Two manned satellites approaching one another at a relative speed of 0.100 m/s intend to dock. The first has a mass of 4.00 ✕ 103 kg, and the second a mass of 7.50 ✕ 103 kg. Assume that the positive direction is directed from the second satellite towards the first satellite.(a) Calculate the final velocity after docking, in the frame of reference in which the first satellite was originally at rest. m/s(b) What is the loss of kinetic energy in this inelastic collision? J(c) Repeat both parts, in the frame of reference in which the second satellite was originally at rest.final velocity m/sloss of kinetic energy JExplain why the change in velocity is different in the two frames, whereas the change in kinetic energy is the same in both.

Question
Asked Sep 19, 2019

Two manned satellites approaching one another at a relative speed of 0.100 m/s intend to dock. The first has a mass of 4.00 ✕ 103 kg, and the second a mass of 7.50 ✕ 103 kg. Assume that the positive direction is directed from the second satellite towards the first satellite.(a) Calculate the final velocity after docking, in the frame of reference in which the first satellite was originally at rest.
 m/s

(b) What is the loss of kinetic energy in this inelastic collision?
 J

(c) Repeat both parts, in the frame of reference in which the second satellite was originally at rest.
final velocity
 m/s
loss of kinetic energy
 J

Explain why the change in velocity is different in the two frames, whereas the change in kinetic energy is the same in both. 

check_circleExpert Solution
Step 1

(a) Considering the case as perfectly inelastic collision. Here first satellite is at rest.

  Write the expression for final velocity.

ти, + mи,
т, + т,
(4x10 kg) (Om/s)+(7.50 x10 kg )(0.100 m/s)
(4x10 kg+7.50x10kg)
0.06 m/s
help_outline

Image Transcriptionclose

ти, + mи, т, + т, (4x10 kg) (Om/s)+(7.50 x10 kg )(0.100 m/s) (4x10 kg+7.50x10kg) 0.06 m/s

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Step 2

(b)  Write the expression fo...

1
AK
тт,
(и — и,)*
_
1 (4x10kg) (7.50x10 kg)
(4x10kg+7.50x10' ke(Om/s-0.100m/s)
-0.5x (2608.6 kg)-0.100m/s)
2
13.04J
help_outline

Image Transcriptionclose

1 AK тт, (и — и,)* _ 1 (4x10kg) (7.50x10 kg) (4x10kg+7.50x10' ke(Om/s-0.100m/s) -0.5x (2608.6 kg)-0.100m/s) 2 13.04J

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