Question

A block $A$ of mass $m$ is in equilibrium after being suspended from the ceiling with the help of a spring of force constant $k$ . The block $B$ of mass $m$ , strikes the block $A$ with a speed $v$ and sticks to it. The value of $v$ for which the spring just attains its natural length is

• A. $\sqrt{\frac{6m{g}^2}{k}}$
• B. $\sqrt{\frac{3m{g}^2}{k}}$
• C. $\sqrt{\frac{2m{g}^2}{k}}$
• D. $\sqrt{\frac{m{g}^2}{2k}}$

Answer 1

Answer: (A)

Let the velocity of the combined system just after the collision is $v^{\prime }$ , then using conservation of linear momentum we get
$mv=2mv\prime$ ,
$\Rightarrow v\prime =\frac{v}{2}$
Initial elongation of the spring is
$x=\frac{mg}{k}$
Now, using the conservation of mechanical energy we get
$\frac{1}{2}2m{\left(v^{\prime }\right)}^2+\frac{1}{2}k{x}^2=2mgx$
$m\left(\frac{v^2}{4}\right)+\frac{m^2{g}^2}{2k}=\frac{2m^2{g}^2}{k}$
$v=\sqrt{\frac{6m{g}^2}{k}}$