I85PREP
⌘ K
AP Physics C: Mechanics study materials
AP Physics C: Mechanics study resources

AP Physics C: Mechanics practice questions

One question at a time. Build your AP Physics C: Mechanics understanding.

First 50 of 200 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Kinematics

A particle moves 4 m east, then 3 m west. What are the magnitude of its displacement and the total distance it travels?

  1. 3 m and 4 m
  2. 7 m and 1 m
  3. 5 m and 7 m
  4. 1 m and 7 m Correct answer.

Explanation

Displacement is the final position minus the initial position, so the particle ends 1 m east of its start. Distance is the total path length traveled, which is 4 m + 3 m = 7 m.

Question 2Kinematics

An object's position along the x-axis changes from x = 2 m at t = 0 to x = 14 m at t = 4 s. What is its average velocity?

  1. 12 m/s
  2. 2.0 m/s
  3. 3.0 m/s Correct answer.
  4. 3.5 m/s

Explanation

Average velocity equals the change in position divided by the time interval: (14 m - 2 m)/(4 s) = 3.0 m/s. The displacement is 12 m, not 14 m, because the initial position must be subtracted.

Question 3Kinematics

On a position-time graph, the instantaneous velocity at a particular time is represented by which quantity?

  1. the y-coordinate at that time
  2. the average slope between the endpoints
  3. the area under the curve up to that time
  4. the slope of the tangent at that time Correct answer.

Explanation

Instantaneous velocity is the time derivative of position, so it equals the slope of the tangent to the position-time curve at that instant. A secant slope gives average velocity, while the y-coordinate gives position.

Question 4Kinematics

A car moves in the +x direction and is slowing down. Which statement about its acceleration is correct?

  1. Acceleration is in the +x direction
  2. Acceleration is negative only if velocity is negative
  3. Acceleration is zero because the speed is changing smoothly
  4. Acceleration is in the -x direction Correct answer.

Explanation

When an object is slowing down, its acceleration points opposite to its velocity. Since the velocity is in the +x direction, the acceleration must be in the -x direction.

Question 5Kinematics

A ball is thrown vertically upward and later returns to its launch height. Ignoring air resistance, which quantity is zero at the highest point?

  1. net force
  2. speed
  3. velocity Correct answer.
  4. acceleration

Explanation

At the highest point the ball momentarily stops before reversing direction, so its velocity is zero. The acceleration due to gravity remains g downward, and therefore the net force is not zero.

Question 6Kinematics

Two identical balls are released from the same height at the same instant. Ball A is dropped from rest, while ball B is launched horizontally. Ignoring air resistance, which ball reaches the ground first?

  1. It depends on the launch speed of ball B
  2. They reach the ground at the same time Correct answer.
  3. Ball A, because it has no horizontal motion
  4. Ball B, because it has horizontal velocity

Explanation

Horizontal and vertical motions are independent when air resistance is negligible. Both balls start with zero vertical velocity from the same height and have the same vertical acceleration g, so their vertical motions are identical.

Question 7Kinematics

A projectile is launched with initial speed 20 m/s at 30° above the horizontal. Taking g = 10 m/s², what is the maximum height above the launch point?

  1. 20 m
  2. 10 m
  3. 5 m Correct answer.
  4. 2.5 m

Explanation

The vertical component of the initial velocity is 20 sin 30° = 10 m/s. At maximum height the vertical velocity is zero, so h = v_y²/(2g) = (10)²/(20) = 5 m.

Question 8Kinematics

On level ground with no air resistance, a projectile launched at 30° above the horizontal has range R. At what other launch angle, with the same initial speed, is the range also R?

  1. 45°
  2. 15°
  3. 90°
  4. 60° Correct answer.

Explanation

For a given initial speed on level ground, the range is the same for launch angles that are complementary-that is, angles summing to 90°. Since 30° + 60° = 90°, the other angle is 60°.

Question 9Kinematics

Train A moves east at 20 m/s relative to the ground, and Train B moves east at 12 m/s relative to the ground. What is the velocity of Train A relative to Train B?

