Question

A billiard ball initially at rest is given an impulse by a cue as shown in figure. The cue is held horizontal at a distance $h$ above the centre line. The ball leaves the cue with speed $v_o$ and acquires a final velocity $\frac{9v_o}{7}$. What is the value of $h$ in terms of the radius of ball?

• A. $h=\frac{4}{5}R$
• B. $h=\frac{2}{5}R$
• C. $h=\frac{2}{3}R$
• D. $h=\frac{3}{5}R$

Answer 1

The correct option is A $h=\frac{4}{5}R$

We first use the conservation of linear momentum

$\Delta P=F\Delta t$

Thus we get

$m{v}_o-0=F\Delta t$

Also using the conservation of angular momentum we have

$\Delta L=\tau \Delta t$

$I_c{\omega }_o-0=Fh\Delta t$

${\omega }_o=\frac{m{v}_{o}h}{I_c}$

Here we see that $\omega \ne vR$. Thus this is a condition of rolling with slipping.

The kinetic frictional force is the only force which acts in horizontal direction and increases the velocity of the ball to $\frac{9}{7}{v}_o$ where pure rolling starts.

Now far the translational motion we have(refer the FBD)

$F_x=m{a}_x$

$\mu N=m{a}_c=\mu mg$

$a_c=\mu g$

Also for rotational motion

$\tau =I_c\alpha$

$-\mu NR=I_c\alpha$

$-\mu mgR=I_c\alpha$

$\alpha =-\frac{\mu mgR}{I_c}$(negative sign means anticlockwise direction)

Now when the ball starts rolling at time , we get the linear velocity of the ball as

$v=v_o+a_{c}t$

$\frac{9}{7}{v}_o=v_o+\mu gt$

$t=\frac{2v_o}{7\mu g}$

Also the angular velocity of the ball is given as

$\omega ={\omega }_o+\alpha t$

$\omega ={\omega }_o-\left(\frac{\mu mgR}{I_c}\right)t=\frac{m{v}_{o}h}{I_c}-\frac{2}{7}\frac{mR{v}_o}{I_c}$

$\omega =\frac{m{v}_o}{I_c}(h-\frac{2}{7}R)$

Now when the pure rolling starts we have

$v=R\omega$

$\frac{9}{7}{v}_o=\frac{m{v}_o}{I_c}(h-\frac{2}{7}R)R$

$\frac{9}{7}=\frac{5}{2}\frac{m}{m{R}^2}(h-\frac{2}{7}R)R$

$\frac{9}{7}=\frac{5}{2R}(h-\frac{2}{7}R)=\frac{5}{2}\frac{h}{R}-\frac{5}{7}$

$h=\frac{4}{5}R$