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Chapter 9: Q110P (page 255)

A 140 g ball with speed 7.8ms strikes a wall perpendicularly and rebounds in the opposite direction with the same speed. The collision lasts 3.80 ms .What are the magnitudes of the (a) impulse and (b) average force on the wall from the ball during the elastic collision?

Short Answer

Expert verified
  1. Magnitude of impulse,Jis2.18kg.ms.
  2. Magnitude of the average force on the wall from the ball, F is 575 N .

Step by step solution

01

Understanding the given information

  1. Mass of the ball, m = 140 g.
  2. Speed of the ball,vi=7.8ms.
  3. Time of collision, t = 3.80 ms .
02

Concept and formula used in the given question

Use the concept of impulse. Using the equation of impulse and find the impulse and then using the same equation of time involved, then find the average force. The equations are given below.

J=m鈭�VFt=m鈭�V

03

(a) Calculate the magnitude of impulse

Magnitude of impulse:

Ball rebounds back with same velocity, we can write,

J=0.14(-v-(v)=0.14脳2v=0.14脳2脳7.8=-2.18kg.ms

Magnitude of impulse is J=2.18kg.ms

04

(b) Calculate the magnitude of impulse average force on the wall from the ball during the elastic collision

Magnitude of the average force on the wall from the ball.

Consider the formulas J=m鈭�VandFt=m鈭�V.

Use value of J here and solve as:

F(3.80脳10-3)=m鈭�VF=2.183.80脳10-3F=574.7NF鈮�575N

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