   Chapter 6, Problem 78AP

Chapter
Section
Textbook Problem

A 60-kg soccer player jumps vertically upwards and heads the 0.45-kg ball as it is descending vertically with a speed of 25 m/s. (a) If the player was moving upward with a speed of 4.0 m/s just before impact, what will be the speed of the ball immediately after the collision if the ball rebounds vertically upwards and the collision is elastic? (b) If the ball is in contact with the player’s head for 20 ms, what is the average acceleration of the ball? (Note that the force of gravity may be ignored during the brief collision time.)

(a)

To determine
The speed of the ball immediately after the collision.

Explanation

Given info: The mass of the soccer player is 60kg . The soccer player jumps vertically upward and heads a ball descending downward. The mass of the ball is 0.45kg . The speed of the soccer player is 4.0ms-1 . The speed of the ball is 25ms-1 . The ball rebounds vertically upward. The collision between the head of the player and the ball is elastic.

Explanation:

Consider upward to be the positive direction and downward to be the negative direction. From conservation of momentum of the ball and player system,

mbvib+mpvip=mbvfb+mpvfp (1)

• mb is mass of the ball
• mp is the mass of the player
• vib is the initial velocity of the ball
• vip is the initial velocity of the player
• vfb is the final velocity of the ball
• vfp is the final velocity of the player

Substitute 60kg for mp , 0.45kg for mb , 4.0ms-1 for vip , 25ms-1 for vib in (1) and rearrange,

(0.45kg)(25ms-1)+(60kg)(4.0ms-1)=(0.45kg)vfb+(60kg)vfp228.75kgms-1=(0

(b)

To determine
The average acceleration of the ball.

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