Two manned satellites approaching one another at a relative speed of 0.400 m/s i
ID: 1371013 • Letter: T
Question
Two manned satellites approaching one another at a relative speed of 0.400 m/s intend to dock. The first has a mass of 3.50 103 kg, and the second a mass of 7.50 103 kg. If the two satellites collide elastically rather than dock, what is their final relative velocity? Adopt the reference frame in which the second satellite is initially at rest and assume that the positive direction is directed from the second satellite towards the first satellite.
I need help with the above question. The answer I have is 0.14545 and it's incorrect. If someone could please show me how to solve, I don't believe I'm using the correct formula.
Explanation / Answer
Let
m1 = 3500 kg
m2 = 7500 kg
u1 = 0.4 m/s
u2 = 0
after the collision velcoity of m1,
v1 = (m1-m2)*u1/(m1+m2)
= (3500 - 7500)*0.4/(3500 + 7500)
= -0.14545 m/s
after the collision velcoity of m2,
v2 = 2*m1*u1/(m1+m2)
= 2*3500*0.4/(3500+7500)
= 0.25454 m/s
final relative velocity
V_relative = v2 - v1
= 0.25454 - (-0.14545)
= 0.4 m/s <<<<<<------------Answer
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