Question

In the figure shown, a block A moving with velocity $10m/s$ on a horizontal surface collides with another block B at rest initially. The coefficient of restitution is $\frac{1}{2}$. Neglect friction everywhere. The distance between the blocks at $5s$ after the collision takes place is $5x$ (in $m$). Then $x$ is

Answer 1

$m\times 10$ = $m{v}_1+$$m{v}_2$
$\Rightarrow$ 10 = $v_1+$$v_2$ ....(i)
and $\frac{1}{2}$$\times 10$ = $v_2-$$v_1$ ....(ii)
From (i) and (ii)
$v_1=$$\frac{5}{2}m/s$; $v_2=$$\frac{15}{2}m/s$
Distance between the two blocks
$s=(-v_1+v_2)t$
$=(-\frac{5}{2}+\frac{15}{2})$$\times 5=25m$

$x=5$