Two manned satellites approaching one another, at a relative speed of 0.100 m/s,

Question

Two manned satellites approaching one another, at a relative speed of 0.100 m/s, intending to dock. The first has a mass of 4.50 103 kg, and the second a mass of 7.50 103 kg. (a) Calculate the final velocity (after docking) in m/s by using the frame of reference in which the first satellite was originally at rest. (Assume the second satellite moves in the positive direction. Include the sign of the value in your answer.) m/s (b) What is the loss of kinetic energy (in J) in this inelastic collision? J (c) Repeat both parts by using the frame of reference in which the second satellite was originally at rest. final velocity (m/s) m/s loss of kinetic energy (J) J

Explanation / Answer

Given that

v1 = 0

v2 = 0.1 m/s

m1 = 4500 kg

m2 = 7500 kg

(a)

let, final velocity = v

apply conservation of momentum,

(m1 + m2)*v = m1*v1 + m2*v2

v = 7500*0.1 / (4500 + 7500)

v = 0.0625 m/s

(b)

loss of kinetic energy,

KEloss = KEi - KEf

KEloss = (1/2)*m2*v2^2 - (1/2)*(m1+m2)*v^2

by solving this,

KEloss = 14.06 J

(c)

Now, second setelite is originally at rest.

apply conservation of momentum,

(m1 + m2)*v = m1*v1 + 0

v = 4500*.1 / 12000 = 0.0375 m/s

KEloss = KEi - KEf

KEloss = (1/2)*m1*v1^2 - (1/2)*(m1+m2)*v^2

by solving this,

KEloss = 14.06 J

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