Question

A spherical ball of radius $r$ initially at rest on a rough horizontal surface is hit horizontally at a distance $x$ above the central line. Due to this sharp impulse the centre of ball acquires a velocity $v_c$. After some time, the ball will start pure rolling with a velocity equal to

• A. $\frac{5}{7}{v}_c\left(\frac{x+r}{r}\right]$
• B. $\frac{2}{7}{v}_c\left(\frac{x+r}{r}\right]$
• C. $\frac{5}{7}{v}_c\left(\frac{x+r}{x}\right]$
• D. $\frac{2}{7}{v}_c\left(\frac{x+r}{x}\right]$

Answer 1

The correct option is A $\frac{5}{7}{v}_c\left(\frac{x+r}{r}\right]$

Let, $I_0$ be linear impulse imparted to ball and $I$ be the moment of Inertia about 'C' and final velocity be $v$.
Initially, ball will gain both linear and angular velocity.
Linear impulse: $I_0=m{v}_c$
and Angular impulse: $I_{0}x=I{\omega }_0$where, ${\omega }_0$ be inital velocity gained.
By using both equation, $m{v}_{c}x=I{\omega }_0$




Apply conservation of angular momentum about lowest point,
$I{\omega }_0+m{v}_{c}r=I\omega +mvr$
where, $\omega =\frac{v}{r}$ at the time of pure rolling
$\Rightarrow m{v}_{c}x+m{v}_{c}r=\left(\frac{2m{r}^2}{5}\right)\times \left(\frac{v}{r}\right)+mvr$
$\Rightarrow v_c\left(x+r\right)=\frac{7}{5}rv\Rightarrow v=\frac{5}{7}{v}_c\left(\frac{x+r}{r}\right)$