Question

Billiard ball A strikes another ball B of the same mass, which is at rest, such that after the impact they move at angles ΘA and ΘB respectively. The velocity of ball A after impact is 4.70 m/s at an angle ΘA = 33.0 ° while ball B moves with speed 4.50 m/s.Figure showing a layout of the problem as described in text What is ΘB (in degrees)? What is the original speed of ball A before impact?

Answer 1

Answer:

7.6427m/s

Explanation:

Given:$v_a=4.7m/s, \theta_a=33.0\textdegree and \ v_b=4.5m/s$

#Applying the conservation of momentum along the x-axis:$mv_i=mv_acos\theta_a+mv_bcos\theta_b$

#And along y-axis:

$0=-sin\theta_a+mv_bsin\theta_b$

#Solving  for $\theta_b$:

$sin\rheta_b=\frac{v_a}{v_b}sin\theta_a=4.7/4.5\times sin33.0\textdegree\\=0.5688\\\therefore \theta_b=34.67\textdegree$

#By substitution in the x-axis equation:

$v_i=4.7cos 33.0\textdegree +4.5cos 34.67\textdegree\\=7.6427m/s$

Hence the original speed of the ball before impact is 7.6427m/s