$A$ block of mass '$m$' moving on a frictionless horizontal surface collides with a spring of spring constant '$K$' and compresses it through a distance '$x$'. The maximum momentum of the block after collision 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.
Statement-$1$: $A$ point particle of mass $m$ moving with speed $u$ collides with a stationary point particle of mass $M$. If the maximum energy loss possible is given as $f \left( \frac{1}{2} m u^2 \right)$, then $f = \left( \frac{m}{M + m} \right)$.
Statement-$2$: Maximum energy loss occurs when the particles get stuck together as a result of the collision.

$A$ block of mass '$m$' with an initial kinetic energy '$E$' moves up an inclined plane of inclination '$\theta$'. If '$\mu$' is the coefficient of friction between the plane and the body, the work done against friction before coming to rest is

$A$ bullet of $10\, \text{g}$, moving with velocity $v$, collides head-on with the stationary bob of a pendulum and recoils with velocity $100\, \text{m/s}$. The length of the pendulum is $0.5\, \text{m}$ and the mass of the bob is $1\, \text{kg}$. The minimum value of $v$ in $\text{m/s}$ so that the pendulum completes a vertical circle is: (Assume the string to be inextensible and $g=10\, \text{m/s}^2$)

$A$ particle of mass $m$ is projected from the ground with an initial speed $u_0$ at an angle $\alpha$ with the horizontal. At the highest point of its trajectory,it makes a completely inelastic collision with another identical particle,which was thrown vertically upward from the ground with the same initial speed $u_0$. The angle that the composite system makes with the horizontal immediately after the collision is :

$A$ toy of mass $20 \,g$ at rest acquires a velocity $(3 \hat{i}-2 \hat{j}) \,m/s$ in $2 \,s$. The power of the toy is: (in $\,W$)