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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?

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Answer:

7.6427m/s

Explanation:

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

#Applying the conservation of momentum along the x-axis:[tex]mv_i=mv_acos\theta_a+mv_bcos\theta_b[/tex]

#And along y-axis:

[tex]0=-sin\theta_a+mv_bsin\theta_b[/tex]

#Solving  for [tex]\theta_b[/tex]:

[tex]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[/tex]

#By substitution in the x-axis equation:

[tex]v_i=4.7cos 33.0\textdegree +4.5cos 34.67\textdegree\\=7.6427m/s[/tex]

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

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