Collision and Impulse — Question 1

Question 1

The rebound impulse exceeds the stopping impulse

A $0.15\,\mathrm{kg}$ ball approaches a fixed wall at $+12\,\mathrm{m/s}$ and leaves at $-8\,\mathrm{m/s}$. Contact lasts $0.020\,\mathrm s$. Other horizontal forces are negligible.

1. Find the wall impulse on the ball and the average contact force.

2. Compare with a ball that stops without rebounding in the same time.

Question 1 - Solution

The rebound impulse exceeds the stopping impulse

(a) Derivation and calculation

Retain signed velocities across the rebound: $$\boxed{J=m(v_f-v_i)=0.15(-8-12)=-3.0\ \mathrm{N\,s}}.$$ $$\boxed{F_{\mathrm{avg}}=J/\Delta t=-3.0/0.020=-150\ \mathrm N}.$$

(b) Result and interpretation

For stopping without rebound, $$\boxed{J_{\mathrm{stop}}=0.15(0-12)=-1.8\ \mathrm{N\,s}},\qquad \boxed{F_{\mathrm{avg,stop}}=-90\ \mathrm N}.$$ Rebounding requires an additional negative impulse of $-1.2$ N s; a difference of unsigned speeds misses the reversal.

Physical check

The rebound impulse magnitude is $3/1.8=5/3$ times the stopping impulse.