  1. 8 m/s east Correct answer.
  2. 32 m/s east
  3. 12 m/s east
  4. 8 m/s west

Explanation

Relative velocity is found by subtracting the observer's velocity: v_A relative to B = v_A - v_B = 20 m/s east - 12 m/s east = 8 m/s east. The trains move in the same direction, so the relative speed is the difference of their speeds.

Question 10Kinematics

A boat heads straight across a river at 4 m/s relative to the water. The river current is 3 m/s downstream. What is the magnitude of the boat's velocity relative to the shore?

  1. 7 m/s
  2. 5 m/s Correct answer.
  3. 1 m/s
  4. 3.5 m/s

Explanation

The boat's velocity relative to the water and the current are perpendicular vectors. Their magnitudes combine by the Pythagorean theorem: sqrt(4² + 3²) = 5 m/s.

Question 11Kinematics

A particle moves along the x-axis. Its velocity-time graph is a rectangle of height 6 m/s from t = 0 to t = 5 s. What is the particle's displacement during this interval?

  1. 11 m
  2. 1.2 m
  3. 60 m
  4. 30 m Correct answer.

Explanation

Displacement is the area under the velocity-time graph. The area of the rectangle is (6 m/s)(5 s) = 30 m.

Question 12Kinematics

On a velocity-time graph, the acceleration at a given time is represented by which quantity?

  1. the average velocity over the interval
  2. the value of velocity at that time
  3. the area under the curve
  4. the slope of the tangent to the curve Correct answer.

Explanation

Acceleration is the time derivative of velocity, so on a velocity-time graph it is the slope of the tangent at that instant. The area under the curve gives displacement, not acceleration.

Question 13Kinematics

A particle starts from rest. Its acceleration is a constant 3 m/s² for 4 s. What is its velocity at t = 4 s?

  1. 7 m/s
  2. 0.75 m/s
  3. 12 m/s Correct answer.
  4. 24 m/s

Explanation

The change in velocity is the area under the acceleration-time graph: Δv = aΔt = (3 m/s²)(4 s) = 12 m/s. Because the particle starts from rest, its final velocity is 12 m/s.

Question 14Kinematics

A car accelerates uniformly from 10 m/s to 30 m/s over a displacement of 80 m. What is its acceleration?

  1. 20 m/s²
  2. 5 m/s² Correct answer.
  3. 0.25 m/s²
  4. 10 m/s²

Explanation

Use v² = u² + 2as, which does not require time: (30)² = (10)² + 2a(80). Solving gives 900 = 100 + 160a, so a = 5 m/s².

Question 15Kinematics

An object starts from rest and accelerates uniformly at 2 m/s² for 6 s. How far does it travel during this time?

  1. 36 m Correct answer.
  2. 24 m
  3. 72 m
  4. 12 m

Explanation

For constant acceleration starting from rest, s = ½at². Substituting gives s = ½(2)(6)² = 36 m.

Question 16Kinematics

A particle's acceleration is a(t) = 6t m/s². At t = 0 its velocity is 2 m/s. What is its velocity at t = 3 s?

  1. 27 m/s
  2. 54 m/s
  3. 18 m/s
  4. 29 m/s Correct answer.

Explanation

Velocity is the integral of acceleration plus the initial velocity: v(t) = 2 + ∫0^t 6t dt = 2 + 3t². At t = 3 s, v = 2 + 3(9) = 29 m/s.

Question 17Kinematics

A particle moves along the x-axis with velocity v(t) = 4t m/s and starts at x = 1 m when t = 0. What is its position at t = 2 s?

  1. 8 m
  2. 5 m
  3. 9 m Correct answer.
  4. 17 m

Explanation

Position is the initial position plus the integral of velocity: x(t) = 1 + ∫0^t 4t dt = 1 + 2t². At t = 2 s, x = 1 + 2(4) = 9 m.

Question 18Kinematics

An object has an initial velocity of 3 m/s. Its acceleration is a constant -2 m/s² from t = 0 to t = 5 s. What is its velocity at t = 5 s?

  1. -7 m/s Correct answer.
  2. 7 m/s
  3. -13 m/s
  4. -10 m/s

Explanation

The change in velocity is the area under the acceleration-time graph: Δv = (-2 m/s²)(5 s) = -10 m/s. Adding this to the initial velocity gives v = 3 - 10 = -7 m/s.

