Two identical blocks $A$ and $B$ each of mass $m$ resting on the smooth horizontal floor are connected by a light spring of natural length $L$ and spring constant $K$. A third block $C$ of mass $m$ moving with a speed $v$ along the line joining $A$ and $B$ collides with $A$.The maximum compression in the spring is

$A$ block of mass $m$ moving with a velocity $v_0$ on a smooth horizontal surface strikes and compresses a spring of stiffness $k$ till mass comes to rest as shown in the figure. This phenomenon is observed by two observers:

$A$: standing on the horizontal surface

$B$: standing on the block 

To an observer $A$, the work done by spring force is 

Explain the elastic potential energy of spring and obtain an expression for this energy.

Two springs have their force constant as $k_1$ and $k_2 (k_1 > k_2)$. when they are  stretched by the same force

Two bodies $A$ and $B$ of mass $m$ and $2\, m$ respectively are placed on a smooth floor. They are connected by a spring of negligible mass. $A$ third body $C$ of mass $m$ is placed on the floor. The body $C$ moves with a velocity $v_0$ along the line joining $A$ and $B$ and collides elastically with $A$. At a certain time after the collision it is found that the instantaneous velocities of $A$ and $B$ are same and the compression of the spring is $x_0$. The spring constant $k$ will be

When a spring is stretched by $2\,\, cm$ , it stores $100\,\, J$ of energy. If it is stretched further by $2\,\, cm$ , the stored energy will be increased by ............. $\mathrm{J}$