A cue ball of mass 260 g has a velocity of 17.4 m/s at 30 degrees when it impact

Question

A cue ball of mass 260 g has a velocity of 17.4 m/s at 30 degrees when it impacts with the eight ball with a velocity of 20 m/s at 240 degrees. Coefficient of restitution is .75. What are the velocities of the balls after the impact? i know its the Conservation of momentum equation but i am confused on how to apply it (Details please). A .35 slug body is moving at 24 ft/s to the left on the X axis. It is acted on by a force on the X axis of F = 8.75 sin (.25?t) lb. What is the velocity of the mass at t = 2.9 s

Explanation / Answer

FOLLOW THIS The force on the +70E-6: Using F = KQ1Q2/r2 +70 and +48: F = kq1q2/r2 = (8.99E9)(70E-6)(48E-6) / (.35)2 = 246.6 N (left - the 48 repels the 70) +70 and

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