Two identical particles are moving with same velocity $v$ as shown in the figure. If the collision is completely inelastic,then:

$A$ body of mass $4 \,kg$ is moving with momentum of $8 \,kg \,m/s$. $A$ force of $0.2 \,N$ acts on it in the direction of motion of the body for $10 \,s$. The increase in kinetic energy in joules is

Consider a drop of rain water having mass $1\,g$ falling from a height of $1\,km$. It hits the ground with a speed of $50\,m s^{-1}$. Take $g = 10\,m s^{-2}$. The work done by the $(i)$ gravitational force and the $(ii)$ resistive force of air is

$A$ simple pendulum of length $L = \frac{10}{3} \text{ m}$ with a bob of mass $M = 3m$ is hanging freely from a rigid support. $A$ bullet of mass $m$ is fired with a velocity $u = 50 \text{ ms}^{-1}$ from the ground at an angle $\theta$ with the horizontal. When the bullet is at its highest point of its trajectory,it collides head-on with the bob of the pendulum and gets embedded in the bob. After collision,if the pendulum moves through a maximum angle of $120^{\circ}$,then the value of $\theta$ is $(g = 10 \text{ ms}^{-2})$.

$A$ body $x$ with a momentum $p$ collides with another identical stationary body $y$ one-dimensionally. During the collision,$y$ gives an impulse $J$ to body $x$. Then,the coefficient of restitution is

$A$ shell at rest on a smooth horizontal surface explodes into two fragments of masses $m_1$ and $m_2$. If just after explosion $m_1$ moves with speed $u$,then the work done by internal forces during the explosion is: