A.2 · IB Physics
Impulse and force–time graphs
A ball hits a wall and bounces back. The change in momentum is fixed by the speeds before and after, so the area under the force–time graph is fixed too. Stretch the contact time and the peak force falls: the idea behind padding, airbags and crumple zones.
Predict first
Start from the default settings (m 0.500 kg, u 10.0 m s⁻¹, v 6.00 m s⁻¹, half-sine pulse, Δt 20 ms). Change Contact time Δt to 40 ms. What happens to the peak force?
The impulse m(u + v) = 8.00 N s is fixed, so the area under the F–t graph is fixed. For this pulse Fmax = J ÷ (0.637 Δt), so doubling Δt halves Fmax: 628 N becomes 314 N.
Check it in the model: Set the Contact time Δt slider from 20 ms to 40 ms and compare the Peak force readout (628 N then 314 N) and the Impulse readout (unchanged).
Positive direction is away from the wall. The contact is shown in slow motion; the clock is in milliseconds.
J = FΔt = Δp = m(v − (−u)) = m(u + v)
Impulse = area under the F–t graph
Favg = Δp ÷ Δt ; Fmax = Favg ÷ (mean/peak ratio of the pulse shape)
Try this
- Choose "Hard wall", throw, then choose "Padded wall (same Δp)" and throw. Compare the two shaded areas, the Impulse readout and the peak forces.
- Set u = 10 and v = 0 (ball sticks), throw, then set v = 10 and throw again. Which collision needs the larger impulse, and by what factor?
- Keep m, u, v and Δt fixed and switch the shape between rectangle, triangle and half-sine. Which has the highest peak force, and why does the average force stay the same?
Common error: saying an airbag or a crumple zone "reduces the impulse". The occupant's change in momentum is the same, so the impulse is unchanged; the longer contact time makes the average and peak forces smaller.