91Ó°ÊÓ

We need to find that which ball, if either, receives the larger impulse?

When an object receives a force for a short time, this force with time is called the impulse,

in force-versus time graph the momentum is same.

Hence, impulse is given by

$impulse=J_x=F_xâ–³t\left(1\right)$

Exerted impulse on the rubble ball is

$J_{steel}=F_{rubber}â–³t$

$F_{rubber}$is the force that is exerted by the rubber.

The impulse that is exerted on the rubber ball is

role="math" localid="1649263813382" $J_{rubber}=F_{steel}â–³t$

Exerted force by the rubber is $F_{steel}$, so both balls exert the same force but in opposite direction

$F_{steel}=-F_{rubber}$

Impulse on the rubber

$$\begin{gathered} J_{rubber}=F_{steel}â–³t \\ =-F_{steel}â–³t \\ =-J_{steel} \end{gathered}$$