Momentum

Impulse and safety applications

Impulse and resultant force Supplement 3 min read

Extended (Supplement)

Impulse

Impulse is the product of a force and the time for which it acts. It equals the change in momentum of the object. \[ \text{Impulse} = F \times \Delta t = \Delta p = mv - mu \]

Quantity	Symbol	Unit
Impulse	\( F\Delta t \)	N·s (≡ kg·m/s)
Force	\( F \)	N
Time of contact	\( \Delta t \)	s
Change in momentum	\( \Delta p \)	kg·m/s

From Newton's Second Law: \[ F = \frac{\Delta p}{\Delta t} \quad \Longrightarrow \quad F \cdot \Delta t = \Delta p \] A large force over a short time can produce the same change in momentum as a small force over a long time. This is the key insight behind crash safety design.

Force–time graph

The area under a force–time graph equals the impulse (and therefore the change in momentum). \[ \text{Impulse} = \text{area under } F\text{–}t \text{ graph} \]

Why longer contact time matters — safety applications

For a fixed change in momentum (e.g. stopping a car), increasing the contact time \( \Delta t \) reduces the force \( F \): \[ F = \frac{\Delta p}{\Delta t} \quad \text{— larger } \Delta t \Rightarrow \text{smaller } F \] This is the engineering principle behind many safety features:

Crumple zones in cars — extend the collision time, reducing peak force on passengers.
Airbags — inflate to increase the time over which a passenger decelerates.
Cushioned landing mats in gymnastics — extend the time of impact with the ground.
Catching a cricket ball — pulling the hands back increases \( \Delta t \) and reduces \( F \).
Helmets — foam liner compresses, extending the time of impact on the skull.

Example 1 — find average force during collision

A 0.15 kg cricket ball is moving at 30 m/s when it is caught and brought to rest in 0.050 s. Find the average force on the ball.

Change in momentum: \( \Delta p = mv - mu = 0.15 \times 0 - 0.15 \times 30 = -4.5\,\text{kg·m/s} \)

Magnitude of impulse = 4.5 N·s

\( F = \dfrac{\Delta p}{\Delta t} = \dfrac{4.5}{0.050} = 90\,\text{N} \)

Example 2 — crumple zone reduces force

A 1200 kg car decelerates from 25 m/s to rest. Without a crumple zone, contact time = 0.080 s. With a crumple zone, contact time = 0.40 s. Compare the forces.

\( \Delta p = 1200 \times 25 = 30\,000\,\text{kg·m/s} \)

Without crumple zone: \( F = 30\,000 / 0.080 = 375\,000\,\text{N} \)

With crumple zone: \( F = 30\,000 / 0.40 = 75\,000\,\text{N} \)

The crumple zone reduces the force by a factor of 5.

Example 3 — find change in momentum from F–t graph

A force–time graph shows a triangular pulse with peak force 600 N over 0.010 s. Find the change in momentum.

Area of triangle = \( \dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2} \times 0.010 \times 600 = 3.0\,\text{N·s} \)

Impulse = 3.0 N·s = change in momentum = 3.0 kg·m/s.

"Explain how an airbag reduces injury" is a 3-mark answer: (1) airbag increases the time over which the passenger decelerates; (2) same change in momentum but larger \( \Delta t \); (3) therefore smaller force acts on the passenger.