plu3.conlpeuugen/Student/mainfr.uni Cutnell, Physics, 10e GENERAL PHYSICS (PHYS

Question

plu3.conlpeuugen/Student/mainfr.uni Cutnell, Physics, 10e GENERAL PHYSICS (PHYS 2001/200 ice Assignment Gradebook ORION Downloadable eTextbook nent FULL SCREEN PRINTER VERSION 4BACK NEXT Chapter 07, Problem 12 A golf ball strikes a hard smooth floor at an angle of 30 7ana, as the drawing shows rebounds at the sane ange. The mass of the ball is .038 kg and ts speed is 39.6 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 onily the vertical component of the ball's momentum changes during impact with the floor, and ignore the weight of the bal.) Number Units the tolerance is +/-2% Question Attempts: Unlimited SAVE FOR LA SUBMIT ANSWER 15

Explanation / Answer

Well, it's not entirely self-explanatory, as it is not clear whether the given angle is w/r/t to the horizontal or the vertical. Assuming it is the vertical, then
Vv = 39.6m/s * cos30.7º = 34.05 m/s
Vh = 39.6m/s * sin30.7º = 20.21 m/s but the horizontal velocity is not affected by impact

Change in horizontal velocity = 0
Change in vertical velocity = 34.05*2 = 68.1 m/s
Therefore the impulse was 68.1m/s * 0.0381kg = 2.594 kg·m/s

If is w/r/t the floor, then the impulse is 2 * 20.21m/s * 0.0381kg = 1.54 kg·m/s

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