AP Prep — Physics 1 Practice

Question 1 of 10

MCQU2Topic 2.5Medium No calc Word Diagram

A block is released from rest on a frictionless ramp inclined at 30 degrees above the horizontal. What is the magnitude of the block's acceleration down the ramp? (Use g = 9.8 m/s2.)

A8.49 m/s2

B4.9 m/s2

C9.8 m/s2

D5.66 m/s2

Explanation

On a frictionless incline the acceleration along the surface is a = g*sin(theta) = 9.8*sin30 = 9.8*0.5 = 4.9 m/s2. Note the result is independent of mass. The distractor 8.49 m/s2 uses cos(theta) instead of sin(theta); 9.8 m/s2 ignores the incline entirely (free fall); 5.66 m/s2 uses tan(theta).

Question 2 of 10

MCQU1Topic 1.2Medium No calc Word

A cyclist travels 120 m north in 20 s, stops and rests for 10 s, then travels 60 m south in 15 s, all along a straight north-south road. What is the cyclist's average velocity for the entire trip?

A4.0 m/s, north

B1.3 m/s, south

C1.7 m/s, north

D1.3 m/s, north

Explanation

Average velocity = (net displacement)/(total time). Net displacement = 120 m north − 60 m south = 60 m north. Total time = 20 + 10 + 15 = 45 s, so average velocity = 60/45 ≈ 1.3 m/s north. The 4.0 m/s value uses total DISTANCE (180 m)/45 s, which gives average speed, not velocity. The 1.7 m/s value forgets to include the 10 s rest in the total time (60 m / 35 s ≈ 1.7 m/s). The 1.3 m/s south option has the correct magnitude but the wrong direction.

Question 3 of 10

MCQU5Topic 5.1Medium No calc Word

A disk initially at rest is given a constant angular acceleration of 2.0 rad/s². Through what angle (in radians) has the disk turned after 5.0 s?

A50 rad

B10 rad

C12.5 rad

D25 rad

Explanation

With ω₀ = 0 and constant α, the angular displacement is θ = ½αt² = ½(2.0 rad/s²)(5.0 s)² = ½(2.0)(25) = 25 rad. The distractor 50 rad drops the factor of ½. The distractor 10 rad computes the final angular velocity ω = αt = 10 rad/s and reports that instead of the angle. 12.5 rad squares t but then omits α.

Question 4 of 10

MCQU8Topic 8.4Medium Calc Word Diagram

A large open tank of water has a small hole in its side. The water surface in the tank is 1.8 m above the hole, and the tank is wide enough that the surface barely moves. Using Torricelli's result, what is the speed of the water as it exits the hole?

A17.6 m/s

B35.3 m/s

C5.9 m/s

D4.2 m/s

Explanation

Torricelli's theorem (from Bernoulli) gives the efflux speed v = √(2gh) = √(2·9.8·1.8) = √35.28 = 5.9 m/s. The 4.2 m/s answer drops the factor of 2, using √(gh). The 35.3 m/s answer forgets to take the square root (that is the value of 2gh). The 17.6 m/s answer uses v = gh, mixing up the kinematic relations.

Question 5 of 10

MCQU7Topic 7.2Easy No calc Word

Two simple pendulums have exactly the same length and swing with the same small amplitude, but pendulum X has twice the mass of pendulum Y. How do their periods compare?

APendulum Y has the larger period, because less mass lets it swing faster.

BThey have equal periods, because a pendulum's period depends only on length and g, not mass.

CPendulum X has the larger period, because its greater mass gives it more inertia and a slower swing.

DIt cannot be determined without knowing each swing's amplitude.

Explanation

The period of a simple pendulum, T = 2π√(L/g), contains no mass term, so identical lengths give identical periods regardless of mass. The 'heavier is slower' and 'lighter is faster' choices wrongly treat the pendulum like a spring or assume mass enters the formula; it does not.

Question 6 of 10

MCQU4Topic 4.4Medium No calc Word Diagram

A 0.5 kg cart moving at 4 m/s collides with a stationary 1.5 kg cart, and the two carts couple together and move as one. What is their common speed just after the collision?

