A uniform circular disc of mass m and radius R is rolling without slipping on a smooth horizontal surface as shown in figure. The disc collides with a rough vertical wall having coefficient of restitution 0.6. If the disc rolls without slipping after collision with the wall, and the coefficient of friction is n/10, then the value of ‘n’ is ____

After hitting the wall both linear velocity and angular velocity will change its direction.

for rolling for final linear velocity and angular velocity, $\mathrm{v}={\mathrm{r\omega }}_{\mathrm{f}}$

As the coefficient of restitution is 0.6, and final linear velocity is 0.6v

$\int \mathrm{N}.\mathrm{dt}=\mathrm{mv}(1+0.6)$...(1)

$\int \mathrm{fdt}=\int \mathrm{\mu Ndt}\Rightarrow {\mathrm{\omega }}_{\mathrm{f}}={\mathrm{\omega }}_{\mathrm{i}}-\left[\frac{1}{\mathrm{I}}\int \mathrm{fdt}\right]$...(2)

from (1) and (2)

$\frac{-0.6\mathrm{v}}{\mathrm{r}}=\frac{\mathrm{v}}{\mathrm{r}}-\left[\frac{1}{\mathrm{i}}\int \mathrm{\mu Ndt}\right]$

$\frac{-1.6\mathrm{v}}{\mathrm{r}}=\left[\frac{2}{{\mathrm{mr}}^2}(\mathrm{\mu }\times 1.6\mathrm{mv})\right]\left( \mathrm{from} \left(1\right)\right)$

we get $\mathrm{\mu }=0.5$