Two balls $A$ and $B$,of masses $M$ and $2M$ respectively,collide with each other. If ball $A$ moves with a speed of $150 \ m \ s^{-1}$ and collides with ball $B$ moving with speed $v$ in the opposite direction,and after the collision ball $A$ comes to rest with the coefficient of restitution being $1$,then the speed of ball $B$ before it collides with ball $A$ is: (in $m \ s^{-1}$)

$A$ ball of mass $1 \ kg$ moving along the $X$-direction collides elastically with a stationary ball of mass $m$. The first ball (mass $= 1 \ kg$) recoils at a right angle to its original direction of motion. If the second ball starts moving at an angle $30^{\circ}$ with the $X$-axis,the value of $m$ must be (in $kg$)

$A$ smooth sphere $A$ of mass $m$ collides elastically with an identical sphere $B$ at rest. The velocity of $A$ before collision is $8 \ m/s$ in a direction making $60^{\circ}$ with the line of centers at the time of impact.
$(i)$ The sphere $A$ comes to rest after collision.
$(ii)$ The sphere $B$ will move with a speed of $8 \ m/s$ after collision.
$(iii)$ The directions of motion of $A$ and $B$ after collision are at right angles.
$(iv)$ The speed of $B$ after collision is $4 \ m/s$.
The correct option is:

$A$ 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:

$A$ body of mass $4\ kg$ collides head-on elastically with another body of mass $2\ kg$ kept at rest in free space. The time of collision is $0.02\ s$ and the average impulsive force acting on each body is $100\ N$. Find the velocity of the $2\ kg$ body after the impact. (in $m/s$)

In a scattering experiment,a particle of mass $2m$ collides with another particle of mass $m$,which is initially at rest. Assuming the collision to be perfectly elastic,the maximum angular deviation $\theta$ of the heavier particle,as shown in the figure,in radians is: