Question

A body of mass $4kg$ collides elastically with another body of mass $2kg$ at rest. If time of collision is $0.02sec$ and average impulse force acted on each body is $100N$, find the velocity of $4kg$ body before and after collision (Assuming collision takes place on a smooth horizontal surface)

Answer 1

Applying impulse-Momentum separately for each bodies

$4v-F\left(\Delta t\right)=4v^{\prime }4v-100\times 0.02=4v^{\prime };100\times 0.02=2v^{\prime \prime };2=2v^{\prime }$

Apply coefficient of restiturion

$-e(v-0)=v^{\prime }-v^{\prime \prime }(\because e=1)\therefore v=v^{\prime \prime }-v^{\prime }$

Solving equation (1) , (2) and (3)

$v=\frac{3}{4}m/s;v^{\prime }=\frac{1}{4}m/s$