Answer to Question #106072 in Mechanics | Relativity for Akshat shivhare

Question

A stationary ball, with a mass of 0.5 kg, is struck by an identical ball moving at
6.0ms .
1
After the collision, the second ball moves 30° to the left of its original
direction. The stationary ball moves 60° to the right of the moving ball’s original
direction. What is the velocity of each ball after the collision?

Answer 1

$\overrightarrow{v}$ - velocity of the second ball before the collision

$|\overrightarrow{v}|=6\frac{m}{s}$

$\overrightarrow{v_1},\overrightarrow{v_2}$ - velocities of the first and second balls after the collision accordingly

$m$ - the mass of each ball

According to the law of impulse conservation,

$m\overrightarrow{v}=m\overrightarrow{v_1}+m\overrightarrow{v_2}\rArr \overrightarrow{v}=\overrightarrow{v_1}+\overrightarrow{v_2}$

So $\overrightarrow{v}$ is the summ of vectors $\overrightarrow{v_1}$ and $\overrightarrow{v_2} \rArr$ according to the triangle law of adding vectors, $\overrightarrow{v}, \overrightarrow{v_1}, \overrightarrow{v_2}$ form a triangle

The angle between $\overrightarrow{v_1}$ and $\overrightarrow{v_2}$ is equal to $30^o+60^o=90^o$

So this triangle is right $\rArr |\overrightarrow{v_1}|=|\overrightarrow{v}|cos60^o=6\frac{m}{s}*\frac{1}{2}=3\frac{m}{s}$

$|\overrightarrow{v_2}|=|\overrightarrow{v}|cos30^o=6\frac{m}{s}*\frac{\sqrt{3}}{2}=3\sqrt{3}\frac{m}{s}\approx5.2\frac{m}{s}$