Momentum and impulse

Key ideas

Momentum measures how hard something is to stop:

$p = mv$

Momentum is a vector (direction matters) with units kg m s⁻¹.

Impulse is the change in momentum, and it equals force × time:

$\Delta p = F\Delta t$

Rearranged, $F = \dfrac{\Delta p}{\Delta t}$ — for a given change in momentum, more time means less force. That single idea explains airbags, crumple zones, bending your knees when landing, and soft catching in cricket.

Worked ExampleThe physics of a soft catch

A $0.16$ kg cricket ball arrives at $25$ m s⁻¹ and is caught, coming to rest in $0.050$ s. Find the impulse and the average force on the hands.

Step 1 — Impulse = change in momentum

$\Delta p = m\Delta v = (0.16)(25 - 0) = 4.0 \text{ kg m s}^{-1}$

Step 2 — Average force

$F = \frac{\Delta p}{\Delta t} = \frac{4.0}{0.050} = 80 \text{ N}$

Ride the catch to stretch the stop to $0.25$ s and the force falls to $\dfrac{4.0}{0.25} = 16$ N — same impulse, five times the time, one-fifth the force. That's exactly why airbags, crumple zones and bent knees work.

Practice question

A $0.50$ kg netball moving at $8.0$ m s⁻¹ is caught and stopped in $0.20$ s. Find the ball's momentum and the average stopping force.

Worked solution: $p = mv = 0.50 \times 8.0 = 4.0$ kg m s⁻¹; $F = \Delta p / \Delta t = 4.0/0.20 = 20$ N.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A $0.42$ kg rugby ball is thrown at $15$ m s⁻¹.

Calculate its momentum.

Merit

A $0.16$ kg cricket ball travelling at $25$ m s⁻¹ is caught and brought to rest in $0.40$ s.

Calculate the average force exerted on the ball while stopping it.

Excellence

A fielder catches a fast cricket ball by drawing their hands back as the ball arrives.

Using the relationship $F = \dfrac{\Delta p}{\Delta t}$ , explain fully, using physics principles, why this reduces the force on their hands. State clearly what stays the same.

Momentum and impulse

Covers AS91171

Key ideas

Momentum measures how hard something is to stop:

$p = mv$

Momentum is a vector (direction matters) with units kg m s⁻¹.

Impulse is the change in momentum, and it equals force × time:

$\Delta p = F\Delta t$

Worked ExampleThe physics of a soft catch

A $0.16$ kg cricket ball arrives at $25$ m s⁻¹ and is caught, coming to rest in $0.050$ s. Find the impulse and the average force on the hands.

Step 1 — Impulse = change in momentum

$\Delta p = m\Delta v = (0.16)(25 - 0) = 4.0 \text{ kg m s}^{-1}$

Step 2 — Average force

$F = \frac{\Delta p}{\Delta t} = \frac{4.0}{0.050} = 80 \text{ N}$

Practice question

A $0.50$ kg netball moving at $8.0$ m s⁻¹ is caught and stopped in $0.20$ s. Find the ball's momentum and the average stopping force.

Worked solution: $p = mv = 0.50 \times 8.0 = 4.0$ kg m s⁻¹; $F = \Delta p / \Delta t = 4.0/0.20 = 20$ N.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A $0.42$ kg rugby ball is thrown at $15$ m s⁻¹.

Calculate its momentum.

Merit

A $0.16$ kg cricket ball travelling at $25$ m s⁻¹ is caught and brought to rest in $0.40$ s.

Calculate the average force exerted on the ball while stopping it.

Excellence

A fielder catches a fast cricket ball by drawing their hands back as the ball arrives.

Using the relationship $F = \dfrac{\Delta p}{\Delta t}$ , explain fully, using physics principles, why this reduces the force on their hands. State clearly what stays the same.