Physics / Unit 6: Momentum and Collisions
6.2 Impulse
Notes
Impulse is a force acting over a time: $J = F\Delta t$. The impulse-momentum theorem says impulse equals the change in momentum: $F\Delta t = \Delta p = m(v_f - v_i)$.
- Unit: $\text{N·s}$, which equals $\text{kg·m/s}$.
- When an object bounces back, the change in momentum is larger than when it just stops, because the velocity reverses.
- Example: a 0.15 kg ball moving at $40 \text{ m/s}$ is hit back at $30 \text{ m/s}$. Taking the original direction as positive, $\Delta p = 0.15(-30 - 40) = -10.5 \text{ kg·m/s}$.
Practice
A 200 N force acts on a cart for 0.5 s. What impulse does it give?
$J = F\Delta t = 200 \times 0.5 = 100 \text{ N·s}$.
A 2 kg ball at rest is kicked and leaves at $12 \text{ m/s}$. The foot touches it for 0.08 s. What average force acts on the ball?
$\displaystyle F = \frac{\Delta p}{\Delta t} = \frac{2 \times 12}{0.08} = 300 \text{ N}$.
Two identical balls hit a wall at the same speed. Ball A sticks to the wall; ball B bounces straight back at the same speed. Which receives the larger impulse?
B's velocity changes from $+v$ to $-v$, a change of $2v$; A's changes by only $v$.
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