Two manned satellites approaching one another at a relative speed of 0.150 m/s i

Question

Two manned satellites approaching one another at a relative speed of 0.150 m/s intend to dock. The first has a mass of 4.00 10^3 kg, and the second a mass of 7.50 10^3 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.

Explanation / Answer

a)

By Conservation of momentum

m1V1+m2V2 = (m1+m2)V

4000*0 +7500*0.15 = (4000+7500)*V

V=0.0978 m/s

b)

Kinetic energy lost due to this inelastic collision is

KELost = (1/2)[m2V22-(m1+m2)V2]

KELost =(1/2)[7500*0.152-(4000+7500)*0.09782]

KELost =29.4 J

c)

By Conservation of momentum

m1V1+m2V2 = (m1+m2)V

4000*0.15 +7500*0 = (4000+7500)*V

V=0.0522 m/s

d)

Kinetic energy lost due to this inelastic collision is

KELost = (1/2)[m1V12-(m1+m2)V2]

KELost =(1/2)[4000*0.152-(4000+7500)*0.05222]

KELost =29.35 J

e)

The two velocities calculated here are the velocity of the center of mass in each of the two different individual reference frames. The loss in KE is the same in both reference frames because the Kinetic energy lost to internal forces Like heat, friction, etc is the same regardless of the coordinate system chosen.

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