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Question
A steel ball of radius R = 20 cm and m = 2 kg is rotating about a horizontal diameter with angular speed w0 = 50 rad/sec. This rotating ball is dropped on a rough horizontal floor and fall freely through a height h = 1.25 m. The coefficient of restitution is e = 1 and coefficient of friction between ball and floor is . Then the distance in m between the point of first and second impact of the ball and floor is
- 4 m/s
- 3 m/s
- 6 m/s
- 8 m/s
The correct answer is: 3 m/s
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