# Momentum and collisions: Unit 3.1(a)

> define linear momentum and impulse as force multiplied by time

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Specification statement (Unit 3.1(a)): define linear momentum and impulse as force multiplied by time

## Explanation

1. **Linear momentum** p = mv. It is a vector (kg m s⁻¹), so its direction matters.
2. **Impulse** = force × time = FΔt = change in momentum Δp. Its unit, N s, is the same as kg m s⁻¹.
3. For a bounce, the change in momentum includes the reversal of direction: from +15 to −10 m s⁻¹ is a change of 25 m s⁻¹, not 5.

## Worked example

A 0.40 kg ball hits a wall at 15 m s⁻¹ and rebounds at 10 m s⁻¹. Contact lasts 0.050 s. Find the impulse and the average force.

1. Take the direction towards the wall as positive. p before = 0.40 × 15 = +6.0 kg m s⁻¹; after = 0.40 × (−10) = −4.0 kg m s⁻¹.
2. Δp = −4.0 − 6.0 = −10 kg m s⁻¹: an impulse of 10 N s away from the wall.
3. Average force = Δp/Δt = 10 / 0.050 = 200 N away from the wall.

Answer: 10 N s; 200 N.

## Check

Which unit is equivalent to kg m s⁻¹?

- N m
- N s
- J s
- W s

The answer and its explanation are shown in the app once you choose an option.

## Apply it

A 60 g tennis ball is served from rest at 50 m s⁻¹. The racket is in contact for 4.0 ms. Calculate the average force on the ball.

The model answer is shown in the app after you have tried it.

## Related

- [Momentum and collisions lesson](https://revise-gray.vercel.app/lesson?subject=wjec-alevel-physics&topic=wjec-alevel-physics.momentum)
- [Momentum and collisions in the library](https://revise-gray.vercel.app/library?topic=wjec-alevel-physics.momentum)