Conservation of Linear Momentum — Worked Solutions

These solutions belong to Conservation of Linear Momentum. Each step states why it is useful. Try the questions first, then use this page to check your reasoning.

Question 1

Two carts collide on a low-friction track. Which condition makes their total horizontal momentum constant?

1. Select both carts and write p_f − p_i = J_external. Why: Internal collision impulses cancel in the combined system.

2. If the net horizontal external impulse is zero, p_f = p_i. Why: Sticking and kinetic-energy changes do not decide whether total momentum stays constant.

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Question 2

A 1 kg cart moves at +4 m/s and a 3 kg cart is stationary. What is their center-of-mass velocity?

1. Use the mass-weighted velocity: v_cm = (m₁v₁ + m₂v₂)/(m₁ + m₂). Why: The center of mass gives each object weight in proportion to its mass.

2. v_cm = [1(4) + 3(0)]/4 = +1 m/s. Why: The ordinary average of velocities would incorrectly treat unequal masses as equal.

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Question 3

An isolated moving system explodes. Which graph must stay horizontal?

1. For the isolated constant-mass system, p_total and M remain constant. Why: Zero net external force prevents a change in total momentum while internal forces redistribute it.

2. v_cm = p_total/M stays constant, so its graph is horizontal. Why: Fragment velocities and kinetic energy can change without changing the system’s center-of-mass motion.

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Question 4

Two riders of masses 40 kg and 60 kg move together at +2.0 m/s on a frictionless surface. They push apart. The 40 kg rider ends at +5.0 m/s. Find the other rider’s velocity and v_cm.

1. Select both riders and calculate p_i = (40 + 60)(2.0) = 200 kg·m/s. Why: Internal pushes cannot change this total on the frictionless surface.

2. Write 40(5.0) + 60v = 200, giving v = 0. Why: The initially moving system need not end with opposite velocity directions.

3. Use v_cm = p_total/M = 200/100 = +2.0 m/s. Why: Zero net external force keeps the center-of-mass velocity constant before, during, and after the push.

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Question 5

Two carts have masses m₁ = 1.0 kg and m₂ = 3.0 kg. A synchronized video gives positions in meters: at t = 0 s, (x₁, x₂) = (0, 6); at 1 s, (2, 5); at 2 s, (4, 4); at 3 s, (3.75, 3.75). The carts meet and stick at about 2 s. Decide whether these data support constant center-of-mass velocity across the interaction. Show the representation you would use, with scales and units, and explain why averaging the two positions without their masses could give a misleading conclusion.

1. Calculate x_cm = (x₁ + 3x₂)/4 at each shared time: 4.50, 4.25, 4.00, and 3.75 m. Why: The unequal masses require a mass-weighted position. Synchronized data make each pair a single system state.

2. Plot x_cm vertically in meters against t horizontally in seconds. The four points lie on x_cm = 4.50 − 0.25t, whose slope is −0.25 m/s before and after contact. Why: A straight center-of-mass position graph supports constant center-of-mass velocity even though the individual cart velocities change during sticking.

3. The unweighted positions average to 3.0, 3.5, 4.0, and 3.75 m, which do not lie on one constant-slope line. Interpret the weighted result within the position uncertainties, and check track level or friction before attributing any real curvature to the interaction. Why: An ordinary average wrongly treats the carts as equal masses. The weighted data support negligible net horizontal external force but do not by themselves prove that all measurement errors vanish.

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