Question

A billiard balls is struck by a cue. The line of action of applied impulse is horizontal and passes through the centre of the ball. The initial velocity $V_0$ of the ball, its radius R, mass m and coefficient of friction $\mu$ between ball and table are known. The ball moves distance $\frac{12V_0^2}{(40+x)\mu g}$ before it ceases to slip on the table. What is the value of x.

Answer 1

Using kinematics equations:

$V^2-V_{\circ }{}^2=2as$

$V=V_{\circ }-at$

$a=-\mu g$

Using torque equations:

$\tau =I\alpha$

$\mu mgR=\frac{2}{5}m{R}^2\alpha$

$\alpha =\frac{5\mu g}{2R}$

$\omega ={\omega }_{\circ }+\alpha \times t$

$\omega =0+$ $\frac{5\mu g}{2R}t$

$\omega =\frac{V}{R}$

$\frac{5\mu g}{2R}t=$ $\frac{V_{\circ }-\mu gt}{R}$

$\frac{7\mu g}{2R}t=$ $\frac{V_{\circ }}{R}$

$t=$ $\frac{2V_{\circ }}{7\mu g}$

$V=V_{\circ }-\frac{2V_{\circ }}{7\mu g}\times \mu g$

$V=\frac{5V_{\circ }}{7}$

$(\frac{5V_{\circ }}{7}{)}^2-V_{\circ }{}^2=-2\mu gs$

$S=\frac{12V_{\circ }{}^2}{49\mu g}$

Hence answer is 49.