(a) Explanation The ball falls from a certain height attaining some velocity with which it strikes the ground. This attained velocity is initial velocity of ball before collision with the ground. Write the expression of motion for the velocity of ball. v 1 2 = u 1 2 + 2 a s 1 (I) Here, v 1 is the final velocity before collision, u 1 is the velocity initially with the ball, a is the acceleration attained by the ball and s 1 is the distance covered by the ball. As ball initially at rest falls from height. Substitute 0 m/s for u 1 iun equation (I) and solve for v 1 . v 1 = 2 a s 1 (II) After the collision with the ground ball again bounces back upto some height with some velocity. This velocity is final velocity after collision. After reaching certain height, ball comes to rest. Write the expression of motion for the velocity of ball. v 2 2 = u 2 2 + 2 a s 2 (III) Here, v 2 is the final velocity after collision, u 2 is the initial velocity with the ball, a is the acceleration attained by the ball and s 2 is the distance covered by the ball. After reaching some height, velocity of ball becomes zero. Substitute 0 m/s for v 2 and solve equation (I) for value of u 2 . u 2 = − ( 2 a s 2 ) (IV) Write the expression for the change in momentum. Δ p = m ( v → − u → ) Here Δ p is the change in the momentum, m is the mass of the ball, v → is the final velocity after collision and u → is the initial velocity of ball (b) To determine The average force floor exerts on the ball.

Physics for Scientists and Engineers with Modern Physics, Technology Update

9th Edition

ISBN: 9781305401969

Author: SERWAY, Raymond A.; Jewett, John W.

Publisher: Cengage Learning

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Chapter 9, Problem 14P

(a)

To determine

The magnitude and direction of impulse delivered to the ball by the floor.

(b)

To determine

The average force floor exerts on the ball.

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Chapter 9, Problem 13P

Chapter 9, Problem 15P

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Physics for Scientists and Engineers with Modern Physics, Technology Update

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