Disc A of mass m collides with stationary disc B of mass 2m as shown in figure. The value of coefficient of restitution for which the two discs move in perpendicular direction after collision is e. Calculate 1/e.

Velocity of disc A in tangential direction is

${\left({\mathrm{v}}_1\right)}_{\mathrm{t}}=\mathrm{usin}{45}^{\circ }=\frac{\mathrm{u}}{\sqrt{2}}....\left(1\right)$

The disc B will move along common normal direction only

Since both the discs move perpendicular to each other, so disc A should move along common tangent direction.

So, ${\mathrm{v}}_{\mathrm{A}}=\frac{\mathrm{u}}{\sqrt{2}}$ along common tangent direction Applying conservation linear momentum along common normal direction, we get

$\mathrm{m}\left(\frac{\mathrm{u}}{\sqrt{2}}\right)+0=\mathrm{m}\left(0\right)+\left(2\mathrm{m}\right){\left({\mathrm{v}}_2\right)}_{\mathrm{n}}$

$\Rightarrow {\left({\mathrm{v}}_2\right)}_{\mathrm{n}}=\frac{\mathrm{u}}{2\sqrt{2}}$

Hence, velocity of disc B along common normal is $\frac{\mathrm{u}}{2\sqrt{2}}$

$\text{ Since, e }=\frac{(\text{ Relative velocity of separation }{)}_{n-\text{ line }}}{(\text{ Relative velocity of approach }{)}_{n\text{-line }}}$

$\Rightarrow \frac{{\left({\mathrm{v}}_2\right)}_{\mathrm{n}}-0}{\mathrm{ucos}{45}^{\circ }}=\frac{\mathrm{u}/2\sqrt{2}}{\mathrm{u}/\sqrt{2}}=\frac{1}{2}\Rightarrow \frac{1}{\mathrm{e}}=2$