Two billiard balls of mass $0.05\,kg$ each,moving in opposite directions with $10\,m/s$,collide and rebound with the same speed. If the time duration of contact is $t = 0.005\,s$,then the force exerted on the balls due to each other is $\dots N$.

What is a head-on collision?

$A$ simple pendulum,made of a string of length $l$ and a bob of mass $m$,is released from a small angle $\theta_0$. It strikes a block of mass $M$,kept on a horizontal surface at its lowest point of oscillations,elastically. It bounces back and goes up to an angle $\theta_1$. Then $M$ is given by

$A$ molecule of mass $m$ moving with velocity $v$ makes $5$ elastic collisions with a wall of a container per second. The change in momentum of the wall per second in $5$ collisions will be:

Body $A$ of mass $4m$ moving with speed $u$ collides with another body $B$ of mass $2m$ at rest. The collision is head-on and elastic in nature. After the collision, the fraction of energy lost by the colliding body $A$ is:

An alpha-particle of mass $m$ undergoes a $1$-dimensional elastic collision with a nucleus of unknown mass $M$ at rest. It is scattered directly backwards,losing $64\%$ of its initial kinetic energy. The mass of the nucleus is .......... $m$.