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 $\dots N$.

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A$: Body $P$ having mass $M$ moving with speed $u$ has a head-on elastic collision with another body $Q$ having mass $m$ initially at rest. If $m << M$,body $Q$ will have a maximum speed equal to $2u$ after the collision.
Reason $R$: During an elastic collision,the momentum and kinetic energy are both conserved.
In the light of the above statements,choose the most appropriate answer from the options given below:

Two balls $X(2 \ kg)$ and $Y(4 \ kg)$ approach each other with equal speeds of $10 \ ms^{-1}$. If the collision is perfectly elastic,then the new velocities of balls $X$ and $Y$ are respectively

$A$ $2\, kg$ block moving with $10\, m/s$ strikes a spring of constant $\pi^2\, N/m$ attached to a $2\, kg$ block at rest kept on a smooth floor. The velocity of the rear $2\, kg$ block after it separates from the spring will be ..... $m/s$.

Two masses $m_A$ and $m_B$ moving with velocities $v_A$ and $v_B$ in opposite directions collide elastically. After the collision,the masses $m_A$ and $m_B$ move with velocities $v_B$ and $v_A$ respectively. The ratio $\frac{m_A}{m_B} = $

The bob $A$ of a simple pendulum is released when the string makes an angle of $45^{\circ}$ 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,what happens after the collision?