College Physics
College Physics
3rd Edition
ISBN: 9780134143323
Author: Knight, Randall Dewey, Jones, Brian, Field, Stuart
Publisher: Pearson,
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Chapter 9, Problem 1CQ
To determine

The ranking of the objects in order from largest to smallest momentum.

Expert Solution & Answer
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Answer to Problem 1CQ

The ranking of the object in order from largest to smallest momentum is as follows.

    p2x>p1x=p3x=p5x>p4x

Explanation of Solution

Write the expression for the momentum.

    px=mvx        (I)

Here, px is gthe momentum, m is the mass and vx is the velocity of the object.

Conclusion:

For object 1:

Substitute 20g for m and 1m/s for vx in equation (I) to find p1x.

    p1x=(20g)(1m/s)=(20g×103kg1g)(1m/s)=0.02kgm/s

For object 2:

Substitute 20g for m and 2m/s for vx in equation (I) to find p2x.

    p2x=(20g)(2m/s)=(20g×103kg1g)(2m/s)=0.04kgm/s

For object 3:

Substitute 10g for m and 2m/s for vx in equation (I) to find p3x.

    p3x=(10g)(2m/s)=(10g×103kg1g)(2m/s)=0.02kgm/s

For object 4:

Substitute 10g for m and 1m/s for vx in equation (I) to find p4x.

    p4x=(10g)(1m/s)=(10g×103kg1g)(1m/s)=0.01kgm/s

For object 5:

Substitute 200g for m and 0.1m/s for vx in equation (I) to find p5x.

    p5x=(200g)(0.1m/s)=(200g×103kg1g)(0.1m/s)=0.02kgm/s

The momentum of second object is largest and fourth object is smallest.

Therefore, the ranking of the object in order from largest to smallest momentum is,

    p2x>p1x=p3x=p5x>p4x

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Chapter 9 Solutions

College Physics

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