Question

Two blocks shown in figure have equal masses. The surface of A is smooth while that of B has a friction coefficient of 0.10 with the floor. Block A is moving at a speed of 10 m/s towards B which is kept at rest. The distance travelled by block B if the collision is perfectly elastic is? $Assumeg=10m/s^2$.

• A. 100 m
• B. 200 m
• C. 50 m
• D. 25 m

Answer 1

The correct option is C 50 m

Using momentum conservation

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

Before collision :
After collision :
$e=\frac{\text{velocity of separation}}{\text{velocity of approach}}=\frac{v_2-v_1}{10}$

For perfectly elastic collision e = 1

$v_2-v_1=10.....\left(ii\right)$

$\text{Acceleration of block B}=\mu g\text{(towards left)}$
$=0.1\times 10=1m/s^2$

$v^2=u^2+2as$

$0={10}^2-2\times 1\times s\Rightarrow s=50m$