A 4.90-kg bowling ball moving at 7.90 m/s collides elastically with a 0.780-kg bowling pin, which is scattered at an angle 0121.1° to the initial direction of the bowling ball and with a speed of 13.0 m/s. Calculate the final velocity (magnitude and direction)of the bowling ball.Magnitude9.81Direction = |4.258 m(measured from initial direction of the bowling ball)%DSubmit AnswerIncompatible units. No conversion found between "m" and the required units. O Tries 0/10 Previous Tries

Answered: Image /qna-images/answer/6e495402-c9b1-42ff-a456-0de95e3f267f.jpg

Answered: A 4.90-kg bowling ball moving at 7.90…

Classical Dynamics of Particles and Systems

5th Edition

ISBN:9780534408961

Author:Stephen T. Thornton, Jerry B. Marion

Publisher:Stephen T. Thornton, Jerry B. Marion

Chapter9: Dynamics Of A System Of Particles

Section: Chapter Questions

Problem 9.30P

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Similar questions

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A block with mass m1 = 0.500 kg is released from rest on a frictionless track at a distance h1, = 2.50 m above the top of a table. It then collides elastically with an object having mass m2 = 1.00 kg that is initially at rest on the table, as shown in Figure P6.71. (a) Determine the velocities of the two objects just after the collision. (b) How high up the track does the 0.500-kg object travel back after the collision? (c) How far away from the bottom of the table does the1.00-kg object land, given that the height of the table h2 = 2.00 m? (d) How far away from the bottom of the table does the 0.500-kg object eventually band? Figure P6.71

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