Question

Two billiard balls are rolling on a flat table one has the velocity component Vx=1m/S and Vy= under root 3m/S and the other has component Vx=2m/S Vy=2m/S if both balls starts moving from same point what is the angle between their paths

Answer 1

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Answer 2

Answer:

The angle between the path of the balls is 0°

Explanation:

Given that,

The velocity component

$v_{x}=1\ m/s$

$v_{y}=2\ m/s$

The velocity component

$v_{x}=2\ m/s$

$v_{y}=2\ m/s$

The velocity of both velocities

$v_{1}=(1\hat{i}+\sqrt{3}\hat{j})$

$v_{1}=(2\hat{i}+2\hat{j})$

The dot product of the velocities

$v_{1}\cdot v_{2}=(1\hat{i}+\sqrt{3}\hat{j})\cdot (2\hat{i}+2\hat{j})$

$v_{1}\cdot v_{2}=2+2\sqrt{3}$

$v_{1}\cdot v_{2}=5.46\ m/s$

The magnitude of the vectors are

$|v_{1}|=\sqrt{1^2+(\sqrt{3})^2}=2$

$|v_{2}|=\sqrt{2^2+2^2}=2.82$

Now, the angle between the path of the balls

$\cos\theta=\dfrac{v_{1}\dotc v_{2}}{|v_{1}|\cdot |v_{2}|}$

$\cos\theta=\dfrac{5.46}{2\times2.82}$

$\cos\theta=0.96\ approx\ 1$

$\cos\theta=cos0^{\circ}$

$\theta=0^{\circ}$

Hence, The angle between the path of the balls is 0°.