College Physics (Instructor's)
College Physics (Instructor's)
11th Edition
ISBN: 9781305965317
Author: SERWAY
Publisher: CENGAGE L
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Chapter 6, Problem 46P

A space probe, initially at rest, undergoes an internal mechanical malfunction and breaks into three pieces. One piece of mass ml = 48.0 kg travels in the positive x-direction at 12.0 m/s, and a second piece of mass m2 = 62.0 kg travels in the xy-plane at an angle of 105° at 15.0 m/s. The third piece has mass m3 = 112 kg. (a) Sketch a diagram of the situation, labeling the different masses and their velocities, (b) Write the general expression for conservation of momentum in the x- and y-directions in terms of m1, m2, m3, v1, v2 and v3 and the sines and cosines of the angles, taking θ to be the unknown angle, (c) Calculate the final x-components of the momenta of m1 and m2. (d) Calculate the final y-components of the momenta of m1 and m2. (e) Substitute the known momentum components into the general equations of momentum for the x- and y-directions, along with the known mass m3. (f) Solve the two momentum equations for v3 cos θ and v3 sin θ, respectively, and use the identity cos2 θ + sin2 θ = 1 to obtain v3. (g) Divide the equation for v3 sin θ by that for v3 cos θ to obtain tan θ, then obtain the angle by taking the inverse tangent of both sides, (h) In general, would three such pieces necessarily have to move in the same plane? Why?

a)

Expert Solution
Check Mark
To determine
Sketch a diagram of the situation labelling the different masses and their velocities.

Answer to Problem 46P

College Physics (Instructor's), Chapter 6, Problem 46P , additional homework tip  1

Explanation of Solution

The diagram of the breakage is,

College Physics (Instructor's), Chapter 6, Problem 46P , additional homework tip  2

The numerical values of the masses and velocities are,

The numerical values of the masses and velocities are,

m1=48.0kgv1=12.0m/sm2=62.0kgv2=15.0m/sm3=112kg

Conclusion:

Thus, the diagram of the breakage is,

College Physics (Instructor's), Chapter 6, Problem 46P , additional homework tip  3

(b)

Expert Solution
Check Mark
To determine
The general expression for the conservation of momentum.

Answer to Problem 46P

The general expression for the conservation of momentum in x-direction is m1v1cos0°+m2v2cos105°+m3v3cosθ=0_ and the general expression for the conservation of momentum in y-direction is m1v1sin0°+m2v2sin105°+m3v3sinθ=0_ .

Explanation of Solution

The general expression for the conservation of momentum in x-direction is,

Σpxf=Σpxim1v1cos0°+m2v2cos105°+m3v3cosθ=0

The general expression for the conservation of momentum in y-direction is,

Σpyf=Σpyim1v1sin0°+m2v2sin105°+m3v3sinθ=0

Conclusion:

Thus, the general expression for the conservation of momentum in x-direction is m1v1cos0°+m2v2cos105°+m3v3cosθ=0_ and the general expression for the conservation of momentum in y-direction is m1v1sin0°+m2v2sin105°+m3v3sinθ=0_ .

(c)

Expert Solution
Check Mark
To determine
The final x-components of the momenta of the masses.

Answer to Problem 46P

The final x-component of the momenta of the mass m1 is 576kgm/s_ and the final x-component of the momenta of the mass m2 is 241kgm/s_ .

Explanation of Solution

The final x-component of the momenta of the mass m1 is,

p1x=m1v1cos0°

Substitute 48.0kg for m1 and 12.0m/s for v1 .

p1x=(48.0kg)(12.0m/s)(1)=576kgm/s

The final x-component of the momenta of the mass m1 is,

p2x=m2v2cos105°

Substitute 62.0kg for m2 , 0.259 for cos105° and 15.0m/s for v2 .

p2x=(62.0kg)(15.0m/s)(259)=241kgm/s

Conclusion:

Thus, the final x-component of the momenta of the mass m1 is 576kgm/s_ and the final x-component of the momenta of the mass m2 is 241kgm/s_ .

(d)

Expert Solution
Check Mark
To determine
The final y-components of the momenta of the masses.

Answer to Problem 46P

The final y-component of the momenta of the mass m1 is 0kgm/s_ and the final y-component of the momenta of the mass m2 is 898kgm/s_ .

Explanation of Solution

The final y-component of the momenta of the mass m1 is,

p1y=m1v1sin0°

Substitute 48.0kg for m1 and 12.0m/s for v1 .

p1y=(48.0kg)(12.0m/s)(0)=0kgm/s

The final y-component of the momenta of the mass m1 is,

p2y=m2v2sin105°

Substitute 62.0kg for m2 , 0.966 for sin105° and 15.0m/s for v2 .

p2x=(62.0kg)(15.0m/s)(0.966)=898kgm/s

Conclusion:

Thus, the final y-component of the momenta of the mass m1 is 0kgm/s_ and the final y-component of the momenta of the mass m2 is 898kgm/s_ .

(e)

Expert Solution
Check Mark
To determine
The final x and y-components of the momenta of mass m3 .

Answer to Problem 46P

The final x and y-components of the momenta of mass m3 are 576kgm/s241kgm/s+(112kg)v3cosθ=0_ and 0+898kgm/s+(112kg)v3sinθ=0_ respectively.

Explanation of Solution

The final x-component of the momenta of mass m3 is 576kgm/s241kgm/s+(112kg)v3cosθ=0

The final y-component of the momenta of mass m3 is 0+898kgm/s+(112kg)v3sinθ=0 .

Conclusion:

Thus, the final x and y-components of the momenta of mass m3 are 576kgm/s241kgm/s+(112kg)v3cosθ=0_ and 0+898kgm/s+(112kg)v3sinθ=0_ respectively.

(f)

Expert Solution
Check Mark
To determine
Solve the two momentum equations to find v3 .

Answer to Problem 46P

The velocity v3 is 8.56m/s_ .

Explanation of Solution

In the x-direction,

v3cosθ=576kgm/s+241kgm/s112kg=2.99m/s

In the y-direction,

v3sinθ=898kgm/s112kg=8.02m/s

Squaring and adding the equations,

v32(cos2θ+sin2θ)=(2.99m/s)2+(8.02m/s)2v3=73.3m2/s2=8.56m/s

Conclusion:

Thus, the velocity v3 is 8.56m/s_ .

(g)

Expert Solution
Check Mark
To determine
Obtain tanθ and thus find θ .

Answer to Problem 46P

The angle θ is 250°_ .

Explanation of Solution

The tangent of the angle is,

tanθ=v3sinθv3cosθ=8.02m/s2.99m/s=2.68

Thus, the angle is,

θ=tan1(2.68)+180°=250°

The angle must be in third quadrant. So, angle 180° is introduced.

Conclusion:

Thus, the angle θ is 250°_ .

(h)

Expert Solution
Check Mark
To determine
Would the three pieces have to move in the same plane.

Answer to Problem 46P

All three pieces have to move in the same plane.

Explanation of Solution

The momentum of the third fragment must be equal in magnitude and must be in the opposite direction to the resultant of the other two fragments momenta. So, all three pieces have to move in the same plane.

Conclusion:

Thus, all three pieces have to move in the same plane.

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Chapter 6 Solutions

College Physics (Instructor's)

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