Collision and Impulse — Question 2

Question 2

Recovering a peak force from triangular contact

A horizontal contact force rises linearly from zero at $t=0$ to $F_{\max}$ at $t=2\,\mathrm{ms}$, then falls linearly to zero at $t=6\,\mathrm{ms}$. It changes a $0.40\,\mathrm{kg}$ object's velocity from $-1$ to $+2\,\mathrm{m/s}$.

1. Find the impulse, peak force and average force.

2. Does the unequal rise and fall time change the triangle-area calculation?

Question 2 - Solution

Recovering a peak force from triangular contact

(a) Derivation and calculation

The momentum change fixes the force-time area: $$\boxed{J=0.40[2-(-1)]=1.20\ \mathrm{N\,s}}.$$ $$J=\tfrac12(0.006)F_{\max}\Rightarrow\boxed{F_{\max}=400\ \mathrm N},\qquad \boxed{F_{\mathrm{avg}}=J/0.006=200\ \mathrm N}.$$

(b) Result and interpretation

The two triangular contributions are $$J=\tfrac12F_{\max}(0.002)+\tfrac12F_{\max}(0.004) =\tfrac12F_{\max}(0.006).$$ The peak’s time affects the intermediate motion but not the total area for the given height and full base.

Physical check

The average force is half the peak for this triangular pulse.