Unit 4 · Topic 4.3 Beta

Conservation of Linear Momentum

Internal forces cancel in pairs, so only external forces change a system's momentum.

Practice 2: Mathematical RoutinesPractice 3: Scientific Questioning and Argumentation

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Question set for this topic

Part 1 · Hook

Why this matters

A rifle kicks back when it fires, a skater drifts backward after throwing a ball, and two pool balls share their motion when they collide. All three follow one rule: with no net external push, a system cannot change its total momentum. Choosing the system decides when the rule applies.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Newton's third law says the force A exerts on B is

  1. equal in size and opposite to the force B exerts on A
  2. larger when A is heavier
  3. always zero
  4. in the same direction as B on A
Show the answer

Third-law pairs are equal and opposite.

  • Correct: equal in size and opposite to the force B exerts on A:
  • larger when A is heavier:
  • always zero:
  • in the same direction as B on A:

2. Impulse equals

  1. the change in momentum
  2. the momentum
  3. the change in kinetic energy
  4. force divided by time
Show the answer

J = Δp.

  • Correct: the change in momentum:
  • the momentum:
  • the change in kinetic energy:
  • force divided by time:

3. The velocity of a system's center of mass is

  1. Σp/Σm
  2. the average of the speeds
  3. Σm/Σp
  4. the fastest object's velocity
Show the answer

v_cm = total momentum / total mass.

  • Correct: Σp/Σm:
  • the average of the speeds:
  • Σm/Σp:
  • the fastest object's velocity:

Part 4 · See it

See it first

Cart A hits cart B on a level track; each pushes the other with equal and opposite forces. Top: a dashed boundary around cart A alone; B's push is external, so A's momentum changes. Bottom: a boundary around both carts; the pushes are internal and the total momentum stays constant.
For cart A alone, B's push is external and A's momentum changes; for both carts, the pushes are internal and the total momentum stays constant. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. Two objects in a system push on each otherthe third law makes the forces, and their impulses, equal and opposite
  2. The internal impulses are equal and oppositeone object's momentum gain is the other's loss, so internal forces cannot change the total
  3. The net external force on the system is zerothe system's total momentum, and its center of mass's velocity, stay constant
  4. An object is left out of the systemits forces become external and the system's momentum can change

Part 6 · Key ideas

Key ideas

  • Internal forces come in third-law pairs, so their impulses cancel: they move momentum between parts but never change the total.
  • Conservation of momentum: when the net external force on a system is zero, its total momentum is constant: Σp_before = Σp_after.
  • The choice of system decides whether momentum is constant. A cart alone in a collision is not; both carts together are.
  • With no net external force, the center of mass moves at constant velocity, whatever the parts do.
  • Justify with the condition: "the net external force on the two-cart system is zero", never "because momentum is conserved".

Part 7 · Misconception

A common mistake

The wrong idea: Momentum is constant because of conservation of momentum.

What actually happens: Momentum is constant for a system on which the net external force is zero. Name the system and say why the external forces add to zero (they balance, or are negligible during a brief collision).

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Diagram

Cart A hits cart B

BeforeA0.50 kgB0.25 kg+0.80 m/sat restAfterA0.50 kgB0.25 kg+0.30 m/svB = ?positive to the right →

The track is level and friction is negligible. Positive is to the right.

1. What is cart B's velocity after the collision, in m/s?

Type a number in m/s.

Show the answer

System: both carts. External forces (gravity, normal) balance and friction is negligible, so p is constant: (0.50)(0.80) = (0.50)(0.30) + (0.25)v_B, giving v_B = +1.0 m/s.

  • Answer: 1.0 m/s

2. Which is the best justification for using momentum conservation here?

  1. The net external force on the two carts is zero: vertical forces balance and friction is negligible
  2. Momentum is conserved because of the law of conservation of momentum, which holds for any collision
  3. The carts exert equal forces on each other, so each cart's momentum is constant
  4. Kinetic energy is conserved in the collision, so momentum is conserved as well
Show the answer

A system's momentum is constant when the net external force on it is zero. State the system and why the external forces add to zero.

