Question

Two billiard balls are rolling on a flat table. One has velocity components $V_x=1m/s,V_y=\sqrt{3}m/s$ and the other has component $V_x=2m/s$ and $V_y=2m/s$. If both the balls start moving from the same point the angle between their path is:

• A. ${60}^{\circ }$
• B. ${45}^{\circ }$
• C. ${22.5}^{\circ }$
• D. ${15}^{\circ }$

Answer 1

The correct option is D ${15}^{\circ }$

Components of the relative velocity of the billiard balls are:
$V_x=1m{s}^{-1}$
$V_y=(2-\sqrt{3})m{s}^{-1}$
$tan\theta =\frac{V_y}{V_x}=\frac{2-\sqrt{3}}{1}$

$i.e.\theta =ta{n}^{-1}\frac{2-\sqrt{3}}{1}={15}^{\circ }$

$\theta$ can also be found by using the formula of $\tan \left(2\theta \right)$.

$\tan \left(2\theta \right)=\frac{2\tan \left(\theta \right)}{1-{\tan }^2\left(\theta \right)}$

$=\frac{2(2-\sqrt{3})}{1-(2-\sqrt{3}{)}^2}$

$=\frac{2(2-\sqrt{3})}{4\sqrt{3}-6}$

$=\frac{2(2-\sqrt{3})}{2\sqrt{3}(2-\sqrt{3})}$

$=\frac{1}{\sqrt{3}}$