An unknown nucleus of mass $m_1$ collides with a stationary ${}^4He$ nucleus of mass $m_2$. After the collision,the two nuclei travel in perpendicular directions relative to each other. If kinetic energy is lost in the collision,the unknown nucleus must be:

$A$ body moving along a straight line collides with another body of the same mass moving in the same direction with half of the velocity of the first body. If the coefficient of restitution between the two bodies is $0.5$,then the ratio of the velocities of the two bodies after collision is (treat the collision as one dimensional).

Two particles of masses $m_1$ and $m_2$ move with initial velocities $u_1$ and $u_2$. On collision,one of the particles gets excited to a higher level after absorbing energy $\varepsilon$. If the final velocities of the particles are $v_1$ and $v_2$,then we must have:

$A$ ball of mass $m$ moving with a velocity of $9 \ m/s$ collides with another ball of the same mass $m$ at rest. After the collision,both balls move at an angle of $30^\circ$ with the original direction of motion. Find the velocity $v$ of both balls after the collision. (in $m/s$)

Two identical balls $A$ and $B$ of equal mass $m$ are lying on a smooth surface as shown in the figure. If ball $A$ hits ball $B$ at rest with a velocity $16 \,ms^{-1}$,then the coefficient of restitution $e$ between $A$ and $B$ so that $B$ just reaches the highest point of the smooth inclined plane of height $5 \,m$ is $\left(g=10 \,ms^{-2}\right)$

$A$ ball is dropped from a height $h$. After it strikes the ground twice,what height will it reach?