  • Correct: The net external force on the two carts is zero: vertical forces balance and friction is negligible: Right: it names the system and the condition.
  • Momentum is conserved because of the law of conservation of momentum, which holds for any collision: Circular: it states the law without the condition that makes it apply.
  • The carts exert equal forces on each other, so each cart's momentum is constant: Each cart's momentum changes. The equal and opposite forces make the changes cancel in the total.
  • Kinetic energy is conserved in the collision, so momentum is conserved as well: Momentum does not follow from kinetic energy; and K need not be conserved here.

3. How does the impulse cart A exerts on cart B compare with the impulse cart B exerts on cart A?

  1. B's impulse on A is larger, because A is heavier
  2. Equal in size and opposite in direction
  3. A's impulse on B is larger, because B speeds up more
  4. Equal in size and in the same direction
Show the answer

Δp_A = (0.50)(0.30 − 0.80) = −0.25 kg·m/s; Δp_B = (0.25)(1.0 − 0) = +0.25 kg·m/s. Equal and opposite, so the total is unchanged.

  • B's impulse on A is larger, because A is heavier: Third-law forces are equal regardless of the masses.
  • Correct: Equal in size and opposite in direction: Right: the forces are a third-law pair acting for the same time.
  • A's impulse on B is larger, because B speeds up more: B's larger velocity change comes from its smaller mass, not from a larger impulse.
  • Equal in size and in the same direction: Third-law forces point in opposite directions.

4. Take cart A alone as the system. Which is correct?

  1. A's momentum is constant, because the track is frictionless
  2. A's momentum is constant, because the collision is brief
  3. A's momentum is not constant, because B exerts an external force on it
  4. A has no momentum after the collision, since its momentum went to B
Show the answer

Whether momentum is constant depends on the system. Cart A alone feels B's external push; the two-cart system does not.

  • A's momentum is constant, because the track is frictionless: Friction is not the only external force. B's push on A changes A's momentum.
  • A's momentum is constant, because the collision is brief: Brief or not, B's push gives A an impulse of −0.25 N·s.
  • Correct: A's momentum is not constant, because B exerts an external force on it: Right: with B outside the system, B's push is external.
  • A has no momentum after the collision, since its momentum went to B: A still moves at 0.30 m/s, so p_A = 0.15 kg·m/s.

5. How does the velocity of the center of mass of the two carts after the collision compare with before?

  1. It increases, because B speeds up
  2. It decreases, because A slows down
  3. It is zero after the collision
  4. It is the same, about +0.53 m/s
Show the answer

v_cm = (0.40 kg·m/s)/(0.75 kg) = 0.53 m/s both times. With no net external force, the center of mass keeps a constant velocity.

  • It increases, because B speeds up: A slows down by the matching amount of momentum; the total is unchanged.
  • It decreases, because A slows down: B gains exactly the momentum A loses.
  • It is zero after the collision: The total momentum is still 0.40 kg·m/s, not zero.
  • Correct: It is the same, about +0.53 m/s: Right: v_cm = Σp/Σm = 0.40/0.75 before and after.

6. A 60. kg skater at rest throws a 3.0 kg ball forward at 8.0 m/s on frictionless ice. What is the skater's recoil speed, in m/s?

Type a number in m/s.

Show the answer

Starting p = 0: 0 = (3.0)(8.0) + (60.)v, so v = −0.40 m/s, a speed of 0.40 m/s backward.

  • Answer: 0.40 m/s

Part 9 · Summary

Summary

Internal forces cancel in pairs, so only external forces change a system's momentum. With zero net external force, total momentum and the center of mass's velocity stay constant. Choosing the system decides which forces are external; always state the condition.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections