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AP Physics 1 – Practice Set

Kinematics, dynamics, energy, momentum, rotation, SHM. MCQ and FRQ style.

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12 questions  ·  45 marks  ·  answers included  ·  AP Physics 1 (College Board)
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This worksheet covers:

  • Kinematics
  • Dynamics
  • Energy
  • Momentum
  • Rotation and simple harmonic motion in AP Physics 1 style
  • Mixing calculation
  • Free-response explanation questions

A calculator is allowed; take $g = 9.81\,\text{m s}^{-2}$.

Questions

Section A – Kinematics and Dynamics

1

A ball is launched horizontally with a speed of $15\,\text{m s}^{-1}$ from the top of a cliff of height $20\,\text{m}$.(a) Calculate the time taken for the ball to reach the ground.(b) Calculate the horizontal distance travelled by the ball before landing.

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(a) $h = \tfrac{1}{2}gt^2 \Rightarrow t = \sqrt{2h/g} = \sqrt{2(20)/9.81} \approx 2.02\,\text{s}$(b) $x = v_0 t = 15 \times 2.02 \approx 30.3\,\text{m}$
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2

A block of mass $5\,\text{kg}$ rests on a frictionless incline at $30^\circ$ to the horizontal.(a) Calculate the block’s acceleration down the incline.(b) Calculate the normal force on the block.

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(a) $a = g\sin30^\circ = 9.81 \times 0.5 \approx 4.90\,\text{m s}^{-2}$(b) $N = mg\cos30^\circ = 5 \times 9.81 \times 0.866 \approx 42.5\,\text{N}$
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3

A $10\,\text{kg}$ object experiences two perpendicular forces: $30\,\text{N}$ directed east and $40\,\text{N}$ directed north.(a) Calculate the magnitude of the net force.(b) Calculate the magnitude of the object’s acceleration.(c) Calculate the angle of the net force measured from east.

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(a) $F_{net} = \sqrt{30^2+40^2} = 50\,\text{N}$(b) $a = F_{net}/m = 50/10 = 5.0\,\text{m s}^{-2}$(c) $\theta = \tan^{-1}(40/30) \approx 53.1^\circ$ north of east
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4

A student pushes a box across the floor at constant velocity, then stops pushing and the box decelerates to rest.Using Newton’s first and second laws, explain why the box decelerates once the student stops pushing, and explain what this tells you about the friction force compared with the push force during the constant-velocity phase.

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While being pushed at constant velocity the net force is zero, so the push force must equal the friction force in magnitude (Newton’s first law). Once the push force is removed, friction is now the only horizontal force, so by Newton’s second law the box has a net force (and acceleration) opposing its motion, causing it to decelerate to rest.
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Section B – Energy and Momentum

5

A spring with spring constant $k = 200\,\text{N m}^{-1}$ is compressed by $0.15\,\text{m}$ and used to launch a block of mass $2\,\text{kg}$ on a frictionless horizontal surface.(a) Calculate the elastic potential energy stored in the spring.(b) Calculate the speed of the block as it leaves the spring.

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(a) $E_p = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(200)(0.15)^2 = 2.25\,\text{J}$(b) By conservation of energy, $\tfrac{1}{2}mv^2 = 2.25 \Rightarrow v = \sqrt{2(2.25)/2} = 1.5\,\text{m s}^{-1}$
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6

A $3\,\text{kg}$ cart moving at $4\,\text{m s}^{-1}$ collides head-on with a $2\,\text{kg}$ cart moving at $2\,\text{m s}^{-1}$ in the opposite direction. The carts stick together after the collision.Calculate the common velocity of the carts immediately after the collision, stating its direction.

Show answer
Taking the $3\,\text{kg}$ cart’s direction as positive: $v_f = \dfrac{m_1u_1+m_2u_2}{m_1+m_2} = \dfrac{(3)(4)+(2)(-2)}{5} = 1.6\,\text{m s}^{-1}$, in the direction of the $3\,\text{kg}$ cart’s original motion.
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7

A $4\,\text{kg}$ box, initially at rest, is pushed with a constant horizontal force of $20\,\text{N}$ across a rough floor (coefficient of kinetic friction $\mu = 0.2$) through a distance of $5\,\text{m}$.(a) Calculate the friction force on the box.(b) Calculate the net work done on the box.(c) Calculate the final speed of the box using the work–energy theorem.

