Question

A billiard ball moving at 6.00 m/s strikes a stationary ball of the same mass. after the collision, the first ball moves at 5.39 m/s at an angle of 26.0° with respect to the original line of motion. assuming an elastic collision (and ignoring friction and rotational motion), find the struck ball's velocity after the collision.

Answer 1

To do this question, we need to know that momentum is conserved, meaning the overall velocity of the two balls has to be the same before and after the collision.  

After collision... 

Ball 1: 4.33m/s *cos 30 = 3.75 m/s (x-component) 
4.33m/s * sin 30 = 2.165 m/s ( y-component) 

Ball 2 (struck ball): 5 m/s - 3.75m/s = 1.25 m/s (x-component)  
-2.165 m/s (y-component) note: it has to be in the opposite direction to conserve momentum 

tan-1(2.165/1.25) = 60 degrees 
Struck ball's velocity = sqrt(1.25^2 + 2.165^2) = 2.5 m/s at 60 degree with respect to the original line of motion.  

Hope you understand!

Answer 2

Answer:

$v = 2.63 m/s$

$\theta = -66.2 degree$

Explanation:

In the initial direction of motion we can use momentum conservation

so here we have

$m_1v_1 = m_1v_{1x} + m_2v_{2x}$

$m(6.00) = m(5.39 cos26) + mv_{2x}$

$6 = 4.84 + v_{2x}$

$v_{2x} = 1.16 m/s$

now by momentum conservation along perpendicular direction we can use as

$0 = m_1v_{1y} + m_2v_{2y}$

$0 = m(5.39 sin26) + mv_{2y}$

$v_{2y} = -2.36 m/s$

So the magnitude of the final speed of the stuck ball is given as

$v = \sqrt{v_{2x}^2 + v_{2y}^2}$

$v = \sqrt{1.16^2 + 2.36^2}$

$v = 2.63 m/s$

direction of motion given as

$tan\theta = \frac{v_{2y}}{v_{2x}}$

$tan\theta = \frac{-2.36}{1.16} = -2.27$

$\theta = -66.2 degree$