Two manned satellites approaching one another, at a reiative speed of 0.150 mys,

Question

Two manned satellites approaching one another, at a reiative speed of 0.150 mys, intending to dock. The first has a mass of $.00 x 3Dl ko, and the second a mass of 7.50 x 10 kg. (a) Calculate the final velocity (after docking) in m/s by using the frame of reference in which the first sstelite was originally at rest. (Assume the second satelite moves in the positive decton Include the sign of the value in your answer.) m/s (b) What is the loss of kinetic energy (in 3) in this inelastic collision? (c) Repeat both parts by ssing the frame of reference in which the second satelite was originaly at rest. final velocity (mVs) loss of kinetic energy() mys Explain why the change in velocity is different in the two frames, whereas the change in kinetic energy is the same in both

Explanation / Answer

Initial momentum = 7.5 * 10^3 * 0.150 = 1125

Total mass = 5 * 10^3 + 7.5 * 10^3 = 1.25 * 10^4 kg

Final momentum = 1.25 * 10^4 * v

(a) 1.25 * 10^4 * v = 1125

v = 0.09 m/s

(b)

Initial KE = ½ * 7.5 * 10^3 * 0.150^2 = 84.375 J

Final KE = ½ * 1.25 * 10^4 * (0.09)^2 = 50.625 J

Loss of KE = 84.375 – 50.625

= 33.75 J

(c) Repeat both parts, in the frame of reference in which the second satellite was originally at rest.

For this problem, initial momentum = 5 * 10^3 * 0.150 = 750

1.25 * 10^4 * v = 750

v = 0.06 m/s

Initial KE = ½ * 5 * 10^3 * 0.150^2 = 56.25 J

Final KE = ½ * 1.25 * 10^4 *0.06^2 = 22.5 J

Loss of KE = 56.25 – 22.5

= 33.75 J

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