The bob $A$ of a simple pendulum is released when the string makes an angle of $45^o$ with the vertical. It hits another bob $B$ of the same material and same mass kept at rest on the table. If the collision is elastic,then

$A$ body of mass $m_1$ undergoes a perfectly elastic collision with a stationary body of mass $m_2$. If the velocity of mass $m_1$ becomes $1/1.5$ times its initial velocity,find the ratio $\frac{m_1}{m_2}$.

$A$ 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:

Six steel balls of identical size are lined up along a straight frictionless groove. Two similar balls moving with a speed $V$ along the groove collide with this row on the extreme left end. Then:

$A$ 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:

$A$ body of mass $m$ is at rest. Another body of the same mass moving with velocity $V$ makes a head-on elastic collision with the first body. After the collision,the first body starts to move with velocity: