Question

Block $A$ is hanging from a vertical spring and is at rest. Block $B$ strikes the block $A$ with velocity $v$ and sticks to it. Then the value of $v$ for which the spring just attains natural length is____

• A. $\sqrt{\frac{60m{g}^2}{k}}$
• B. $\sqrt{\frac{6m{g}^2}{k}}$
• C. $\sqrt{\frac{10m{g}^2}{k}}$
• D. None of these

Answer 1

Answer: (B)

The initial extension in spring is $X_0=\frac{mg}{k}$ .
Just after collision of B with A the speed of combined mass is $\frac{v}{2}$ . For the spring to just attain natural length the combined mass must rise up by $X_0=\frac{mg}{k}$ and comes to rest.

Applying conservation of energy between initial and final states,
$\frac{1}{2}2m{\left[\frac{v}{2}\right]}^2+\frac{1}{2}k{\left[\frac{mg}{k}\right]}^2=2mg\left[\frac{mg}{k}\right]$
Solving we get, $v=\sqrt{\frac{6m{g}^2}{k}}$