Question

A billiard ball moving at 5.00 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 4.33 m/s at an angle of 30.0degrees with respect to the original line of motion. Assuming an elastic collision, find the struck ball's velocity and direction after the collision.

Answer 1

All Balls are identical so have same mass M.

$P_{xi}=5M\\ P_{xf}=4.33\cos{30\degree}M+ V\sin{\theta}M\\$

Applying conservation of linear momentum along original line of motion.

$$V\cos{\theta}=5-4.33\times{\cfrac{\sqrt{3}}{2}}......Eq[1]$$

Applying conservation of linear momentum along perpendicular direction.

$$P_{yi}=P_{yf}\\ 0=4.33M\sin{30\degree}-MV\sin{\theta}\\ V\sin{\theta}= \cfrac{4.33}{2}......Eq[2]\\$$

on solving Eq[1] & Eq[2], we get,

$$V=\sqrt{\bigg(5-4.33\cfrac{\sqrt{3}}{2}\bigg)^2+\bigg(\cfrac{4.33}{2}\bigg)^2}\\ V=2.5m/sec$$

and

$$\theta=\tan^{-1}\bigg(\cfrac{4.33}{10-4.33\times{\sqrt{3}}}\bigg)\\ \theta=\tan^{-1}(1.72)=59.8\degree\approx60\degree$$

Hence, Velocity of second ball is 2.5 m/sec and its direction from original line of motion is 60⁰.