Question

Two billiard balls are rolling on a flat table. One has velocity compoments, Vx= 1 m/s, Vy=root3 m/s and other has components Vx= 2 m/s and Vy= 2 m/s. If both the balls start moving from the same point, what is the angle between their paths?

Answer 1

Here,
$\overrightarrow{V_1}=\left(1\overrightarrow{i}+\sqrt{3}\overrightarrow{j}\right)$

$\overrightarrow{V_2}=\left(2\overrightarrow{i}+2\overrightarrow{j}\right)$

Dot product of the two vectors are
$\overrightarrow{V_1}.\overrightarrow{V_2}=\left(1\overrightarrow{i}+\sqrt{3}\overrightarrow{j}\right).\left(2\overrightarrow{i}+2\overrightarrow{j}\right)=1.\left(2\right)+1\sqrt{3}.\left(2\right)=5.47$

Magnitude of the vectors are

$$\begin{gathered} \left|V_1\right|=\sqrt{1^2+{\left(\sqrt{3}\right)}^2}=1.82 \\ \left|V_2\right|=\sqrt{2^2+2^2}=2.83 \end{gathered}$$

Now the angle between the path of the balls

$$\begin{gathered} \cos \theta =\frac{V_1.V_2}{\left|V_1\right|.\left|V_2\right|}=\frac{5.47}{1.82\times 2.83}=\frac{5.47}{5.16}=1 \\ or,\theta =0^o \end{gathered}$$