A space probe, initially at rest, undergoes an internal mechanical malfunction and breaks into three pieces. One piece of mass m l = 48.0 kg travels in the positive x -direction at 12.0 m/s, and a second piece of mass m 2 = 62.0 kg travels in the xy -plane at an angle of 105° at 15.0 m/s. The third piece has mass m 3 = 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 m 1 , m 2 , m 3 , v 1 , v 2 and v 3 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 m 1 and m 2 . (d) Calculate the final y -components of the momenta of m 1 and m 2 . (e) Substitute the known momentum components into the general equations of momentum for the x - and y -directions, along with the known mass m 3 . (f) Solve the two momentum equations for v 3 cos θ and v 3 sin θ , respectively, and use the identity cos 2 θ + sin 2 θ = 1 to obtain v 3 . (g) Divide the equation for v 3 sin θ by that for v 3 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 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 cos2θ + sin2θ = 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
To determine
Sketch a diagram of the situation labelling the different masses and their velocities.
Answer to Problem 46P
Explanation of Solution
The diagram of the breakage is,
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,
(b)
Expert Solution
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=Σpxi⇒m1v1cos0°+m2v2cos105°+m3v3cosθ=0
The general expression for the conservation of momentum in y-direction is,
Σpyf=Σpyi⇒m1v1sin0°+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
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
576kg⋅m/s_ and the final x-component of the momenta of the mass
m2 is
−241kg⋅m/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)=576kg⋅m/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)=−241kg⋅m/s
Conclusion:
Thus, the final x-component of the momenta of the mass
m1 is
576kg⋅m/s_ and the final x-component of the momenta of the mass
m2 is
−241kg⋅m/s_.
(d)
Expert Solution
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
0kg⋅m/s_ and the final y-component of the momenta of the mass
m2 is
898kg⋅m/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)=0kg⋅m/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)=898kg⋅m/s
Conclusion:
Thus, the final y-component of the momenta of the mass
m1 is
0kg⋅m/s_ and the final y-component of the momenta of the mass
m2 is
898kg⋅m/s_.
(e)
Expert Solution
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
576kg⋅m/s−241kg⋅m/s+(112kg)v3cosθ=0_ and
0+898kg⋅m/s+(112kg)v3sinθ=0_ respectively.
Explanation of Solution
The final x-component of the momenta of mass
m3 is
576kg⋅m/s−241kg⋅m/s+(112kg)v3cosθ=0
The final y-component of the momenta of mass
m3 is
0+898kg⋅m/s+(112kg)v3sinθ=0.
Conclusion:
Thus, the final x and y-components of the momenta of mass
m3 are
576kg⋅m/s−241kg⋅m/s+(112kg)v3cosθ=0_ and
0+898kg⋅m/s+(112kg)v3sinθ=0_ respectively.
The angle must be in third quadrant. So, angle
180° is introduced.
Conclusion:
Thus, the angle
θ is
250°_.
(h)
Expert Solution
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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A toy car having mass
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vi = 4.65 m/s.
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