Question

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

• A. 20 m
• B. 10 m
• C. 25 m
• D. Cannot be determined because masses are not given.

Answer 1

Answer: (C)

25 m

$m\times 10=m{v}_1+m{v}_2$
$\Rightarrow 10=v_1+v_2....\left(i\right)$

and $\frac{1}{2}\times 10=v_2-v_1....\left(ii\right)$

From$\left(i\right)$ and $\left(ii\right)$

$v_1=\frac{5}{2}m/s;v_2=\frac{15}{2}m/s$

Distance between the two blocks

$S=(-v_1+v_2).t$

$=(-\frac{5}{2}+\frac{15}{2})\times 5=25m$