Elastic and Inelastic Collisions — Worked Solutions

These solutions belong to Elastic and Inelastic Collisions. Each step states why it is useful. Try the questions first, then use this page to check your reasoning.

Question 1

Two carts bounce apart after contact. Must the collision be elastic?

1. Elastic means K_total,f = K_total,i. Why: The collision type is defined by total kinetic energy, not by whether the objects separate.

2. A rebound can leave less kinetic energy, so compare measured endpoint energies. Why: Only sticking identifies the perfectly inelastic subtype; bouncing does not prove elasticity.

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Question 2

A mass m at speed v sticks to a stationary mass 3m with negligible external impulse. What is the shared final speed?

1. With negligible external impulse, mv = (m + 3m)v_f. Why: The stationary target contributes no initial momentum and the stuck masses share v_f.

2. Solve 4mv_f = mv to obtain v_f = v/4. Why: The combined moving mass is four times the initially moving mass.

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Question 3

An isolated collision keeps total momentum but decreases total kinetic energy. What is the best conclusion?

1. Momentum can remain constant while total kinetic energy decreases. Why: The momentum balance and energy classification track different quantities.

2. Classify the collision as inelastic and include the converted internal energy. Why: Energy can become deformation, sound, and thermal energy without being destroyed.

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Question 4

A 0.40 kg cart at +3.0 m/s sticks to a stationary 0.80 kg cart. Find the final speed and the kinetic-energy decrease.

1. Include both carts and write 0.40(3.0) = (0.40 + 0.80)v_f. Why: Sticking gives one final velocity, with negligible external impulse.

2. Solve v_f = +1.0 m/s. Why: The initial momentum is +1.2 kg·m/s and the combined mass is 1.20 kg.

3. K_i = ½(0.40)(3.0²) = 1.8 J; K_f = ½(1.20)(1.0²) = 0.60 J. Why: The stationary cart has no initial kinetic energy, but both masses contribute afterward.

4. The decrease is 1.2 J and remaining fraction is 1/3. Why: Converted internal energy completes the energy ledger while momentum stays +1.2 kg·m/s.

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Question 5

Two separate trials use the same initial total kinetic-energy interval, 1.16–1.24 J. In trial A the final interval is 1.12–1.22 J; in trial B it is 0.80–0.90 J. These intervals state the possible values given the measurement uncertainty. In both trials the carts separate after contact and the measured total momenta agree before and after. Which trial supports a clear inelastic classification? What can you claim about trial A? Choose one additional measurement improvement that would make the energy classification more useful and explain its connection to the calculation.

1. Compare energy intervals, not just central values: A overlaps the initial interval from 1.16 to 1.22 J; B has no overlap and even its largest final value is below the smallest initial value. Why: A kinetic-energy difference must exceed the stated measurement uncertainty to support a clear decrease.

2. Trial B supports an inelastic collision. Trial A is consistent with an elastic model, but these data cannot rule out a small kinetic-energy decrease or prove exact elasticity. Why: Separating carts do not establish elasticity, and momentum agreement is compatible with either collision type.

3. For example, calibrate the video distance scale and estimate endpoint velocities from several frames immediately outside contact, repeating trials with the same initial setup. Why: K contains v², so improved velocity estimates reduce a major source of energy uncertainty. Measurements far from contact can include frictional changes that are not part of the collision.

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