A body of mass $M$ moves with velocity $v$ and collides elastically with a another body of mass $m$ ($M>>m$) at rest then the velocity of body of mass $m$ is
$v$
$2v$
$v/2$
$Zero$
A sphere of mass $m $ moving with a constant velocity $u$ hits another stationary sphere of the same mass. If $e$ is the coefficient of restitution, then the ratio of the velocity of two spheres after collision will be
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 $60^o$ with the line of centres 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 $A$ and $B$ after collision are at right angles.
$(iv)$ The speed of $B$ after collision is $4\ m/s$ . The correct option is
A tennis ball is released from height $h $ above ground level. If the ball makes inelastic collision with the ground, to what height will it rise after third collision
As per the given figure, a small ball $P$ slides down the quadrant of a circle and hits the other ball $Q$ of equal mass which is initially at rest. Neglecting the effect of friction and assume the collision to be elastic, the velocity of ball $Q$ after collision will be $............\,m/s$ $:\left( g =10\,m / s ^2\right)$
Two identical balls $A$ and $B$ are released from the positions shown in figure. They collide elastically on horizontal portion $MN$ . The ratio of the heights attained by $A$ and $B$ after collision will be : (neglect friction)