Show answer
(a) $f = \mu mg = 0.2 \times 4 \times 9.81 \approx 7.85\,\text{N}$(b) $W_{net} = (F-f)d = (20-7.85)(5) \approx 60.8\,\text{J}$(c) $\tfrac{1}{2}mv^2 = W_{net} \Rightarrow v = \sqrt{2(60.8)/4} \approx 5.51\,\text{m s}^{-1}$
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8

Explain why total momentum is conserved in a collision between two carts on a track, even though kinetic energy may not be conserved. Refer to Newton’s third law in your explanation.

Show answer
During the collision the carts exert equal and opposite forces on each other (Newton’s third law) for the same time interval, so they experience equal and opposite impulses. Since the external horizontal forces on the system are negligible, the total momentum of the two-cart system is unchanged, even though some kinetic energy may be converted to heat, sound or deformation.
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Section C – Rotation and Simple Harmonic Motion

Extended response
9

A uniform disk of mass $4\,\text{kg}$ and radius $0.25\,\text{m}$ (moment of inertia $I = \tfrac{1}{2}MR^2$) starts at rest and a constant torque of $6\,\text{N m}$ is applied about its central axis.(a) Calculate the moment of inertia of the disk.(b) Calculate the angular acceleration of the disk.

Show answer
(a) $I = \tfrac{1}{2}(4)(0.25)^2 = 0.125\,\text{kg m}^2$(b) $\alpha = \tau/I = 6/0.125 = 48\,\text{rad s}^{-2}$
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10

A figure skater spinning with moment of inertia $4\,\text{kg m}^2$ at an angular velocity of $2\,\text{rad s}^{-1}$ pulls in her arms, reducing her moment of inertia to $1\,\text{kg m}^2$.(a) Calculate her new angular velocity, assuming no external torque acts on her.(b) State and explain what happens to her rotational kinetic energy as she pulls her arms in.

Show answer
(a) Conservation of angular momentum: $I_1\omega_1 = I_2\omega_2 \Rightarrow \omega_2 = \dfrac{(4)(2)}{1} = 8\,\text{rad s}^{-1}$(b) Rotational kinetic energy increases (from $\tfrac{1}{2}(4)(2)^2=8\,\text{J}$ to $\tfrac{1}{2}(1)(8)^2=32\,\text{J}$) because the skater does positive work pulling her arms inward against the centripetal requirement, converting internal (muscular) energy into rotational kinetic energy; angular momentum, not kinetic energy, is what is conserved.
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11

A $0.5\,\text{kg}$ block is attached to a spring of spring constant $50\,\text{N m}^{-1}$ and undergoes simple harmonic motion on a frictionless horizontal surface with amplitude $0.10\,\text{m}$.(a) Calculate the period of oscillation.(b) Calculate the maximum speed of the block.

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(a) $T = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.5/50} \approx 0.628\,\text{s}$(b) $\omega = \sqrt{k/m} = \sqrt{100} = 10\,\text{rad s}^{-1}$, so $v_{max} = A\omega = 0.10 \times 10 = 1.0\,\text{m s}^{-1}$
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12

A simple pendulum and a mass–spring system both oscillate with simple harmonic motion.Explain why the period of a simple pendulum does not depend on its mass, while the period of a mass–spring system does not depend on the amplitude of oscillation. Support your explanation with reference to the relevant period equations.

Show answer
For a simple pendulum, $T = 2\pi\sqrt{L/g}$: the restoring force ($mg\sin\theta$) and the inertia ($ma$) both depend on mass in the same way, so mass cancels out and does not appear in the period equation. For a mass–spring system, $T = 2\pi\sqrt{m/k}$: amplitude does not appear because both the restoring force ($-kx$) and the resulting acceleration scale linearly with displacement, so a larger amplitude produces a proportionally larger restoring force and acceleration, keeping the time for one cycle unchanged.
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AP Physics 2 – Practice SetAP Physics C: Mechanics – Practice SetAP Physics C: E&M – Practice Set

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