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A bowling ball with a mass of 9kg is thrown down a lane with a constant speed of 3 m/s. The ball hits the 1.5kg pin, initially at rest, at the end of the lane. After the collision the pin moves with a speed of 5 m/s. How fast is the ball moving after the impact?​

User TheRemix
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8.7k points

1 Answer

8 votes

Answer:

M1 V1 = M1 V2 + M2 V3 conservation of momentum

V2 = (M1 V1 - M2 V3) / M1 where V2 = speed of M1 after impact

V2 = (3 * 9 - 1.5 * 5) / 9 = (27 - 7.5) / 9 = 2.17 m/s

Note: All speeds are in the same direction and have the same sign

User Netero
by
9.2k points
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