Impulse and Change in Momentum wrote impulse as $F\Delta t$ and found it as the area under a force–time graph. Real collisions don't have a constant force: a bat's push rises from zero to a peak and falls back in a millisecond or two. Evaluating Integrals with FTC Part 2 handles any shape.
Key idea
$$\vec{J} = \int_{t_1}^{t_2} \vec{F}\,dt = \Delta\vec{p} = m\vec{v}_2 - m\vec{v}_1.$$ The impulse is the area under the $F$–$t$ curve, in N·s (the same as kg·m/s). The average force over the contact is $F_{\text{avg}} = J/\Delta t$.
Why it works
Newton's second law in its original form says $\vec{F} = \dfrac{d\vec{p}}{dt}$: force is the rate of change of momentum. Integrate both sides over the collision: $\int \vec{F}\,dt = \int d\vec{p} = \Delta\vec{p}$. Nothing about the force's shape was assumed, so a spike, a smooth bump or a wobbly curve all work, as long as you integrate the real force.
Worked example: A smooth push
A bat pushes a $0.145\ \text{kg}$ baseball, initially at rest, with $F(t) = 1000\sin(\pi t/0.01)$ newtons for $0 \le t \le 0.01\ \text{s}$. Find the impulse and the ball's speed.
- $$J = \int_0^{0.01} 1000\sin\left(\dfrac{\pi t}{0.01}\right)dt = 1000\cdot\dfrac{0.01}{\pi}\left[-\cos\dfrac{\pi t}{0.01}\right]_0^{0.01} = \dfrac{2(1000)(0.01)}{\pi}$$
- $J = 6.37\ \text{N·s}$.
- $v = J/m = 6.37/0.145 = 43.9\ \text{m/s}$.
Answer: $J = 6.37\ \text{N·s}$; $v = 43.9\ \text{m/s}$.
Worked example: A bounce
A $0.1\ \text{kg}$ ball hits a wall at $15\ \text{m/s}$ and rebounds at $10\ \text{m/s}$ after $0.005\ \text{s}$ of contact, with a half-sine force $F_{\max}\sin(\pi t/T)$. Find the peak force.
- With the rebound direction positive, $\Delta p = 0.1(10) - 0.1(-15) = 2.5\ \text{N·s}$: the speeds add, because the direction reverses.
- Half-sine impulse: $\dfrac{2F_{\max}T}{\pi} = 2.5$, so $F_{\max} = \dfrac{\pi(2.5)}{2(0.005)} = 785\ \text{N}$.
- The average force is $2.5/0.005 = 500\ \text{N}$, about 64% of the peak.
Answer: $F_{\max} \approx 785\ \text{N}$.
Common mistake
Multiplying the peak force by the collision time. For the baseball that gives $1000 \times 0.01 = 10\ \text{N·s}$, 57% too much, because the force is near its peak only briefly. The right rectangle uses the average force, which you can only find after integrating. Check with the graph: a rectangle drawn at the peak height always encloses more area than the curve under it.
This is the physics of padding: a soft surface spreads the same $\Delta p$ over a longer time, so the area stays the same while the peak force drops.