Two balls A and B , of masses M and 2 M respectively collide each other. If the ball A moves with a speed of…

Applying conservation of linear momentum before and after collsion,

$$\begin{gathered} M\left(150\right)+2Mv=M\left(0\right)+2M\left(v^{\text{'}}\right) \\ \Rightarrow 150+2v=2v^{\text{'}} \\ \Rightarrow v^{\text{'}}=75+v ....\left(1\right) \end{gathered}$$

Now, coefficient of restitution is given by $e=\frac{v^{\text{'}}-0}{150-v}=1$

$\Rightarrow 150-v=v^{\text{'} } ...\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$, we have

$$\begin{gathered} 150-v=75+v \\ \Rightarrow 75=2v \\ \Rightarrow v=37.5 \mathrm{m} {\mathrm{s}}^{-1} \end{gathered}$$

Thus, speed of ball $B$ before it collides with ball $A$ is $37.5 \mathrm{m} {\mathrm{s}}^{-1}$.