AP Physics C: Mechanics FRQ Practice

1. Kinematics and Newton's Laws

Question:

A block of mass 2.0 kg is sliding on a horizontal, frictionless surface with an initial velocity of 5.0 m/s. At time t = 0, a force of 10.0 N is applied to the block in the same direction as the velocity.

(a) Determine the acceleration of the block.

(b) Calculate the velocity of the block at time t = 3.0 s.

(c) What is the distance traveled by the block during the 3.0 seconds?

Response Guidelines:

• Part (a):
  The acceleration can be calculated using Newton’s second law:
  F = ma
  Where:
  F = 10.0 N (force)
  m = 2.0 kg (mass of the block)
  a = F/m = 10.0 N / 2.0 kg = 5.0 m/s²
  The acceleration of the block is 5.0 m/s².
• Part (b):
  The velocity of the block at time t = 3.0 s can be found using the kinematic equation:
  v = v₀ + at
  Where:
  v₀ = 5.0 m/s (initial velocity)
  a = 5.0 m/s² (acceleration)
  t = 3.0 s
  v = 5.0 + (5.0)(3.0) = 20.0 m/s
  The velocity at time t = 3.0 s is 20.0 m/s.
• Part (c):
  The distance traveled by the block can be found using the kinematic equation:
  d = v₀t + (1/2)at²
  Where:
  v₀ = 5.0 m/s (initial velocity)
  a = 5.0 m/s² (acceleration)
  t = 3.0 s
  d = (5.0)(3.0) + (1/2)(5.0)(3.0)² = 15.0 + 22.5 = 37.5 m
  The distance traveled by the block during the 3.0 seconds is 37.5 meters.

2. Work and Energy

Question:

A 5.0 kg object is dropped from rest from a height of 10.0 m. Neglecting air resistance, calculate the following:

(a) The speed of the object just before it hits the ground.

(b) The work done by the gravitational force.

(c) The time it takes for the object to reach the ground.

Response Guidelines:

• Part (a):
  To find the speed of the object just before it hits the ground, use the conservation of mechanical energy. The initial potential energy is converted to kinetic energy at the moment before the object hits the ground:
  mgh = (1/2)mv²
  Where:
  m = 5.0 kg (mass of the object)
  g = 9.8 m/s² (acceleration due to gravity)
  h = 10.0 m (height)
  v = √(2gh) = √(2 × 9.8 × 10.0) ≈ 14.0 m/s
  The speed of the object just before it hits the ground is approximately 14.0 m/s.
• Part (b):
  The work done by the gravitational force is equal to the change in gravitational potential energy, which is:
  W = mgh
  Where:
  m = 5.0 kg
  g = 9.8 m/s²
  h = 10.0 m
  W = (5.0)(9.8)(10.0) = 490 J
  The work done by the gravitational force is 490 joules.
• Part (c):
  The time it takes for the object to reach the ground can be found using the kinematic equation:
  h = (1/2)gt²
  Where:
  h = 10.0 m
  g = 9.8 m/s²
  t = √(2h/g) = √(2 × 10.0 / 9.8) ≈ 1.43 s
  The time it takes for the object to reach the ground is approximately 1.43 seconds.

3. Rotational Motion and Torque

Question:

A solid disk of mass 4.0 kg and radius 0.5 m rotates about an axis through its center with an angular velocity of 6.0 rad/s. A force of 12.0 N is applied tangentially to the edge of the disk.

(a) Calculate the moment of inertia of the disk.

(b) Determine the torque exerted by the force.

(c) Find the angular acceleration of the disk if the force is applied for 3.0 s.

Response Guidelines:

• Part (a):
  The moment of inertia for a solid disk rotating about an axis through its center is given by:
  I = (1/2)mr²
  Where:
  m = 4.0 kg (mass of the disk)
  r = 0.5 m (radius of the disk)
  I = (1/2)(4.0)(0.5)² = 1.0 kg·m²
  The moment of inertia of the disk is 1.0 kg·m².
• Part (b):
  The torque exerted by the force is given by:
  τ = F × r
  Where:
  F = 12.0 N (force applied)
  r = 0.5 m (radius of the disk)
  τ = (12.0)(0.5) = 6.0 N·m
  The torque exerted by the force is 6.0 N·m.
• Part (c):
  The angular acceleration can be found using the rotational form of Newton’s second law:
  τ = Iα
  Where:
  τ = 6.0 N·m (torque)
  I = 1.0 kg·m² (moment of inertia)
  α = τ / I = 6.0 / 1.0 = 6.0 rad/s²
  The angular acceleration of the disk is 6.0 rad/s².

Grading Rubric for AP Physics C: Mechanics FRQs

1. Kinematics and Newton's Laws

• Part (a):
  0-1 points: Incorrect or incomplete calculation of acceleration.
  2 points: Correct calculation of acceleration with proper formula and units.
• Part (b):
  0-1 points: Incorrect or incomplete calculation of velocity.
  2 points: Correct calculation of velocity with proper formula and units.
• Part (c):
  0-1 points: Incorrect or incomplete calculation of distance.
  2 points: Correct calculation of distance with proper formula and units.

2. Work and Energy

• Part (a):
  0-1 points: Incorrect or incomplete calculation of speed.
  2 points: Correct calculation of speed with proper formula and units.
• Part (b):
  0-1 points: Incorrect or incomplete calculation of work.
  2 points: Correct calculation of work with proper formula and units.
• Part (c):
  0-1 points: Incorrect or incomplete calculation of time.
  2 points: Correct calculation of time with proper formula and units.

3. Rotational Motion and Torque

• Part (a):
  0-1 points: Incorrect or incomplete calculation of moment of inertia.
  2 points: Correct calculation of moment of inertia with proper formula and units.
• Part (b):
  0-1 points: Incorrect or incomplete calculation of torque.
  2 points: Correct calculation of torque with proper formula and units.
• Part (c):
  0-1 points: Incorrect or incomplete calculation of angular acceleration.
  2 points: Correct calculation of angular acceleration with proper formula and units.