Impulse and Momentum

By the end of this lesson, you will be able to calculate momentum and impulse, apply the impulse-momentum theorem, and explain how extending collision time reduces force.

Linear Momentum

$$\vec{p} = m\vec{v}$$

• Momentum is a vector; same direction as velocity
• Units: kg·m/s
• Larger mass or velocity → more momentum

Impulse

$$\vec{J} = \vec{F}_{\text{avg}}\Delta t = \Delta\vec{p}$$

Impulse-momentum theorem: the impulse equals the change in momentum.

$\vec{J}$ has units of N·s = kg·m/s (same as momentum).

Force-Time Graphs

The area under a $F$-$t$ graph = impulse = $\Delta p$.

A large force for a short time can produce the same change in momentum as a small force for a long time.

Applications of Extending Contact Time

Airbags, crumple zones, padding: increase $\Delta t$ → reduce $F_{\text{avg}}$ for the same $\Delta p$.

Example: A 70 kg person decelerated from 15 m/s to 0:

$\Delta p = 70 \times 15 = 1050$ N·s

Without airbag ($\Delta t = 0.01$ s): $F = 1050/0.01 = 105,000$ N — fatal.

With airbag ($\Delta t = 0.15$ s): $F = 1050/0.15 = 7,000$ N — survivable.

Example 1 — Calculating Impulse

A 0.5 kg ball hits a wall at 20 m/s and bounces back at 15 m/s. Find the impulse.

$\Delta p = p_f - p_i = 0.5(-15) - 0.5(20) = -17.5$ N·s

Impulse = 17.5 N·s (directed away from wall).

Example 2 — Finding Average Force

The ball in Example 1 is in contact with the wall for 0.05 s. Find average force.

$F_{\text{avg}} = \Delta p/\Delta t = 17.5/0.05 = 350$ N

Summary

Momentum: $\vec{p} = m\vec{v}$; vector; units kg·m/s.

Impulse: $\vec{J} = \vec{F}\Delta t = \Delta\vec{p}$; same units as momentum.

Extending $\Delta t$: for the same $\Delta p$, larger $\Delta t$ → smaller average force.

Force-time graph: area = impulse = $\Delta p$.