Question

For the situation shown in the given figure, block $A$ of mass $10\, kg$ is at the ground. The block $B$ is released from rest when the light spring is unstretched. The minimum mass of the block $B$ in kg such that block $B$ may leave the contact of ground is ______.

Answer 1

When the block $A$ just starts to leave the surface, then.
$k x_{0}=M g$
For minimum value of $m, x_{0}$ should be maximum elongation in the spring. At the instant of maximum elongation, the velocity of block $B$ will be zero.
$\therefore$ Loss in gravitational potential energy of block $B=$ Gain in elastic potential energy by spring
$\therefore m g x_{0}=\frac{1}{2} k x_{0}^{2}$
$\therefore x_{0}=\frac{2 m g}{k}$
or $\frac{M g}{k}=\frac{2 m g}{k}$
$ \Rightarrow m=\frac{M}{2}=5\, kg$