m/s collides head-on with a 2 kg mass moving at 6 m/s in
the opposite direction. The collision is perfectly inelastic. What is the loss of kinetic
energy?
Answer: 30 J
Momentum conservation: (3)(4) + (2)(-6) = 12-12=0, so final velocity = 0. Initial KE =
0.5*3*16 + 0.5*2*36 = 24+36=60 J. Final KE = 0. Loss = 60 J.
5. A uniform rod of mass M and length L is pivoted at one end. It is released from
rest in the horizontal position. What is the angular acceleration at the instant of
release?
Answer: 3g / (2L)
Torque due to gravity about pivot: = Mg(L/2). Moment of inertia about end: I =
(1/3)ML^2. = /I = (MgL/2)/(ML^2/3) = (3g)/(2L).
Page 2
, 6. A physical pendulum consists of a uniform disk of mass M and radius R pivoted at
a point on its edge. Find the period of small oscillations.
Answer: 2 (3R/(2g))
Distance from pivot to CM = R. I about pivot = I_CM + MR^2 = (1/2)MR^2 + MR^2 =
(3/2)MR^2. T = 2 (I/(mgd)) = 2 ((3/2)MR^2/(MgR)) = 2 (3R/(2g)).
7. A satellite orbits Earth in a circular orbit of radius r. The gravitational force is
proportional to 1/r^2. If the orbit radius is increased to 4r, what is the ratio of the
new orbital period to the original?
Answer: 8
Kepler's third law: T^2 r^3. So T r^(3/2). For r -> 4r, T_new = (4^(3/2)) T_old = 8
T_old. Ratio = 8.
8. In an Atwood machine with two masses m1 and m2 (m1 > m2) hanging over a
pulley of moment of inertia I and radius R, the string does not slip. What is the
acceleration of mass m1?
Answer: g (m1 - m2) / (m1 + m2 + I/R^2)
Equations: m1g - T1 = m1a, T2 - m2g = m2a, (T1 - T2)R = I = I a/R. Solve for a: a =
g(m1 - m2) / (m1 + m2 + I/R^2).
9. A particle's position on the x-axis is given by x(t) = A t^4 + B t^3, where A and B
are constants. At t = 0, the particle is at rest. At what non-zero time t does the
particle again come to rest?
Answer: t = -3B/(4A)
Velocity v(t)=dx/dt=4A t^3+3B t^2 = t^2(4A t+3B). v=0 at t=0 and when 4A t+3B=0 =>
t=-3B/(4A). Thus correct is A.
10. Two blocks of masses m1 and m2 (m1 > m2) are connected by a light string over
a frictionless pulley. The system is released from rest on a frictionless incline with
angles 1 and 2 for m1 and m2 respectively. For the system to accelerate with m1
moving downward, which condition must hold?
Answer: m1 sin 1 > m2 sin 2
The net force along the incline is m1g sin1 - m2g sin2 (assuming m1 moves downward).
For acceleration downward, net force positive, so m1 sin1 > m2 sin2. Option A.
Page 3