Question 19Kinematics

A velocity-time graph shows positive velocity increasing with time, but the slope of the graph is becoming less steep. Which description of the acceleration is correct?

  1. Negative and increasing in magnitude
  2. Positive and decreasing in magnitude Correct answer.
  3. Zero because velocity is positive
  4. Positive and increasing in magnitude

Explanation

Acceleration is the slope of the velocity-time graph. The slope is positive because velocity is increasing, but the slope is becoming less steep, so the acceleration is positive and decreasing in magnitude.

Question 20Kinematics

A ball is thrown straight upward. Ignoring air resistance, what is its acceleration at the highest point of its motion?

  1. g upward
  2. g downward Correct answer.
  3. zero
  4. g downward only after it starts falling

Explanation

Gravity acts throughout the entire flight, including at the highest point. At that instant the velocity is zero, but the acceleration is still g directed downward.

Question 21Kinematics

A runner completes exactly one lap of a circular track of radius 50 m in 100 s. Taking π ≈ 3.14, what are the average speed and the magnitude of the average velocity?

  1. 0 m/s and 3.1 m/s
  2. 3.1 m/s and 0 m/s Correct answer.
  3. 3.1 m/s and 3.1 m/s
  4. 0 m/s and 0 m/s

Explanation

The path length is the circumference, 2πr ≈ 314 m, so average speed is 314 m / 100 s ≈ 3.1 m/s. After one complete lap the displacement is zero, so the magnitude of average velocity is zero.

Question 22Kinematics

A ball is launched horizontally at 8 m/s from a height of 20 m. Taking g = 10 m/s², how far horizontally does it travel before hitting the ground?

  1. 32 m
  2. 20 m
  3. 16 m Correct answer.
  4. 8 m

Explanation

The vertical fall time satisfies h = ½gt², so t = sqrt(2h/g) = sqrt(40/10) = 2 s. The horizontal distance is x = v_x t = (8 m/s)(2 s) = 16 m.

Question 23Kinematics

A projectile is launched with speed v at an angle θ above the horizontal. What is the vertical component of its initial velocity?

  1. v sin θ Correct answer.
  2. v tan θ
  3. v cos θ
  4. v / sin θ

Explanation

The initial velocity vector has magnitude v and makes an angle θ with the horizontal. Its vertical component is v sin θ, while its horizontal component is v cos θ.

Question 24Kinematics

A car moving at speed v can stop in distance d with constant braking acceleration. If its initial speed is doubled while the braking acceleration is unchanged, what is the new stopping distance?

  1. d/2
  2. 2d
  3. d
  4. 4d Correct answer.

Explanation

With constant acceleration, stopping distance satisfies v² = 2ad, so d is proportional to v². Doubling v makes v² four times larger, so the stopping distance becomes 4d.

Question 25Kinematics

An object's velocity is +4 m/s for 3 s, then -2 m/s for 2 s. What is its total displacement during these 5 s?

  1. 8 m Correct answer.
  2. 4 m
  3. 20 m
  4. 16 m

Explanation

Displacement is the signed area under the velocity-time graph. For the first interval, (4 m/s)(3 s) = 12 m; for the second, (-2 m/s)(2 s) = -4 m. The total displacement is 12 - 4 = 8 m.

Question 26Kinematics

A particle moves along the x-axis with position x(t) = 4t² − 2t, where x is in meters and t is in seconds. What is the particle's instantaneous velocity at t = 1 s?

  1. −2 m/s
  2. 2 m/s
  3. 8 m/s
  4. 6 m/s Correct answer.

Explanation

Instantaneous velocity is the time derivative of position, so v(t) = dx/dt = 8t − 2. Substituting t = 1 s gives v = 8(1) − 2 = 6 m/s. The derivative must be taken before substituting, since the velocity is changing with time.

Question 27Kinematics

A particle moving along the x-axis has velocity v(t) = 3t² − 6t, with v in m/s and t in seconds. At what time is the particle's acceleration equal to zero?

