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. $\sqrt{\frac{8m{g}^2}{k}}$

Answer 1

Answer: (B)

Applying the law of linear momentum conservation, $mv+0=2mV\Rightarrow V=\frac{v}{2}$
Final $K.E.=\frac{1}{2}\times 2m\frac{v^2}{2}=\frac{m{v}^2}{4}$
Applying the law of energy conservation when the spring attains its natural length,
$\frac{m{v}^2}{4}+0=mgx+\frac{1}{2}k{x}^{2}R$
$\because kx=mg$
$\Rightarrow x=\frac{mg}{k}$
$\Rightarrow \frac{m{v}^2}{4}=mg\times \frac{mg}{k}+\frac{1}{2}k\frac{(mg{)}^2}{k^2}$
$\Rightarrow \frac{v^2}{4}=\frac{m{g}^2}{k}+\frac{m{g}^2}{2k}$
$\Rightarrow \frac{v^2}{2}=\frac{3m{g}^2}{k}$
$\Rightarrow v=\sqrt{\frac{6m{g}^2}{k}}$