Question

A spherical ball of radius $r$ initially at rest on a rough horizontal surface is hit horizontally at a point at a distance $x$ above the central line. Due to this sharp impulse the centre of the ball acquires a velocity $v_{ c }$. After some time, the ball will start pure rolling with a velocity equal to $\frac{n}{7} v_{c}\left[\frac{x+r}{r}\right]$. The value of $n$ is___.

Answer 1

Initially ball will gain both linear and angular velocity.
Linear Impulse: $I_{0}=m v_{c}$
Angular Impulse: $I_{0} x=I \omega_{0}$
$ \Rightarrow m v_{c} x=I \omega_{0}$

$I$ is moment of Inertia about $C$.
Apply conservation of angular momentum about lowest point.
$I \omega_{0}+m v_{c} r=I \omega+m v r$
where
$\omega=\frac{v}{r}$ at the time of pure rolling
$\Rightarrow m v_{c} x+m v_{c} r=\frac{2}{5} m r^{2} \frac{v}{r}+m v r$
$\Rightarrow v_{C}(x+r)=\frac{7}{5} r v$
$\Rightarrow v=\frac{5}{7} v_{c}\left[\frac{x+r}{r}\right]$