  1. 2 s
  2. 0 s
  3. 1 s Correct answer.
  4. 3 s

Explanation

Acceleration is the time derivative of velocity, so a(t) = dv/dt = 6t − 6. Setting a = 0 gives 6t = 6, hence t = 1 s. At that instant the velocity is momentarily at a minimum, not zero, which is why the times when v = 0 are not the answer.

Question 28Kinematics

A ball is thrown straight upward with an initial speed of 15 m/s from the top of a cliff that is 20 m above the ground. Taking g = 10 m/s² and neglecting air resistance, how long does the ball take to reach the ground below?

  1. 4.0 s Correct answer.
  2. 5.0 s
  3. 2.0 s
  4. 3.0 s

Explanation

With upward positive, the vertical displacement when the ball lands is −20 m, so −20 = 15t − 5t². Rearranging gives t² − 3t − 4 = 0, which factors as (t − 4)(t + 1) = 0. The physically meaningful root is t = 4.0 s; the negative root corresponds to a time before the launch.

Question 29Kinematics

A particle moves in a straight line with initial speed v₀, and the only force on it produces an acceleration a = −kv, where k is a positive constant and v is the instantaneous speed. How far does the particle travel before coming to rest?

  1. Infinite distance, because the speed approaches zero only as t → ∞
  2. v₀/k Correct answer.
  3. v₀²/(2k)
  4. kv₀

Explanation

Using a = v dv/dx, the equation becomes v dv/dx = −kv, so dv/dx = −k and the speed decreases linearly with distance. Integrating from v₀ to 0 gives Δx = v₀/k. Although the particle never exactly stops in finite time, its total travel distance is finite.

Question 30Force and Translational Dynamics

A hockey puck slides across a frictionless horizontal surface at constant velocity. Which statement best describes the net force on the puck?

  1. The net force is zero because the velocity is constant. Correct answer.
  2. The net force is upward, balancing the normal force.
  3. The net force is equal to the puck's weight.
  4. The net force points in the direction of motion to maintain the speed.

Explanation

Constant velocity means zero acceleration, so Newton's second law requires the net force to be zero. The normal force and weight cancel vertically, and no horizontal force is needed to maintain motion.

Question 31Force and Translational Dynamics

A 4.0 kg cart experiences a net force of 12 N to the right. What is its acceleration?

  1. 3.0 m/s² to the right Correct answer.
  2. 8.0 m/s² to the right
  3. 48 m/s² to the right
  4. 0.33 m/s² to the right

Explanation

Newton's second law gives a = F_net/m. Substituting 12 N and 4.0 kg gives 3.0 m/s², in the direction of the net force.

Question 32Force and Translational Dynamics

A book rests on a table. Which force is the Newton's third-law partner of the gravitational force exerted by Earth on the book?

  1. The net force on the book while it is at rest
  2. The normal force exerted by the table upward on the book
  3. The gravitational force exerted by the book on Earth Correct answer.
  4. The normal force exerted by the book downward on the table

Explanation

A third-law partner has the same type of force, equal magnitude, opposite direction, and acts on the other object. Earth pulls the book downward, so the book pulls Earth upward with a gravitational force.

Question 33Force and Translational Dynamics

A block is at rest on a horizontal floor. Which statement correctly describes the normal force on the block?

  1. It is perpendicular to the surface and may differ from the weight. Correct answer.
  2. It points along the surface to oppose any attempted sliding.
  3. It is always equal to the block's weight, even when other vertical forces act.
  4. It is the third-law reaction to the block's weight.

Explanation

The normal force is a contact force perpendicular to the surface. Its magnitude adjusts to keep the block from passing through the floor, so it can equal, exceed, or be less than the weight depending on other forces.

Question 34Force and Translational Dynamics

A crate remains at rest on a horizontal floor while a small horizontal push is applied. As the push increases but the crate stays at rest, what happens to the static friction force?

  1. It decreases steadily as the applied force grows.
  2. It stays equal to the crate's weight the whole time.
  3. It grows to match the applied force up to a limit. Correct answer.
  4. It stays constant at the kinetic friction value.

Explanation

Static friction is adjustable: it matches the applied force to keep the crate at rest, up to the maximum value μ_sN. Once the applied force exceeds that maximum, the crate slips.

