You know from Newton's second law that a net force changes velocity, and from Softening a Collision that stretching a collision over more time softens it. Impulse puts those two ideas into one equation.
Why it works
Start from $\Sigma F = ma$ and write the acceleration as $a = \dfrac{\Delta v}{\Delta t}$. Multiply both sides by $\Delta t$: $$\Sigma F\,\Delta t = m\Delta v = \Delta p.$$ The left side, force times the time it acts, is called the impulse $J$. A big force for a short time and a small force for a long time can give the same impulse, and so the same change in momentum.
Key idea
Impulse equals change in momentum: $J = F_{\text{avg}}\Delta t = \Delta p = m v_f - m v_i$. Its unit, $\text{N·s}$, is the same as $\text{kg·m/s}$. When the force changes during the push, the impulse is the area under the force–time graph.
Worked examples
Worked example: Average force from a change in momentum
A $0.43\ \text{kg}$ soccer ball at rest is kicked to $20\ \text{m/s}$. The foot touches it for $0.010\ \text{s}$. What average force acts on the ball?
- $$\Delta p = (0.43\ \text{kg})(20\ \text{m/s}) - 0 = 8.6\ \text{kg·m/s}.$$
- $$F_{\text{avg}} = \dfrac{\Delta p}{\Delta t} = \dfrac{8.6\ \text{kg·m/s}}{0.010\ \text{s}} = 860\ \text{N}.$$
Answer: $860\ \text{N}$
Worked example: A bounce versus a stop
A $0.50\ \text{kg}$ ball hits a wall at $6.0\ \text{m/s}$. Find the impulse from the wall (a) if it stops dead and (b) if it bounces straight back at $4.0\ \text{m/s}$. Take "toward the wall" as positive.
- (a) $$J = m(v_f - v_i) = (0.50)(0 - 6.0) = -3.0\ \text{N·s}.$$
- (b) After the bounce the velocity is $-4.0\ \text{m/s}$: $$J = (0.50)(-4.0 - 6.0) = -5.0\ \text{N·s}.$$
- The bounce needs the bigger impulse: the wall must stop the ball and then throw it back.
Answer: (a) $-3.0\ \text{N·s}$; (b) $-5.0\ \text{N·s}$
Worked example: Reading the area
A $2.0\ \text{kg}$ cart rolls left at $1.0\ \text{m/s}$. It gets the push in the graph above, to the right. What is its velocity afterward (right positive)?
- Area: two triangles, $2 \times \tfrac{1}{2}(0.1)(20) = 2.0\ \text{N·s}$, plus a rectangle, $(0.2)(20) = 4.0\ \text{N·s}$. So $J = 6.0\ \text{N·s}$.
- $v_f = v_i + \dfrac{J}{m} = -1.0 + \dfrac{6.0}{2.0} = +2.0\ \text{m/s}$.
Answer: $2.0\ \text{m/s}$ to the right
Common mistake
Thinking a bounce changes momentum less than a stop. Speed drops less in a bounce, but momentum is a vector: going from $+6$ to $-4$ is a change of $10$, while going from $+6$ to $0$ is a change of only $6$. Always subtract signed velocities, $v_f - v_i$. Check: a ball that reverses must have an impulse larger than $mv_i$.
This is the physics of airbags, gym mats and catching an egg with soft hands: the change in momentum is fixed by the speeds, so the only way to lower the force is to make $\Delta t$ longer. The Momentum and Collisions lab of the Science Lab drops an egg on a pillow and on a floor so you can compare the two force graphs.