Question

A moving block having mass $m$ collides with stationary block of mass $4m$, initial velocity of moving block is $v_o$, if coefficient of restitution is $0.25$, find speed of lighter block after collision.

• A. Zero
• B. $\frac{v_o}{3}$
• C. $v_o$
• D. $\frac{v_o}{3}$

Answer 1

The correct option is A Zero


Conserving the momentum we get,
$m{v}_o+0=m{v}_1+4m{v}_2$
$v_o=v_1+4v_2$ -----(i)
Definition of coefficient of restitution $e$
$e=\frac{\text{Speed of seperation}}{\text{Speed of approach}}=\frac{v_2-v_1}{v_o}=\frac{1}{4}$
$4v_2-4v_1=v_o$ ----(ii)

Solving (i) and (ii)
$v_1+4v_2=v_o$
$\underset{–}{4v_2-4v_1=v_o}$
$5v_1=0$
$v_1=0$