Change in Momentum and Impulse — Worked Solutions

These solutions belong to Change in Momentum and Impulse. Each step states why it is useful. Try the questions first, then use this page to check your reasoning.

Question 1

A 0.20 kg ball changes velocity from +6.0 m/s to −4.0 m/s. What is its momentum change?

1. Write Δp = m(v_f − v_i), with right positive. Why: A momentum change subtracts signed velocities, not their magnitudes.

2. Δp = 0.20[−4.0 − (+6.0)] = −2.0 kg·m/s. Why: The ball reverses, so its velocity change is −10 m/s; the impulse is leftward.

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Question 2

A force–time triangle has base 0.20 s and height 8.0 N above zero. What impulse does it deliver?

1. Find the signed force–time area: ½ × duration × peak force. Why: Area under net force versus time is impulse, and the region is triangular.

2. J = ½(0.20 s)(8.0 N) = +0.80 N·s. Why: The area is above zero, so its sign is positive. Force height alone cannot give impulse.

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Question 3

A momentum–time graph is horizontal at +5 kg·m/s. What is the net external force?

1. Net force is the slope of momentum against time. Why: F_net = rate of change of p.

2. A horizontal line has zero slope, so F_net = 0. Why: The nonzero momentum value describes continuing motion, not an unbalanced force.

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Question 4

A 2.0 kg object is moving downward at 3.0 m/s. A hand brings it to rest in 0.40 s. Find the average upward hand force, taking g = 9.8 m/s².

1. Choose upward positive and write Δp = m(v_f − v_i) = 2.0[0 − (−3.0)] = +6.0 N·s. Why: The object’s initial downward velocity must have a negative sign.

2. Divide by time: F_net,avg = 6.0/0.40 = +15 N. Why: Net external impulse changes momentum.

3. Draw upward hand force and downward gravity, giving F_hand − mg = 15 N. Why: The requested contact force is not the same as the net force.

4. F_hand = 15 + 2.0(9.8) = 34.6 N upward. Why: The hand must both offset gravity and produce the upward momentum change.

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Question 5

A 0.50 kg cart approaches a bumper at +2.0 m/s and leaves at −1.0 m/s. Contact lasts about 0.020 s. You have a motion sensor and can record the bumper force at either 50 samples/s or 2,000 samples/s. Choose a force setting and describe what measurements would let you test the impulse–momentum theorem, including one check for a systematic error. A softer bumper produces the same endpoint velocities over 0.040 s. Predict how its net impulse and average net force compare with the first bumper; justify the signs and ratios.

1. Choose 2,000 samples/s: a 0.020 s contact contains about 40 sampling intervals, while 50 samples/s gives only about one. Why: The short pulse needs enough samples to estimate its area and changing shape. One interval can miss the peak or most of the pulse, depending on sample timing.

2. Measure signed velocities just outside contact and signed net force throughout the same interval. Compare m(v_f − v_i) = 0.50(−1.0 − 2.0) = −1.5 N·s with the force–time area. Check the sensor’s zero before and after the trial and account for any horizontal force other than the measured contact force. Why: Matched intervals test one interaction. A force offset produces a systematic area error; contact force equals net horizontal force only when other horizontal forces are negligible or included.

3. The first average net force is −1.5/0.020 = −75 N. The softer bumper still gives −1.5 N·s, but its average net force is −1.5/0.040 = −37.5 N. Why: Equal masses and endpoint velocities give equal signed momentum changes. Twice the contact duration halves average force magnitude at fixed impulse.

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