$A$ heavy body moving with a velocity of $6\,ms^{-1}$ collides elastically with a light body (whose mass is half of the mass of the heavy body) at rest. The velocity of the light body will be (in $ms^{-1}$):

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} = $

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

Ball $A$ moving at $12 \ m/s$ collides elastically with ball $B$ at rest as shown in the figure. If both balls have the same mass,what is the final velocity of ball $A$ after the collision (in $m/s$)? (Assume the collision is two-dimensional and the angle of incidence is $60^\circ$ with the vertical axis).

$A$ light particle moving horizontally with a speed of $12\ m/s$ strikes a very heavy block moving in the same direction at $10\ m/s$. The collision is one-dimensional and elastic. After the collision,the particle will

$A$ steel sphere of radius $1.2 \ cm$ collides with a second steel sphere at rest. If the collision is elastic and after the collision the first sphere continues to move in its initial direction with a velocity of $\frac{7}{9}$ times its initial velocity,then the radius of the second sphere is (in $cm$)