$A$ heavy body moving with a velocity $30\, m/s$ and another small object at rest undergo an elastic collision. The latter will move with a velocity of .............. $m/s$.

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:

Consider an elastic collision of a particle of mass $m$ moving with a velocity $u$ with another particle of the same mass at rest. After the collision, the projectile and the struck particle move in directions making angles $\theta_1$ and $\theta_2$ respectively with the initial direction of motion. The sum of the angles, $\theta_1 + \theta_2$, is equal to: (in $^\circ$)

$A$ particle of mass $1\, kg$ moving with velocity $1\, m/s$,collides elastically with another particle of mass $m$ initially at rest. In the collision,the particle of mass $1\, kg$ loses $\frac{3}{4}$ of its kinetic energy $(K.E.)$. The value of $m$ is:

$A$ 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:

$A$ smooth sphere $A$ moves on a frictionless horizontal surface with an angular velocity $\omega$ and a linear velocity $v$ of its center of mass. It undergoes an elastic collision with another identical sphere $B$ at rest. Neglecting friction everywhere,if the angular speeds after the collision are $\omega_A$ and $\omega_B$ respectively,then: