Edexcel A-Level Physics 9PH0
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Mechanics
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Further Mechanics
Conservation of momentum in two dimensions
- 1 thing to know
- Papers 1 and 3
Overview
Recall questions on resolving momenta into components, why magnitudes cannot simply be added, and applying conservation to an explosion or a glancing collision.
What you need to know
Answer each one before you reveal it.
Questions from this subtopic
Each one with the model answer underneath.
- 1
A tennis ball of mass 0.058 kg leaves a racket at 45 m s⁻¹ from rest after a contact time of 5.0 ms. What average force acted on it?
Model answer
The change of momentum is 0.058 × 45 = 2.61 kg m s⁻¹, so the average force is 2.61 ÷ (5.0 × 10⁻³) = 522 N, or 520 N to 2 significant figures.
- 2
State the principle of conservation of linear momentum:
Model answer
The total linear momentum of a system stays constant provided no resultant external force acts on it.
- 3
How does conservation of momentum follow from Newton's third law?
Model answer
Two interacting bodies exert equal and opposite forces on each other for the same length of time, so they receive equal and opposite impulses and their momentum changes cancel exactly.
- 4
How is conservation of momentum applied to a collision that is not along a single line?
Model answer
Momentum is conserved independently in two perpendicular directions, so the momenta are resolved into components and the totals before and after are equated in each direction separately.
- 5
Why can momenta in two dimensions not simply be added as magnitudes?
Model answer
Momentum is a vector, so only components along a common axis may be added arithmetically; otherwise vector addition is required.
- 6
A stationary object explodes into exactly two fragments. What can be said about their momenta?
Model answer
The two fragments must carry equal magnitudes of momentum in exactly opposite directions, so that the total remains zero.
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