$A$ steel sphere of radius $1.2 \ cm$ collides with a second steel sphere at rest. If the collision is elastic and after the collision the first sphere continues to move in its initial direction with a velocity of $\frac{7}{9}$ times its initial velocity,then the radius of the second sphere is (in $cm$)

$A$ 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:

$A$ billiard table has length and width as shown in the figure. $A$ ball is placed at point $A$. At what angle $\theta$ should the ball be projected so that after colliding with two walls,the ball will fall into the pocket $B$? Assume that all collisions are perfectly elastic and neglect friction.

$A$ particle of mass $m$ moving with a velocity $u$ makes an elastic one-dimensional collision with a stationary particle of mass $m$. They are in contact for a total time $T$. The contact force increases linearly from $0$ to $F_0$ in time $\frac{T}{4}$, remains constant for a further time $\frac{T}{2}$, and decreases linearly from $F_0$ to $0$ in the final time $\frac{T}{4}$, as shown in the graph. The magnitude of $F_0$ is:

$A$ moving particle collides with a stationary particle of mass $\frac{1}{n}$ times the mass of the moving particle. The fraction of its kinetic energy transferred to the stationary particle is:

$A$ ball of mass $m$ suspended from a rigid support by an inextensible massless string is released from a height $h$ above its lowest point. At its lowest point, it collides elastically with a block of mass $2m$ at rest on a frictionless surface. Neglect the dimensions of the ball and the block. After the collision, the ball rises to a maximum height of