Question

Two balls of masses 2 kg and 4 kg are moving towards each other with velocities $4m{s}^{-1}$ and $2m{s}^{-1}$ respectively on a frictionless surface. After the collision, the 2 kg ball returns back with a velocity of $2m{s}^{-1}$.

Determine (a) the velocity of 4 kg block after the collision. (b) the loss in kinetic energy in collision (c) coefficient of restitution.

Answer 1

1. In every collision momentum is always conserved when the external force is not there.
   So, here we will apply the conservation of momentum
   $m_1\overrightarrow{v_1}+m_2\overrightarrow{v_2}=m_1\overrightarrow{v{\text{'}}_1}+m_2\overrightarrow{v{\text{'}}_2}............\left(1\right)$
   Here are the masses of ball 2 kg and 4 kg respectively and $\overrightarrow{v_1}$ and $\overrightarrow{v_2}$ are the velocity of balls 2 kg and 4 kg respectively before collision while $\overrightarrow{v{\text{'}}_1}$ and $\overrightarrow{v{\text{'}}_2}$ are velocities for same but after collision
2. Lets put the values in equation (1)
   $$\begin{gathered} 2\left(4\right)+4(-2)=2(-2)+4\overrightarrow{v{\text{'}}_2} \\ \therefore \overrightarrow{v{\text{'}}_2}=1m{s}^{-1} \end{gathered}$$
   Here, velocities in the direction of $\overrightarrow{v_1}$is taken as positive while opposite to that of $\overrightarrow{v_1}$ is taken negatively.
   Now initial kinetic energy,
   $$\begin{gathered} K_1=\frac{1}{2}{m}_1{v_1}^2+\frac{1}{2}{m}_2{v_2}^2 \\ \therefore K_1=\frac{1}{2}\left(2\right){\left(4\right)}^2+\frac{1}{2}\left(4\right){\left(1\right)}^2\therefore K_2=6J \end{gathered}$$
   Hence, loss in kinetic energy is $K_2-K_1=12-6=6J$
   Now coefficient of restitution,
   $e=\frac{2-1}{4-2}$
   So $e=0.5$

And, the Velocity of a 4 kg ball after the collision is $1m{s}^{-1}$ in the direction of the initial velocity of the 2 kg ball.