Question

two balls are moving on a table one has the velocity components u x is equals to 1 metre per second u y is equals to under root 3 metre per second and the other has components v X is equals to 2 metre per second and v y is equals to 2 metre per second if both starts moving from the same point angle between their direction of motion is​

Answer 1

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

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

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

The dot 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}

Hope it helps......