Collision and Impulse — Question 6

Question 6

Designing contact under a peak-force limit

A pad must stop a $2\,\mathrm{kg}$ object moving at $6\,\mathrm{m/s}$. Its opposing force is nonnegative in magnitude and never exceeds $300\,\mathrm N$. Other forces along the motion are negligible.

1. Find a universal lower bound on stopping time.

2. If the force pulse must be triangular, find its minimum duration and explain the difference.

Question 6 - Solution

Designing contact under a peak-force limit

(a) Derivation and calculation

Stopping requires impulse magnitude $$J_{\mathrm{req}}=m v_0=2(6)=12\ \mathrm{N\,s}.$$ $$12=\int_0^T F(t)\,dt\le300T\Rightarrow\boxed{T\ge0.040\ \mathrm s}.$$ An ideal rectangular pulse at the cap attains this universal bound.

(b) Result and interpretation

A triangular pulse has only half the rectangle’s area: $$12=\tfrac12F_{\mathrm{peak}}T\le\tfrac12(300)T \Rightarrow\boxed{T\ge0.080\ \mathrm s}.$$ Constraining the pulse shape lowers its maximum average force at the same peak.

Physical check

The two minimum times differ by exactly the ratio of rectangle and triangle areas.