Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 8, Problem 42P

(a)

To determine

To sketch: The position and velocity vector of the two particles in x-y plane.

(a)

Expert Solution
Check Mark

Answer to Problem 42P

Answer: the position vector of the two particles in x-y plane is shown in Figure I.

Explanation of Solution

The position vector shows the location of particle with respect to x and y axis in x-y plane. The velocity vector shows the magnitude as well as the direction of the particle’s velocity in x-y plane.

Given information:

The mass of particle is m1=2kg and m2=3kg and the location is r1=(1i^+2j^)m , r2=(4i^3j^)m and the velocity is v1=(3i^+0.5j^)m/s and v2=(3i^2j^)m/s .

The graph of position vector is shown in Figure I.

Principles of Physics: A Calculus-Based Text, Chapter 8, Problem 42P , additional homework tip  1

Figure I

The graph of velocity vector is shown in Figure II.

Principles of Physics: A Calculus-Based Text, Chapter 8, Problem 42P , additional homework tip  2

Figure II

Conclusion: Therefore, the position vector of two particles in x-y plane is shown in Figure I and velocity vector i s shown in Figure II.

(b)

To determine

The position of the centre of the mass.

(b)

Expert Solution
Check Mark

Answer to Problem 42P

Solution: The position of the centre of the mass is 2.i^1j^ .

Explanation of Solution

Given information:

The mass of particle is m1=2kg and m2=3kg and the location is r1=(1i^+2j^)m , r2=(4i^3j^)m and the velocity is v1=(3i^+0.5j^)m/s and v2=(3i^2j^)m/s .

Formula to calculate the centre of mass of the system is,

rCM=m1r1+m2r2m1+m2 (I)

  • rCM is the center of mass of the system.

Substitute 2kg for m1 , 3kg for m2 , (1i^+2j^)m for r1 and (4i^3j^)m for r2 in equation (I).

rCM=(2kg)(1i^+2j^)m+(3kg)(4i^3j^)m2kg+3kg=2i^1j^

Conclusion:

Therefore, the position of the centre of the mass is 2.i^1j^ .

(c)

To determine

The velocity of the centre of the mass.

(c)

Expert Solution
Check Mark

Answer to Problem 42P

The velocity of the centre of the mass is 3i^1j^ .

Explanation of Solution

Given information:

The mass of particle is m1=2kg and m2=3kg and the location is r1=(1i^+2j^)m , r2=(4i^3j^)m and the velocity is v1=(3i^+0.5j^)m/s and v2=(3i^2j^)m/s .

Formula to calculate the centre of mass of the system is,

vCM=m1v1+m2v2m1+m2 (II)

  • rCM is the center of mass of the system.

Substitute 2kg for m1 , 3kg for m2 , (3i^+0.5j^)m/s for v1 and (3i^2j^)m/s for v2 in equation (II).

vCM=(2kg)(3i^+0.5j^)m/s+(3kg)(3i^2j^)m/s2kg+3kg=3i^1j^

The velocity vector is shown in Figure III.

Principles of Physics: A Calculus-Based Text, Chapter 8, Problem 42P , additional homework tip  3

Figure III

Conclusion:

Therefore, the velocity of the centre of the mass is 3i^1j^ .

(d)

To determine

The total linear momentum of the system.

(d)

Expert Solution
Check Mark

Answer to Problem 42P

The total linear momentum of the system is (15i^5j^)kgm/s .

Explanation of Solution

Given information:

The mass of particle is m1=2kg and m2=3kg and the location is r1=(1i^+2j^)m , r2=(4i^3j^)m and the velocity is v1=(3i^+0.5j^)m/s and v2=(3i^2j^)m/s .

Formula to calculate the linear momentum of the system is,

p=m1v1+m2v2

  • p is the linear momentum of the system.

Substitute 2kg for m1 , 3kg for m2 , (3i^+0.5j^)m/s for v1 and (3i^2j^)m/s for v2 in  above equation.

p=(2kg)(3i^+0.5j^)m/s+(3kg)(3i^2j^)m/s=(15i^5j^)kgm/s

Conclusion:

Therefore, the total linear momentum of the system is (15i^5j^)kgm/s .

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Chapter 8 Solutions

Principles of Physics: A Calculus-Based Text

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