A golf ball strikes a hard, smooth floor at an angle of 37.3 A golf ball strikes

Question

A golf ball strikes a hard, smooth floor at an angle of 37.3

A golf ball strikes a hard, smooth floor at an angle of 37.3 degree and, as the drawing shows, rebounds at the same angle. The mass of the ball is 0.0357 kg, and its speed is 58.3 m/s just before and after striking the floor. What is the magnitude of the impulse applied to the golf ball by the floor? (Hint: Note that only the vertical component of the ball's momentum changes during impact with the floor, and ignore the weight of the ball.)

Explanation / Answer

impulse =change in momentum

       = 0.0357 x (58.3sin37.2i - 58.3cos37.3j - (58.3sin37.2i + 58.3cos37.3j))

        = 0.0357x 2 x 58.3 cos37.3 = 3.31 kg.m/s

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