$A$ ball of mass $1\,kg$ moving with a velocity of $4\,m/s$ collides with a stationary ball of mass $M$. The collision is oblique. After the collision,the first ball moves at a right angle to its initial direction with a velocity of $3\,m/s$. The momentum of the second ball (in $kg\cdot m/s$) after the collision would be nearly:

$A$ spring-block system is resting on a frictionless floor as shown in the figure. The spring constant is $2.0 \,N \,m^{-1}$ and the mass of the block is $2.0 \,kg$. Ignore the mass of the spring. Initially, the spring is in an unstretched condition. Another block of mass $1.0 \,kg$ moving with a speed of $2.0 \,m \,s^{-1}$ collides elastically with the first block. The collision is such that the $2.0 \,kg$ block does not hit the wall. The distance, in metres, between the two blocks when the spring returns to its unstretched position for the first time after the collision is. . . .

$A$ man who is running has half the kinetic energy of a boy of half his mass. The man speeds up by $1 \, m/s$ and then has the same kinetic energy as the boy. The original speed of the man was:

Assertion $(A)$: In an elastic collision of two billiard balls,the total kinetic energy $(KE)$ is conserved during the short time of collision of the balls (i.e.,when they are in contact).
Reason $(R)$: Energy spent against friction does not follow the law of conservation of energy.

$A$ child swings on a swing such that the minimum and maximum heights from the ground are $0.75 \ m$ and $2 \ m$ respectively. The maximum velocity of the swing is ...... $ms^{-1}$.

$A$ block $C$ of mass $m$ is moving with velocity $v_0$ and collides elastically with block $A$ of mass $m$ which is connected to another block $B$ of mass $2m$ through a spring of spring constant $k$. What is $k$ if $x_0$ is the compression of the spring when the velocity of $A$ and $B$ is the same?