$A$ particle of mass $m$ is moving with a horizontal speed of $6 \, m/s$ as shown in the figure. If $m << M$,then for a one-dimensional elastic collision,the speed of the lighter particle after the collision will be:

$A$ moving body with a mass $m_1$ and velocity $u$ strikes a stationary body of mass $m_2$. The masses $m_1$ and $m_2$ should be in the ratio $\frac{m_1}{m_2}$,so as to decrease the velocity of the first body to $\frac{2u}{3}$ and give a velocity of $v$ to $m_2$,assuming a perfectly elastic impact. Then,the ratio $\frac{m_1}{m_2}$ is

The bob $A$ of a pendulum,released from a horizontal position to the vertical,hits another bob $B$ of the same mass at rest on a table as shown in the figure. If the length of the pendulum is $1\,m$,calculate:
$(a)$ The height to which bob $A$ will rise after the collision.
$(b)$ The speed with which bob $B$ starts moving.
Neglect the size of the bobs and assume the collision to be elastic.

There is a head-on elastic collision between two particles moving in the same direction with speeds $5 \,m/s$ and $3 \,m/s$ respectively. After the collision,the velocity of the first particle becomes $4 \,m/s$ in the same direction. What is the velocity of the second particle?

The coefficient of restitution for a perfectly elastic collision is

In an elastic collision of two billiard balls,which of the following quantities remain conserved during the short time of collision of the balls? (i.e.,when they are in contact)
$(a)$ Kinetic energy.
$(b)$ Total linear momentum.
Give reason for your answer in each case.