Question

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

Answer 1

Step 1: Given data

Particle one velocity in the x-direction $V_{x1}=1\frac{\text{m}}{\text{s}}$

Particle one velocity in the y-direction $V_{y1}=\sqrt{3}\frac{\text{m}}{\text{s}}$

Particle second velocity in the x-direction $V_{x2}=2\frac{\text{m}}{\text{s}}$

Particle second velocity in the y-direction $V_{y2}=2\frac{\text{m}}{\text{s}}$

Step 2: Calculate the angle between their path

1. The problem specifies two velocity components, namely the x-component and the y-component.
2. This means that the balls have been moving in the XY plane. We will determine the relative velocity of one billiard ball in relation to the other one.
3. We will next utilize fundamental vector knowledge to calculate the angle of such a relative velocity, and then apply trigonometry characteristics to obtain the real angle.
4. Firstly we find the angle of velocity component first ball and further the angle of velocity component of the second ball. Then the angle between the paths of the two balls is equivalent to the angle between their velocity vectors.

Mathematical representation is below:

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{V_{y1}}{V_{x1}}\right)-{\tan }^{-1}\left(\frac{V_{y2}}{V_{x2}}\right) \\ ={\tan }^{-1}\left(\frac{\sqrt{3}{\displaystyle \frac{\text{m}}{\text{s}}}}{1{\displaystyle \frac{\text{m}}{\text{s}}}}\right)-{\tan }^{-1}\left(\frac{2{\displaystyle \frac{\text{m}}{\text{s}}}}{2{\displaystyle \frac{\text{m}}{\text{s}}}}\right) \\ ={\tan }^{-1}\left(\sqrt{3}\right)-{\tan }^{-1}\left(1\right) \\ ={60}^{°}-45° \\ =15° \end{gathered}$$

Hence, the angle between their path is $15°$.