Question

A solid sphere rolling on a rough horizontal surface with a linear speed $\nu$ collides elastically with a fixed, smooth vertical wall. Find the speed of the sphere after it has began pure rolling in the backwards direction.

• A. $\frac{2}{7}\nu$
• B. $\frac{3}{7}\nu$
• C. $\frac{4}{7}\nu$
• D. None

Answer 1

Answer: (B)

$\frac{3}{7}\nu$

We have the torque as Force $F_f$ into radius $r$.
Thus we get this torque as $F_{f}r=I\alpha$.
The moment of inertia for the solid sphere is $\frac{2}{5}m{r}^2\alpha$
We get this $\alpha$ as $\frac{5F_f}{2mr}$
This is the angular acceleration before collision.
Now after collision we have the equation as-
${\omega }_f=-\omega +\alpha t=-\omega +\frac{5F_f}{2mr}t$
$r{\omega }_f=-r\omega +\frac{5}{2}\frac{F_f}{m}t$
$v_f=-v+\frac{5}{2}\frac{F_f}{m}t$
Thus we get-
$v_f=-v+\frac{5}{2}(v-v_f)$
Rearranging we get -
$v_f=\frac{3}{7}v$