Question 35Force and Translational Dynamics

A ball is thrown upward and is moving upward through the air. Ignoring air resistance, what forces act on the ball while it is in the air?

  1. Gravity downward only Correct answer.
  2. Gravity downward and an upward force from the throw
  3. No forces, because the ball is moving upward
  4. Gravity downward and a decreasing upward force

Explanation

After the ball leaves the hand, gravity is the only force acting if air resistance is ignored. Upward motion does not require an upward force; the ball is simply moving upward while accelerating downward.

Question 36Force and Translational Dynamics

A light rope passes over a frictionless pulley and supports two identical hanging masses. The system is released from rest. What is the acceleration of each mass?

  1. Each mass accelerates downward at g.
  2. Each mass has zero acceleration because the forces balance. Correct answer.
  3. Each mass accelerates upward at g.
  4. The masses accelerate in opposite directions with magnitude g.

Explanation

The masses are equal, so their weights are equal, and the tension is the same throughout a massless rope over a frictionless pulley. Each mass therefore has zero net force and zero acceleration.

Question 37Force and Translational Dynamics

A car travels at constant speed around a circular track. Which direction is the net force on the car?

  1. Tangent to the circle in the direction of motion
  2. Radially outward from the center of the circle
  3. Vertically upward to balance the car's weight
  4. Radially inward toward the center of the circle Correct answer.

Explanation

At constant speed on a circular path, the acceleration is centripetal and points toward the center. The net force must point in the same direction as that acceleration.

Question 38Force and Translational Dynamics

A 5.0 kg mass and a 3.0 kg mass hang from opposite ends of a light rope over a frictionless pulley. Using g = 10 m/s², what is the magnitude of the acceleration of the masses?

  1. 1.25 m/s²
  2. 5.0 m/s²
  3. 2.5 m/s² Correct answer.
  4. 10 m/s²

Explanation

For an Atwood machine, a = (m₂ − m₁)g/(m₁ + m₂). With 5.0 kg and 3.0 kg, this gives (2.0)(10)/8.0 = 2.5 m/s².

Question 39Force and Translational Dynamics

A 2.0 kg block slides down a 30° incline. The coefficient of kinetic friction is 0.20. Using g = 10 m/s², what is the block's acceleration down the incline?

  1. 6.7 m/s²
  2. 3.3 m/s² Correct answer.
  3. 1.7 m/s²
  4. 5.0 m/s²

Explanation

The component of gravity down the incline is mg sinθ, and kinetic friction up the incline is μ_k mg cosθ. Dividing the net force by m gives a = g(sinθ − μ_k cosθ) = 3.3 m/s².

Question 40Force and Translational Dynamics

A 60 kg person stands on a scale in an elevator accelerating upward at 2.0 m/s². Using g = 10 m/s², what does the scale read?

  1. 480 N
  2. 720 N Correct answer.
  3. 840 N
  4. 600 N

Explanation

For upward acceleration, the scale force must exceed the weight: N − mg = ma. Thus N = m(g + a) = 60(10 + 2.0) = 720 N.

Question 41Force and Translational Dynamics

Block A (2.0 kg) and block B (3.0 kg) are in contact on a frictionless horizontal floor. A horizontal force of 10 N pushes block A, which then pushes block B. What force does block A exert on block B?

  1. 4 N
  2. 6 N Correct answer.
  3. 10 N
  4. 2 N

Explanation

The two blocks accelerate together with a = F/(m_A + m_B) = 10/5.0 = 2.0 m/s². The force A exerts on B is the only horizontal force on B, so it is m_B a = 3.0 × 2.0 = 6.0 N.

Question 42Force and Translational Dynamics

A block remains at rest on an incline whose angle is slowly increased. It just begins to slide when the angle reaches 35°. What is the coefficient of static friction between the block and the incline?

  1. 1.43
  2. 0.35
  3. 0.57
  4. 0.70 Correct answer.

Explanation

At the threshold of slipping, the static friction force reaches its maximum μ_sN. Balancing forces along the incline gives mg sinθ = μ_s mg cosθ, so μ_s = tanθ = tan35° ≈ 0.70.

Question 43Force and Translational Dynamics

A curve of radius 50 m is banked so that a car moving at 20 m/s can negotiate it without relying on friction. Using g = 10 m/s², what banking angle is required?

