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)

$\sqrt{\frac{6m{g}^2}{k}}$

$Initialelongationoftheblockisgivenbymg=k{x}_0\Rightarrow x_0=\frac{mg}{k}Aftercollisionthevelocityofthetwoblocksystembecomesmv=2m{v}_1\Rightarrow v_1=\frac{v}{2}nowconservationofenergy\frac{1}{2}2m{v_1}^2+\frac{1}{2}k{x_0}^2=2mg{x}_0\Rightarrow \frac{1}{2}2m{\left(\frac{v}{2}\right)}^2+\frac{1}{2}k{x_0}^2=2mg{x}_0\Rightarrow \frac{m{v}^2}{4}=\frac{2m^2{g}^2}{k}-\frac{m^2{g}^2}{2k}=\frac{3m^2{g}^2}{2k}\Rightarrow v=\sqrt{\frac{6m{g}^2}{k}}$