Chapter 07, Problem 12 Your answer is partially correct. Try again. A golf ball

Question

Chapter 07, Problem 12 Your answer is partially correct. Try again. A golf ball strikes a hard, smooth floor at an angle of 22.7 and, as the drawing shows, rebounds The mass of the ball is 0.0508 kg, and its speed is 51.7 m/s just before and after striking the floor. What is the magnitude of the impulse applied to the golf ball by the floor? (Hinti Note that enly the vertical cmpoent of the ba's momentum changes during impact with the floor, and ignore the weight of the ball.) Number T3.42 Units Tkg-m/s M the tolerance is +/-2% LINK TO TEXT Version 4.24.1. Privacy Policy 2000-2017)ohn Wiley & Sons, Inc, All Rights Reserved. A Division of John Wiley & Sons, Inc

Explanation / Answer

Impulse = Change in momentum

Resolve the velocity vector into its rectangular components. Now, since speed before and after striking the floor remains the same, hence the horizontal component (v cos th ) doesn’t change.

Hence, only the vertical components come into play.

Therefore, Initial momentum = mv sin th.

And, final momentum = mv sin th.

Thus, Impulse = mv sin th – mv sin th = 2mv sin th.

The magnitude of the impulse applied to the golf ball by the floor

Del P = 2mv sin th = 2 x 0.0508 x 51.7 x sin 22.7 ° = 2.03 kg-m/s

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