  1. 22°
  2. 64°
  3. 51°
  4. 39° Correct answer.

Explanation

For a frictionless banked curve, the horizontal component of the normal force supplies centripetal force, giving tanθ = v²/(rg). Substituting v = 20 m/s, r = 50 m, and g = 10 m/s² gives tanθ = 0.80, so θ ≈ 39°.

Question 44Force and Translational Dynamics

A 0.50 kg mass hangs at rest from a vertical spring with spring constant 100 N/m. Using g = 10 m/s², how much is the spring stretched from its natural length?

  1. 2 cm
  2. 20 cm
  3. 0.5 cm
  4. 5 cm Correct answer.

Explanation

At equilibrium, the spring force balances the weight: kx = mg. Thus x = mg/k = (0.50)(10)/100 = 0.050 m = 5.0 cm.

Question 45Force and Translational Dynamics

A 0.20 kg ball attached to a string moves in a vertical circle of radius 0.50 m. At the top of the path, its speed is 4.0 m/s. Using g = 10 m/s², what is the tension in the string at that point?

  1. 2.0 N
  2. 4.4 N Correct answer.
  3. 8.4 N
  4. 6.4 N

Explanation

At the top of the vertical circle, both tension and weight point downward toward the center. Thus T + mg = mv²/r, so T = (0.20)(4.0²)/0.50 − (0.20)(10) = 6.4 − 2.0 = 4.4 N.

Question 46Force and Translational Dynamics

A falling object experiences quadratic air drag. As its speed increases toward terminal speed, what happens to the net force and acceleration?

  1. Net force remains constant, and acceleration remains constant.
  2. Net force becomes upward, and acceleration becomes upward in the direction of motion.
  3. Net force increases, and acceleration increases.
  4. Net force decreases toward zero, and acceleration decreases toward zero. Correct answer.

Explanation

As speed increases, quadratic drag grows and opposes gravity more strongly, so the net downward force decreases. Since a = F_net/m, acceleration decreases toward zero as terminal speed is approached.

Question 47Force and Translational Dynamics

A 2.0 kg object has two horizontal forces acting on it: 6.0 N east and 8.0 N north. What is the magnitude of its acceleration?

  1. 5.0 m/s² Correct answer.
  2. 10 m/s²
  3. 7.0 m/s²
  4. 1.0 m/s²

Explanation

The two forces are perpendicular, so the net force magnitude is √(6.0² + 8.0²) = 10 N. Newton's second law then gives a = 10 N / 2.0 kg = 5.0 m/s².

Question 48Force and Translational Dynamics

A 2.0 kg mass is at x = 0 and a 3.0 kg mass is at x = 5.0 m. They are connected by a light rod. Where is the center of mass of the system?

  1. 3.0 m Correct answer.
  2. 2.5 m
  3. 3.5 m
  4. 2.0 m

Explanation

The center of mass is x_cm = (m₁x₁ + m₂x₂)/(m₁ + m₂). Substituting gives (2.0 × 0 + 3.0 × 5.0)/5.0 = 3.0 m.

Question 49Force and Translational Dynamics

A crate sits on the flat bed of a truck. The truck accelerates forward at 2.0 m/s² and the crate does not slip. What is the direction and cause of the friction force on the crate?

  1. Static friction acts forward on the crate. Correct answer.
  2. Kinetic friction acts forward on the crate.
  3. Static friction acts backward on the crate.
  4. No friction acts because the crate moves with the truck.

Explanation

The crate accelerates forward with the truck, so the net horizontal force on it must be forward. Static friction from the truck bed provides that forward force when the crate does not slip.

Question 50Force and Translational Dynamics

A 0.50 kg ball is attached to a 1.0 m string and moves in a horizontal circle at constant speed. The string makes a 30° angle with the vertical. Using g = 10 m/s², what is the tension in the string?

  1. 2.9 N
  2. 4.3 N
  3. 5.8 N Correct answer.
  4. 10 N

Explanation

The vertical component of tension supports the ball's weight, so T cos30° = mg. Thus T = (0.50)(10)/cos30° ≈ 5.8 N.