A block of mass ${\mathrm{m}}_1 =150\mathrm{kg}$ is at rest on a very long frictionless table, one end of which is terminated in a wall. Another block of mass ${\mathrm{m}}_2$ is placed between the first block and the wall, and set in motion towards ${\mathrm{m}}_1$ with constant speed ${\mathrm{u}}_2$ . Assuming that all collisions are completely elastic, find the value of ${\mathrm{m}}_2$ for which both blocks move with the same velocity after ${\mathrm{m}}_2$ has collided once with ${\mathrm{m}}_1$ and once with the wall. (The wall has
effectively infinite mass.)

After the collision, let us consider

${\mathrm{v}}_1$ = speed of mass ${\mathrm{m}}_1$ towards left
${\mathrm{v}}_2$ = speed of mass ${\mathrm{m}}_2$ towards right

From conservation of momentum, we have

$0 + {\mathrm{m}}_2{\mathrm{u}}_2 = {\mathrm{m}}_1{\mathrm{v}}_1 - {\mathrm{m}}_2{\mathrm{v}}_2$                (1)

The mass ${\mathrm{m}}_2$ rebounds elastically from the wall and its velocity gets reversed after the collision with the wall. According to the problem, the mass ${\mathrm{m}}_2$ has the same speed as that of mass ${\mathrm{m}}_1$ after its collision with the wall, that is, ${\mathrm{v}}_2 = {\mathrm{v}}_1$ .
Equation (1) becomes

${\mathrm{m}}_2{\mathrm{u}}_2 = \left({\mathrm{m}}_1 - {\mathrm{m}}_2\right){\mathrm{v}}_1$                 (2)

Since the collision is elastic, then

$$\begin{gathered} \frac{1}{2}{\mathrm{m}}_2{{\mathrm{u}}_2}^2 = \frac{1}{2}{m}_1{{\mathrm{v}}_1}^2 + \frac{1}{2}{\mathrm{m}}_2{{\mathrm{V}}_1}^2 \left(\therefore {\mathrm{v}}_1 = {\mathrm{v}}_2\right) \\ {\mathrm{m}}_2{{\mathrm{u}}_2}^2 = \left({\mathrm{m}}_1 + {\mathrm{m}}_2\right){{\mathrm{v}}_1}^2 \left(3\right) \end{gathered}$$From Eq. (2), we get

${\mathrm{v}}_1 = \frac{{\mathrm{m}}_2{\mathrm{u}}_2}{{\mathrm{m}}_1 - {\mathrm{m}}_2}$

Substituting this value of ${\mathrm{v}}_1$ in Eq. (3), we get

$$\begin{gathered} {\mathrm{m}}_2{{\mathrm{u}}_2}^2 = \frac{\left({\mathrm{m}}_1 + {\mathrm{m}}_2\right){\left({\mathrm{m}}_2{\mathrm{u}}_2\right)}^2}{{\left({\mathrm{m}}_1 - {\mathrm{m}}_2\right)}^2} \\ \Rightarrow \left({\mathrm{m}}_1 - {\mathrm{m}}_2\right) = \left({\mathrm{m}}_1 + {\mathrm{m}}_2\right)\left({\mathrm{m}}_2\right) \\ \Rightarrow {{\mathrm{m}}_1}^2 + {{\mathrm{m}}_2}^2 - 2{\mathrm{m}}_1{\mathrm{m}}_2 = {\mathrm{m}}_1{\mathrm{m}}_2 + {{\mathrm{m}}_2}^2 \\ \Rightarrow {{\mathrm{m}}_1}^2 = 3{\mathrm{m}}_1{\mathrm{m}}_2 \\ \Rightarrow {\mathrm{m}}_1 = \frac{{\mathrm{m}}_1}{3} = \frac{150}{3} = 50\mathrm{kg} \end{gathered}$$