$A$ body falling on the ground from a height of $10 \, m$,rebounds to a height of $2.5 \, m$. The ratio of the velocities of the body just before and after the collision will be:

$A$ body of mass $2 \ kg$ makes an elastic collision with another body at rest and continues to move in the original direction with one-fourth of its original speed. The mass of the second body which collides with the first body is ............... $kg$.

$A$ point mass $M$ moving with a certain velocity collides with a stationary point mass $M / 2$. The collision is elastic and in one-dimension. Let the ratio of the final velocities of $M$ and $M / 2$ be $x$. The value of $x$ is

$A$ billiard ball of mass $M$,moving with velocity $v_1$ collides with another ball of the same mass but at rest. If the collision is elastic,the angle of divergence after the collision is (in $^{\circ}$)

$A$ simple pendulum,made of a string of length $l$ and a bob of mass $m$,is released from a small angle $\theta_0$. It strikes a block of mass $M$,kept on a horizontal surface at its lowest point of oscillations,elastically. It bounces back and goes up to an angle $\theta_1$. Then $M$ is given by

The rebound coefficient between a tennis ball and a racket is defined as $e = v_2/v_1$,where $v_1$ is the incoming speed of the ball and $v_2$ is the speed of the ball after rebound while the racket is at rest. $A$ tennis ball falls from height $H$ to a racket at rest and bounces back to $0.8 H$. $A$ tennis player is using the racket to hit an incoming tennis ball traveling at $150 \ km/hr$ and the racket is moving at $100 \ km/hr$ towards the ball. What is the speed of the ball after being hit? (Assume the mass of the racket $\gg$ that of the ball).