The questions are on the impulse momentum theorem test. Take it under real conditions before reading the mark scheme.
Quick answer: Linear momentum is the product of mass and velocity, p = mv, a vector in kg m/s. This mark scheme gives the model answers and the marks for the 8 questions of the AP Physics 1 test on impulse momentum theorem, from definitions to sticking or bouncing, with the equivalent answers to accept.
Model answers with the mark scheme. Award method marks where a signed momentum change is set up correctly even if the arithmetic is slightly wrong, and apply the unit mark in each calculation.
Level: AP Physics 1 (algebra-based) · Curriculum: AP P1 Unit 4
Question 1 Definitions (2 marks)
a. Impulse is the product of the net force and the time for which it acts, J = FΔt (or the area under a force–time graph).
b. Impulse equals the change in momentum, J = Δp.
Marking: 1 mark for each statement; the first must include both force and time.
Question 2 Bullet and car (2 marks)
a. Bullet: p = 0.0040 × 400 = 1.6 kg m/s. Car: p = 1500 × 1.0 = 1500 kg m/s.
b. 1500 ÷ 1.6 = 9.4 × 10² times larger.
Marking: 1 mark for both momenta, 1 mark for the ratio.
Question 3 Goalkeeper’s catch (3 marks)
a. Δp = 0.43 × (0 − 20) = −8.6 kg m/s.
b. Impulse = −8.6 N s (magnitude 8.6 N s against the motion).
c. F = −8.6 ÷ 0.040 = −2.2 × 10² N.
Marking: 1 mark each; allow the magnitude if the direction is stated.
Question 4 Skater and a graph (3 marks)
a. Area = ½ × 2 × 40 + 2 × 40 + ½ × 1 × 40 = 40 + 80 + 20 = 140 N s.
b. v = J/m = 140 ÷ 70 = 2.0 m/s.
Marking: (a) 1 mark for method (splitting the area), 1 mark for the result; (b) 1 mark.
Question 5 Airbags (3 marks)
The driver’s momentum must drop from its initial value to zero, so the impulse needed, Δp, is fixed. Since Favg = Δp/Δt, an airbag that increases the stopping time gives a smaller average force on the driver than a rigid surface, reducing injury.
Marking: 1 mark for fixed change in momentum or impulse; 1 mark for the time of contact increasing; 1 mark for lower force as the consequence.
Question 6 Rebounding ball (3 marks)
a. Δp = 0.25 × (−6.0 − 8.0) = −3.5 kg m/s.
b. F = −3.5 ÷ 0.020 = −1.8 × 10² N (175 N away from the wall).
Marking: (a) 1 mark for a correct subtraction with signs, 1 mark for the answer; (b) 1 mark.
Question 7 Car acceleration (2 marks)
a. Δp = 1500 × (20 − 8.0) = 1.8 × 10⁴ kg m/s.
b. F = 18 000 ÷ 5.0 = 3.6 × 10³ N.
Marking: 1 mark each.
Question 8 Sticking or bouncing (2 marks)
The rubber ball receives the larger impulse. The clay ball’s velocity changes from v to 0, whereas the rubber ball’s changes from v to −v, a change of 2v, so Δp (and the impulse) is about twice as large.
Marking: 1 mark for naming rubber; 1 mark for comparing velocity changes.
