$A$ body of mass $2\,kg$ makes an elastic collision with a second body of mass $m$ at rest and continues to move in the original direction but with one-fourth of its original speed. What is the mass of the second body? ................ $kg$

$A$ 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?

Four smooth steel balls of equal mass at rest are free to move along a straight line without friction. The first ball is given a velocity of $0.4\, m/s$. It collides head-on with the second elastically,the second one similarly with the third and so on. The velocity of the last ball is .............. $m/s$.

$A$ 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:

$A$ body of mass $M_1$ collides elastically with another mass $M_2$ at rest. There is maximum transfer of energy when

Consider the collision depicted in the figure to be between two billiard balls with equal masses $m_{1} = m_{2}$. The first ball is called the cue, while the second ball is called the target. The billiard player wants to 'sink' the target ball in a corner pocket, which is at an angle $\theta_{2} = 37^{\circ}$. Assume that the collision is elastic and that friction and rotational motion are not important. Obtain $\theta_{1}$.