$A$ 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

Body $A$ of mass $4m$ moving with speed $u$ collides with another body $B$ of mass $2m$ at rest. The collision is head-on and elastic in nature. After the collision, the fraction of energy lost by the colliding body $A$ is:

$A$ smooth sphere $A$ of mass $m$ collides elastically with an identical sphere $B$ at rest. The velocity of $A$ before collision is $8 \ m/s$ in a direction making $60^{\circ}$ with the line of centers at the time of impact.
$(i)$ The sphere $A$ comes to rest after collision.
$(ii)$ The sphere $B$ will move with a speed of $8 \ m/s$ after collision.
$(iii)$ The directions of motion of $A$ and $B$ after collision are at right angles.
$(iv)$ The speed of $B$ after collision is $4 \ m/s$.
The correct option is:

When two particles collide elastically,then impulsive force can't change

$A$ ball of mass $m$ moving with velocity $V$ makes a head-on elastic collision with a ball of the same mass moving with velocity $2V$ towards it. Taking the direction of $V$ as positive,the velocities of the two balls after the collision are:

$A$ 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 $^{\circ}$)