$A$ moving body with mass $m_1$ strikes a stationary mass $m_2$. What should be the ratio $\frac{m_1}{m_2}$ so as to decrease the velocity of the first body by $1.5$ times after the collision?

$A$ ball of $0.4 \, kg$ mass and a speed of $3 \, m/s$ has a head-on,completely elastic collision with a $0.6 \, kg$ mass initially at rest. Find the speeds of both balls after the collision:

$A$ ball moving with velocity $V$ undergoes a perfectly elastic collision with an identical ball moving in the opposite direction with velocity $2V$. Taking the direction of $V$ as positive,find the velocities of both balls after the collision.

$A$ particle of mass $m$ moving with velocity $u$ makes an elastic one-dimensional collision with a stationary particle of mass $m$. They are in contact for a very short time $T$. Their force of interaction increases from zero to $F_0$ linearly in time $T/2$, and decreases linearly to zero in further time $T/2$. The magnitude of $F_0$ is

There are $n$ elastic balls placed on a smooth horizontal plane. The masses of the balls are $m, \frac{m}{2}, \frac{m}{2^2}, \ldots, \frac{m}{2^{n-1}}$ respectively. If the first ball hits the second ball with velocity $v_0$, then the velocity of the $n^{\text{th}}$ ball will be,

$A$ ball falls from a height of $5\,m$ and strikes the roof of a lift. If at the time of collision,the lift is moving in the upward direction with a velocity of $1\,m/s$,then the velocity with which the ball rebounds after collision will be (assume $e = 1$):