The problems are on the impulse momentum theorem worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Linear momentum is the product of mass and velocity, p = mv, a vector in kg m/s. These worked answers correct the 20 problems of the AP Physics 1 worksheet on impulse momentum theorem, from momentum of moving objects to comparing two ways to find impulse: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the formula, the substitution with signs and the final result in bold. Check your sign convention first; if your magnitude is right but the sign differs, make sure you chose the positive direction consistently.
Level: AP Physics 1 (algebra-based) · Curriculum: AP P1 Unit 4
Problem 1 Momentum of moving objects
1. p = mv = 1200 × 15 = 1.8 × 10⁴ kg m/s.
2. p = 0.058 × 50 = 2.9 kg m/s.
3. The car has the larger momentum: 18 000 ÷ 2.9 ≈ 6.2 × 10³ times larger.
Problem 2 Impulse from a force
1. J = FΔt = 250 × 0.040 = 10 N s.
2. It points in the direction of the force. 1 N s = 1 (kg m/s²) × s = 1 kg m/s.
Problem 3 Change in momentum
1. Δp = 2.0 × (7.0 − 3.0) = 8.0 kg m/s.
2. Impulse equals the change in momentum: 8.0 N s in the direction of motion.
3. F = Δp/Δt = 8.0 ÷ 0.50 = 16 N.
Problem 4 Matching units
1. 1 N s = 1 (kg m/s²) s = 1 kg m/s.
2. Both N s and kg m/s; N/s is the unit of a rate of change of force, not impulse. N s and kg m/s.
Problem 5 Speed from impulse
1. v = J/m = 9.0 ÷ 0.45 = 20 m/s.
2. v = 9.0 ÷ 0.90 = 10 m/s, half as fast for double the mass.
Problem 6 Stopping a cricket ball
1. Δp = 0.16 × (0 − 30) = −4.8 kg m/s.
2. F = Δp/Δt = −4.8 ÷ 0.020 = −2.4 × 10² N, a force of 240 N against the ball’s motion.
Problem 7 Bouncing off a wall
1. Δp = 0.15 × (−12 − 20) = −4.8 kg m/s.
2. F = −4.8 ÷ 0.010 = −4.8 × 10² N: 480 N directed away from the wall (opposite to the initial motion).
Problem 8 Area under a graph
1. Area of the triangle = ½ × 2.0 × 60 = 60 N s.
2. Triangle 60 + rectangle (3.0 × 60 = 180) + triangle (½ × 1.0 × 60 = 30) = 270 N s.
3. v = J/m = 270 ÷ 30 = 9.0 m/s.
Problem 9 Airbag in a crash
Δp = 70 × (0 − 14) = −980 kg m/s.
1. F = −980 ÷ 0.10 = −9.8 × 10³ N.
2. F = −980 ÷ 0.020 = −4.9 × 10⁴ N.
3. 49 000 ÷ 9800 = 5 times larger.
Problem 10 Impulse of gravity
1. J = mgΔt = 0.20 × 9.81 × 1.5 = 2.9 N s (2.94 N s).
2. v = J/m = 2.943 ÷ 0.20 = 15 m/s (14.7 m/s), which matches gt = 9.81 × 1.5 = 14.7 m/s.
Problem 11 Force that reverses
1. Area = 40 × 3.0 = +120 N s.
2. Area = −20 × 4.0 = −80 N s.
3. Total = +40 N s. Δv = J/m = 40 ÷ 10 = 4.0 m/s, so vf = 2.0 + 4.0 = +6.0 m/s.
Problem 12 Clay and rubber
1. Clay: Δp = 0.30 × (0 − 6.0) = −1.8 kg m/s. Rubber: Δp = 0.30 × (−6.0 − 6.0) = −3.6 kg m/s.
2. The rubber ball receives twice the impulse, because its velocity changes by 12 m/s while the clay ball’s changes by only 6.0 m/s. The wall must reverse the rubber ball, not just stop it.
Problem 13 Volleyball bump
1. Δp = 0.27 × (0 − 12) = −3.2 kg m/s (−3.24).
2. F = −3.24 ÷ 0.050 = −65 N.
3. F = −3.24 ÷ 0.15 = −22 N, one third of the first force.
Problem 14 Accelerating car
1. Δp = 1200 × (25 − 10) = 1.8 × 10⁴ kg m/s.
2. Δt = Δp/F = 18 000 ÷ 3000 = 6.0 s.
Problem 15 Bouncing tennis ball
1. Before: v = √(2 × 9.81 × 1.8) = 5.94 m/s downward. After: v = √(2 × 9.81 × 1.0) = 4.43 m/s upward. 5.9 m/s and 4.4 m/s.
2. Take up as positive: J = 0.060 × (4.43 − (−5.94)) = 0.060 × 10.37 = 0.62 N s upward.
3. Net force = 0.622 ÷ 0.012 = 51.9 N, and the floor also supports the weight of 0.59 N, so the floor force is about 52 N.
Problem 16 Hose against a wall
1. In one second 3.0 kg changes velocity from 8.0 m/s to 0: |Δp| = 3.0 × 8.0 = 24 kg m/s, so F = 24 ÷ 1.0 = 24 N.
2. Now Δv = 16 m/s, so |Δp| = 3.0 × 16 = 48 kg m/s and F = 48 N.
3. A bounce reverses the velocity, so the change in momentum per second is twice as great, requiring twice the force.
Problem 17 Jumping from the floor
1. J = Δp = 60 × 2.8 = 1.7 × 10² N s (168 N s).
2. Fnet = 168 ÷ 0.30 = 5.6 × 10² N.
3. The floor must also support the weight: N = Fnet + mg = 560 + 589 = 1.1 × 10³ N.
4. h = v²/2g = 2.8² ÷ 19.62 = 0.40 m.
Problem 18 Trolley data table
1. Area by trapezoids = 0.1 × (0/2 + 2 + 6 + 8 + 6 + 2 + 0/2) = 0.1 × 24 = 2.4 N s.
2. Δv = J/m = 2.4 ÷ 0.80 = 3.0 m/s, so vf = 0.50 + 3.0 = 3.5 m/s.
3. Where the real curve is convex (a rounded peak), straight segments between points cut off part of the area under the curve, so the estimate is slightly too small.
Problem 19 Spot the error
1. Momentum is a vector. The ball reverses direction, so p changes from +0.80 to −0.80 kg m/s; only the speed is unchanged.
2. J = 0.20 × (−4.0 − 4.0) = −1.6 N s.
3. F = −1.6 ÷ 0.025 = −64 N, directed away from the wall.
Problem 20 Comparing two ways to find impulse
1. Δp = 0.50 × (−0.40 − 0.60) = −0.50 N s, magnitude 0.50 N s.
2. Difference = (0.50 − 0.48) ÷ 0.50 = 4 %.
3. Friction or a small tilt in the track changes the velocity, and the sensor may miss the first and last part of the contact. Using a level track, a higher sampling rate and several repeats reduce the difference.
