Two solid rubber balls $A$ and $B$ having masses $200 \, g$ and $400 \, g$ respectively are moving in opposite directions. The velocity of $A$ is $0.3 \, m/s$. After the collision,both balls come to rest. Determine the velocity of $B$ in $m/s$.

$A$ wooden block of mass $m$ moves with velocity $v$ and collides with another block of mass $4m$,which is at rest. After the collision,the block of mass $m$ comes to rest. The coefficient of restitution will be:

$A$ body falls on a surface of coefficient of restitution $0.6$ from a height of $1 \ m$. Then the body rebounds to a height of (in $m$)

An electron collides with a free molecule initially in its ground state. The collision leaves the molecule in an excited state that is metastable and does not decay to the ground state by radiation. Let $K$ be the sum of the initial kinetic energies of the electron and the molecule,and $p$ be the sum of their initial momenta. Let $K^{\prime}$ and $p^{\prime}$ represent the same physical quantities after the collision. Then,

$A$ ball is dropped from a height $h_0$. It undergoes $n$ collisions with the ground. If the velocity of the ball after $n$ rebounds is $u_n$ and it reaches a height $h_n$,which formula gives the coefficient of restitution $e$?

$A$ ball of mass $m$ moving with velocity $v$ collides head-on with a second ball of mass $m$ at rest. If the coefficient of restitution is $e$,the velocity of the first ball after collision is $v_1$,and the velocity of the second ball after collision is $v_2$,then: