Linear Momentum — Worked Solutions

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

Question 1

Two moving carts have a total momentum of zero. What must be true?

1. Write p_total = p₁ + p₂. Why: Total momentum is a vector sum, so direction belongs in each term.

2. Equal and opposite momenta give p_total = 0 even with both carts moving. Why: Cancellation of vectors does not make each vector zero; kinetic energies remain positive.

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Question 2

A cart moves left. You choose right as positive. Which describes its momentum?

1. Choose right as positive and write p = mv. Why: The sign convention lets us represent leftward velocity with a negative component.

2. Since m is positive and v is negative, p is negative and points left. Why: Multiplication by mass preserves the velocity direction; no continuing force is needed for steady motion.

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Question 3

A light cart has mass m and speed v. A heavier cart has mass 2m and speed v/2 in the same direction. Compare their momenta.

1. For the light cart, p = mv; for the heavy cart, p = (2m)(v/2). Why: Compare using the same direction and reference frame.

2. Cancel the factor of two to get p_heavy = mv = p_light. Why: Momentum depends on the product of mass and velocity, while kinetic energy has a different speed dependence.

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Question 4

A 0.40 kg cart and a 0.60 kg cart start at rest on a level low-friction track. A spring between them is released. The lighter cart moves left at 3.0 m/s. Find the other cart’s velocity and compare their kinetic energies.

1. Choose right positive and include both carts and the spring. Why: The spring forces are internal and horizontal external impulse is negligible.

2. Write 0 = 0.40(−3.0) + 0.60v and solve v = +2.0 m/s. Why: Initial momentum is zero, so the outgoing cart momenta must cancel.

3. Calculate K_light = ½(0.40)(3.0²) = 1.8 J and K_heavy = ½(0.60)(2.0²) = 1.2 J. Why: Kinetic energies add as scalars; the lighter cart has greater speed for equal momentum magnitude.

4. The spring supplied 3.0 J of kinetic energy in the ideal model. Why: The increase in total kinetic energy comes from spring potential energy, not an increase in total momentum.

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Question 5

Two carts travel on a level track in one camera frame. Cart A has mass 0.40 kg and velocity +3.0 m/s; cart B has mass 0.60 kg and velocity −1.0 m/s. Draw their momentum arrows to scale and show the system total. A second camera rides rightward at 1.0 m/s relative to the track. Without changing the physical motion, what total momentum would that camera measure? Explain why comparing one cart from each camera would be a poor test of momentum conservation.

1. In the track frame, p_A = 0.40(3.0) = +1.2 kg·m/s and p_B = 0.60(−1.0) = −0.60 kg·m/s; the total arrow is +0.60 kg·m/s. Why: Use signed components from the same frame. For an arrow scale of 0.20 kg·m/s per division, the arrows have lengths six right, three left, and three right for the total.

2. Subtract the camera velocity from both velocities: v_A′ = +2.0 m/s and v_B′ = −2.0 m/s. The new total is 0.40(2.0) + 0.60(−2.0) = −0.40 kg·m/s. Why: Momentum depends on reference frame. The whole-system change between frames is −M v_camera = −1.0 kg·m/s, which agrees with the two totals.

3. Do not mix a track-frame cart momentum with a camera-frame cart momentum in one system sum. Why: Such a sum is not the momentum of a single system state in any stated frame. Test conservation by using one frame consistently before and after an interaction.

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