COLLEGE PHYSICS-CONNECT ACCESS
COLLEGE PHYSICS-CONNECT ACCESS
5th Edition
ISBN: 9781260486834
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 7, Problem 10P
To determine

Rank the momentum from smallest to largest.

Expert Solution & Answer
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Answer to Problem 10P

The momentum from smallest to largest is (e)=(d)<(a)=(b)<(c)_

Explanation of Solution

Write the expression for kinetic energy.

  Ekinetic=12mv2        (I)

Here, m is the mass, v is the velocity.

Write the expression for momentum.

  p=mv        (II)

Conclusion:

Substitute, 8kg for m, and 400J for Ekinetic in equation (I) to find the velocity.

  400J=12(8kg)v2va=10m/s

Substitute, 8kg for m, and 10m/s for va in equation (II) to find pa.

  pa=(8kg)(10m/s)=80kgm/s

Substitute, 2kg for m, and 1600J for Ekinetic in equation (I) to find the velocity.

  1600J=12(2kg)v2vb=40m/s

Substitute, 2kg for m, and 40m/s for vb in equation (II) to find pb.

  pb=(2kg)(40m/s)=80kgm/s

Substitute, 4kg for m, and 1600J for Ekinetic in equation (I) to find the velocity.

  1600J=12(4kg)v2vc=28.3m/s

Substitute, 4kg for m, and 28.3m/s for vc in equation (II) to find pc.

  pc=(4kg)(28.3m/s)=113.2kgm/s

Substitute, 16kg for m, and 100J for Ekinetic in equation (I) to find the velocity.

  100J=12(16kg)v2vd=3.54m/s

Substitute, 16kg for m, and 3.54m/s for vb in equation (II) to find pd.

  pd=(16kg)(3.54m/s)=56.6kgm/s

Substitute, 1kg for m, and 1600J for Ekinetic in equation (I) to find the velocity.

  1600J=12(1kg)v2ve=56.6m/s

Substitute, 1kg for m, and 56.57m/s for vc in equation (II) to find pc.

  pc=(1kg)(56.57m/s)=56.6kgm/s

Hence the momentum from smallest to largest is (e)=(d)<(a)=(b)<(c)_.

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Chapter 7 Solutions

COLLEGE PHYSICS-CONNECT ACCESS

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