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v0.1.1065
NotesPhysics HLTopic 1.2Conservation of momentum & collisions
Back to Physics HL Topics
1.2.82 min read

Conservation of momentum & collisions

IB Physics • Unit 1

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Contents

  • What 'conservation of momentum' means
  • Working it out — total p before = total p after
  • Exam-style question
The big idea: Momentum is mass × velocity: p = mv (unit kg m s⁻¹).

In any collision or explosion with no outside push, the total momentum before = total momentum after.

The objects swap momentum between them, but the total never changes.
Spot it — and mind the signs: Velocity has a direction, so give one way + and the other −. A cart moving left at 2 m s⁻¹ has velocity −2 m s⁻¹.

Add up m × v for every object before, and it must equal the same total after.
Two kinds of collision: Momentum is always conserved in both — but kinetic energy is not.

- Elastic: kinetic energy is also conserved (KE before = KE after). Things bounce cleanly. - Inelastic: some KE turns into heat/sound. If they stick together it is perfectly inelastic (most KE lost).

Momentum of one object is mass × velocity. This one is given in the data booklet:

Given in the data booklet — momentum of a single object.
momentum (kg m s⁻¹)
mass (kg)
velocity — carries a sign for direction (m s⁻¹)

Conservation of momentum then says the total before equals the total after. (This is a principle — it is not printed as an equation, so you write it out yourself.)

total momentum before = total momentum after (u = before, v = after)
the two masses (kg)
velocities before the collision (m s⁻¹)
velocities after the collision (m s⁻¹)
How to use it: 1. Pick a positive direction.

2. Write m × v for everything before and add them.

3. Set that equal to the total after, then solve for the unknown.

Keep every minus sign.

IB-style question — find the velocity after a collision

A 2.0 kg trolley moving right at 3.0 m s⁻¹ hits a stationary 1.0 kg trolley. They do not stick: the 2.0 kg trolley continues at 1.0 m s⁻¹. Find the velocity of the 1.0 kg trolley afterwards.

Solution

  1. Total momentum before = total momentum after:
  2. Put in the numbers (the stationary one has u₂ = 0):
  3. Tidy each side:
  4. Solve for v₂ — keep the unit:

Final answer

The 1.0 kg trolley moves off at 4.0 m s⁻¹ (in the same direction, +).

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How this is tested: Collisions appear as a full Paper 2 calculation and as quick Paper 1A multiple-choice questions.

- What they ask: read cart velocities off a v–t graph, then 'show that' a collision is elastic (compare KE before and after), or find a common speed after two things stick. - Classic trap: assuming kinetic energy is conserved. Momentum is always conserved; KE is conserved only if the collision is elastic. When objects stick, KE is lost.
Testing for elastic — use kinetic energy: Kinetic energy is the energy of motion. It is given in the booklet:
Given in the data booklet — kinetic energy of a moving mass.
kinetic energy (J)
mass (kg)
speed (m s⁻¹)

Work out the total KE before and the total KE after. If they're equal, the collision is elastic.

IB-style question — show the collision is elastic

Two trolleys each of mass 1.5 kg are tracked on a v–t graph. Trolley X moves at 5.0 m s⁻¹ toward a stationary trolley Y. After they collide, X moves at 1.0 m s⁻¹ and Y at 4.0 m s⁻¹ (same direction). Show that the collision is elastic.

Solution

  1. Total KE before — only X is moving:
  2. Total KE after — both move:
  3. Work that out:
  4. Before = after, so no KE was lost:

Final answer

KE before (18.75 J) = KE after (18.75 J), so the kinetic energy is conserved — the collision is elastic.

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the condition that must be met for the total momentum of a system of objects to be conserved during a collision. [1 mark]

Related Physics HL Topics

Continue learning with these related topics from the same unit:

1.1.1Velocity and displacement
1.1.2Acceleration
1.1.3Displacement from a velocity–time graph
1.1.4The suvat equations
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