$A$ 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 the normal reaction $N$ between the block and the spring on the block is

Two blocks each of mass $m$  are connected to a spring of spring constant $k.$  If both are given velocity $v$  in opposite directions, then the maximum elongation of the spring is

A vertical spring with force constant $k$ is fixed on a table. A ball of mass $m$ at a height $h$ above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance $d.$ The net work done in the process is

This question has Statement $1$ and Statement $2$. Of the four choices given after the Statements, choose the one that best describes the two Statements.

If two springs $S_1$ and $S_2$ of force constants $k_1$ and $k_2$, respectively, are stretched by the same force, it is found that more work is done on spring $S_1$ than on spring $S_2$.
STATEMENT 1 : If stretched by the same amount work
done on $S_1$, Work done on $S_1$ is more than $S_2$
STATEMENT2: $k_1 < k_2$

$A$ small block of mass $m$ is placed on $a$ wedge of mass $M$ as shown, which is initially at rest. All the surfaces are frictionless . The spring attached to the other end of wedge has force constant $k$. If $a'$ is the acceleration of $m$ relative to the wedge as it starts coming down and $A$ is the acceleration acquired by the wedge as the block starts coming down, then  Maximum velocity of $M$ is:

The system of the wedge and the block connected by a massless spring as shown in the figure is released with the spring in its natural length. Friction is absent. maximum elongation in the spring will be