A2.0 m/s

B4.0 m/s

C1.0 m/s

D1.33 m/s

Explanation

For a perfectly inelastic collision, momentum is conserved: m₁v₁ = (m₁ + m₂)v_f. So v_f = (0.5·4)/(0.5 + 1.5) = 2.0/2.0 = 1.0 m/s. The 2.0 answer averages the two velocities (4 + 0)/2, which is only valid for equal masses. The 1.33 value divides by only the target mass 1.5 instead of the total mass. The 4.0 answer ignores that the added mass must slow the system.

Question 7 of 10

MCQU6Topic 6.1Medium No calc

A solid uniform disk of mass 4.0 kg and radius 0.25 m spins about its central axis at 12 rad/s. The moment of inertia of a solid disk about this axis is I = ½MR². What is the disk's rotational kinetic energy?

A9.0 J

B36 J

C0.75 J

D18 J

Explanation

Rotational KE = ½Iω² with I = ½MR² = ½(4.0)(0.25²) = 0.125 kg·m². So KE = ½(0.125)(12²) = ½(0.125)(144) = 9.0 J. The 18 J trap uses I = MR² (dropping the ½ in the moment of inertia). The 36 J trap drops the ½ in the moment of inertia AND the ½ in the KE formula. The 0.75 J trap forgets to square ω, using ½Iω.

Question 8 of 10

MCQU2Topic 2.2Medium No calc Word

A crate sitting on a level floor has a weight of 100 N. A person presses straight down on the top of the crate with an additional force of 30 N. What is the magnitude of the normal force the floor exerts on the crate?

A30 N

B70 N

C100 N

D130 N

Explanation

The crate is in vertical equilibrium, so the upward normal force balances all downward forces: N = weight + applied push = 100 N + 30 N = 130 N. The distractor 70 N comes from subtracting the push instead of adding it; 100 N comes from ignoring the extra downward push. A downward applied force increases the normal force.

Question 9 of 10

MCQU1Topic 1.1Easy No calc Word

A hiker walks 300 m due east and then 400 m due west along a straight, level trail. What is the magnitude and direction of the hiker's displacement for the whole walk?

A100 m, west

B700 m, east

C100 m, east

D500 m, west

Explanation

Displacement is a vector equal to the net change in position. Taking east as positive: (+300 m) + (−400 m) = −100 m, i.e. 100 m west. The 700 m option is the total DISTANCE traveled (a scalar that adds path lengths), not the displacement. The 100 m east option has the direction reversed, and 500 m comes from incorrectly combining the legs at a right angle when the motion is actually one-dimensional.

Question 10 of 10

MCQU3Topic 3.2Medium Calc Word

A worker pulls a sled a horizontal distance of 8.0 m by applying a constant 50 N force directed 30° above the horizontal. How much work does the worker's force do on the sled?

A3.5×10² J

B1.2×10³ J

C4.0×10² J

D2.0×10² J

Explanation

Work by a constant force is W = Fd·cosθ, where θ is the angle between the force and the displacement. W = (50)(8.0)cos30° = (50)(8.0)(0.866) = 3.5×10² J. The 4.0×10² J choice ignores the angle entirely (Fd). The 2.0×10² J choice uses sinθ instead of cosθ. Only the component of force along the displacement does work.

Free Response 1 · Section II

FRQMathematical RoutinesU2 Calc

A block of mass m₁ = 3.0 kg rests on a horizontal tabletop. A light, inextensible string runs horizontally from the block, passes over an ideal frictionless pulley at the edge of the table, and connects to a second block of mass m₂ = 2.0 kg that hangs freely. The coefficient of kinetic friction between m₁ and the tabletop is μ = 0.25. The system is released from rest and the hanging block descends. Use g = 9.8 m/s².

(a) In words, describe the free-body diagram for each block, naming every force that acts on it and its direction.

(b) Starting from Newton's second law applied to each block, derive a symbolic expression for the magnitude of the acceleration a of the system in terms of m₁, m₂, μ, and g.

(c) Calculate the numerical value of a.

(d) Calculate the tension in the string.

Free response is self-scored — work it out, then reveal the model answer and scoring checklist to compare.

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