COLLEGE PHYSICS LL W/ 6 MONTH ACCESS
COLLEGE PHYSICS LL W/ 6 MONTH ACCESS
2nd Edition
ISBN: 9781319414597
Author: Freedman
Publisher: MAC HIGHER
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Chapter 7, Problem 57QAP
To determine

The velocities of the balls after the collision

Expert Solution & Answer
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Answer to Problem 57QAP

The velocity of first ball after collision is 1 m/s towards left and second ball is 2 m/s towards right.

Explanation of Solution

Given info:

A 2.00 kg ball moves at 3.00 m/s towards the right. It collides elastically with a 4.00 kg ball that is initially at rest.

Formula used:

Using the formula, conservation of linear momentum

  m1u1+m2u2=m1v1+m2v2 (1)

Here, m1and m2 are the masses of first and second balls respectively and u1and u2 are the velocities of the both balls before collision and v1and v2 are the velocities after collision.

Also, for elastic collision

  e=v2v1u1u2 (2)

Calculation:

We have, m1=2.00kg, m2=4.00kg and u1=3.00m/s and u2=0

Substituting the given values in equ.(1), we get

  2kg×3m/s+4kg×0=2kg×v1+4kg×v2v1+2v2=3(3)

Now from equation (2)

  1=v2v130(e=1,forelasticcollision)v2v1=3(4)

Adding equation (3) and (4), we get

  (v1+2v2)+(v2v1)=3+33v2=6v2=2m/s (towards right)

Substituting in equation (3), we get

  (v1+2×2)=3v1=34=1(towardsleft)

Conclusion:

Thus, the velocity of first ball after collision is 1 m/s towards left and second ball is 2 m/s towards right.

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COLLEGE PHYSICS LL W/ 6 MONTH ACCESS

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