AP Physics 1: Algebra-Based FRQ Practice

1. Kinematics and Free Fall

Question:

A ball is dropped from a height of 20 meters. Neglect air resistance.

(a) Calculate the time it takes for the ball to reach the ground.

(b) Calculate the velocity of the ball just before it hits the ground.

(c) If the ball were thrown upwards with an initial velocity of 5 m/s, calculate the time it would take for the ball to return to the original height.

Response Guidelines:

• Part (a): Use the kinematic equation:
  d=12gt2d = \frac{1}{2} g t^2
  Where:
  • d=20 md = 20 \, \text{m} (distance fallen)
  • g=9.8 m/s2g = 9.8 \, \text{m/s}^2 (acceleration due to gravity)
    Solve for time tt:
    t=2dg=2×209.8≈2.02 secondst = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2 \times 20}{9.8}} \approx 2.02 \, \text{seconds}
• Part (b): Use the kinematic equation:
  v=gtv = g t
  Where vv is the final velocity, t=2.02 secondst = 2.02 \, \text{seconds}, and g=9.8 m/s2g = 9.8 \, \text{m/s}^2.
  v=9.8×2.02≈19.8 m/sv = 9.8 \times 2.02 \approx 19.8 \, \text{m/s}
• Part (c): For the upward motion, use the equation:
  vf=vi+atv_f = v_i + a t
  Where vf=0 m/sv_f = 0 \, \text{m/s} (final velocity at the highest point), vi=5 m/sv_i = 5 \, \text{m/s}, and a=−9.8 m/s2a = -9.8 \, \text{m/s}^2.
  Solving for time tt:
  0=5+(−9.8)t0 = 5 + (-9.8) t
  t=59.8≈0.51 secondst = \frac{5}{9.8} \approx 0.51 \, \text{seconds} (time to reach the highest point)
  The total time for the upward and downward motion is twice this time:
  ttotal=2×0.51≈1.02 secondst_{\text{total}} = 2 \times 0.51 \approx 1.02 \, \text{seconds}

2. Newton's Second Law and Friction

Question:

A 5.0 kg block is pushed along a horizontal surface with a constant force of 30 N. The coefficient of kinetic friction between the block and the surface is 0.4.

(a) Calculate the frictional force acting on the block.

(b) Determine the acceleration of the block.

(c) If the force is increased to 40 N, calculate the new acceleration of the block.

Response Guidelines:

• Part (a):
  The frictional force can be calculated using:
  Ff=μkNF_f = \mu_k N
  Where:
  • μk=0.4\mu_k = 0.4 (coefficient of kinetic friction)
  • N=mg=5.0×9.8=49.0 NN = mg = 5.0 \times 9.8 = 49.0 \, \text{N} (normal force)
    Ff=0.4×49.0=19.6 NF_f = 0.4 \times 49.0 = 19.6 \, \text{N}
• Part (b):
  Use Newton’s second law:
  Fnet=maF_{\text{net}} = ma
  The net force is the applied force minus the frictional force:
  Fnet=30−19.6=10.4 NF_{\text{net}} = 30 - 19.6 = 10.4 \, \text{N}
  Now, solve for acceleration:
  a=Fnetm=10.45.0=2.08 m/s2a = \frac{F_{\text{net}}}{m} = \frac{10.4}{5.0} = 2.08 \, \text{m/s}^2
• Part (c):
  When the applied force is increased to 40 N, the new net force is:
  Fnet=40−19.6=20.4 NF_{\text{net}} = 40 - 19.6 = 20.4 \, \text{N}
  Now, solve for the new acceleration:
  a=Fnetm=20.45.0=4.08 m/s2a = \frac{F_{\text{net}}}{m} = \frac{20.4}{5.0} = 4.08 \, \text{m/s}^2

3. Work and Energy

Question:

A 2.0 kg object is launched vertically with an initial velocity of 10 m/s.

(a) Calculate the initial kinetic energy of the object.

(b) Determine the maximum height the object reaches.

(c) At the maximum height, calculate the potential energy of the object.

Response Guidelines:

• Part (a):
  The kinetic energy is given by the formula:
  KE=12mv2KE = \frac{1}{2} m v^2
  Where:
  • m=2.0 kgm = 2.0 \, \text{kg}
  • v=10 m/sv = 10 \, \text{m/s}
    KE=12×2.0×102=100 JKE = \frac{1}{2} \times 2.0 \times 10^2 = 100 \, \text{J}
• Part (b):
  At the maximum height, the object’s velocity becomes 0 m/s. Using conservation of energy (initial kinetic energy equals the work done to lift the object), we can solve for the height hh:
  KEinitial=PEmaxKE_{\text{initial}} = PE_{\text{max}}
  Where:
  • KEinitial=100 JKE_{\text{initial}} = 100 \, \text{J}
  • PEmax=mghPE_{\text{max}} = mgh
    100=2.0×9.8×h100 = 2.0 \times 9.8 \times h
    h=1002.0×9.8=5.10 mh = \frac{100}{2.0 \times 9.8} = 5.10 \, \text{m}
• Part (c):
  At the maximum height, all the kinetic energy has been converted into potential energy. Therefore, the potential energy at the maximum height is:
  PE=mgh=2.0×9.8×5.10=100 JPE = mgh = 2.0 \times 9.8 \times 5.10 = 100 \, \text{J}

Grading Rubric for AP Physics 1: Algebra-Based FRQs

1. Kinematics and Free Fall

• Part (a):
  • 0-1 points: Incorrect or incomplete calculation of time.
  • 2 points: Correct time calculation with proper formula and correct use of units.
• Part (b):
  • 0-1 points: Incorrect or incomplete calculation of velocity.
  • 2 points: Correct velocity calculation with proper formula and correct use of units.
• Part (c):
  • 0-1 points: Incorrect or incomplete calculation of time for upward motion.
  • 2 points: Correct time calculation for upward motion, including total time.

2. Newton's Second Law and Friction

• Part (a):
  • 0-1 points: Incorrect or incomplete identification of the frictional force.
  • 2 points: Correct identification of frictional force with proper formula and correct use of units.
• Part (b):
  • 0-1 points: Incorrect or incomplete calculation of acceleration.
  • 2 points: Correct acceleration calculation with proper formula and correct use of units.
• Part (c):
  • 0-1 points: Incorrect or incomplete calculation of new acceleration.
  • 2 points: Correct acceleration calculation with proper formula and correct use of units.

3. Work and Energy

• Part (a):
  • 0-1 points: Incorrect or incomplete calculation of initial kinetic energy.
  • 2 points: Correct kinetic energy calculation with proper formula and correct use of units.
• Part (b):
  • 0-1 points: Incorrect or incomplete calculation of maximum height.
  • 2 points: Correct maximum height calculation with proper formula and correct use of units.
• Part (c):
  • 0-1 points: Incorrect or incomplete calculation of potential energy at maximum height.
  • 2 points: Correct potential energy calculation at maximum height.

Sample Grading Breakdown (for one FRQ)

• Kinematics and Free Fall: 6 points
• Newton's Second Law and Friction: 6 points
• Work and Energy: 6 points
• Total: 18 points (for all three FRQs combined)