Relate impulse to the change in momentum of an object using $\vec{J} = \vec{F}\Delta t = \Delta \vec{p}$, and interpret force-time graphs.

What this dot point is asking

You need to define momentum and impulse, state and apply the impulse-momentum theorem, find impulse as the area under a force-time graph, and use these ideas to explain safety features such as crumple zones and airbags.

Momentum

Because momentum is a vector, direction matters. A ball bouncing back reverses its momentum, so its change in momentum is larger than if it simply stopped.

Impulse and the impulse-momentum theorem

Newton's second law can be written in terms of momentum: the net force equals the rate of change of momentum,

Rearranging gives the impulse-momentum theorem:

Impulse and momentum change have the same units, N s = kg m s$^{-1}$. The key insight is the trade-off between force and time: for a fixed change in momentum, increasing the contact time decreases the average force.

Force-time graphs

When a force varies during a collision, the impulse is the area under the force-time graph:

For a triangular force pulse, the area is $\tfrac{1}{2}(\text{base})(\text{height})$. The average force is the constant force that would give the same area (same impulse) over the same time.

Why safety features work

Crumple zones, airbags, helmet padding and crash mats all work the same way. In a collision the change in momentum (the impulse) is essentially fixed - the occupant must be brought to rest. By extending the contact time $\Delta t$, these devices reduce the average force $F = \Delta p / \Delta t$ on the body. A longer, gentler deceleration replaces a short, violent one, reducing injury.

The same principle explains why a cricketer "gives" with the ball when catching, and why falling onto a soft surface hurts less than onto concrete.