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(6%) Problem 15: Two manned satellites are approaching one another at a relative speed of 0.225 m/s, intending to dock. The f

What is the change in Kinetic Energy, in joules, in this frame of reference?

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Answer #1

Let the satellites are with masses m1 = 3750 kg, m2 = 9800 kg

are approaching with relative velocity is v =0.225 m/s

   let the m2 is at rest and m1 is approaching m2


taking +VE X direction as positive and -ve X direction as -ve for the velocity and momentum


   m1 ----------- m2


by conservation of momentum

   m1*v1 + m2*v2 = (m1+m2)v
v is the velocity after docking


   3750*0.225 +9800*0 = (3750+9800)v

   v = 0.0622 m/s

now the change in kinetic energy is k2-k1 = dk.e


   k1 = 0.5*m1*v1^2 = 0.5*3750*0.225^2 J = 94.92 J

   k2 = 0.5(m1+m2)*v^2 = 0.5(3750+9800)0.0622^2 J = 26.21 J


so the change in kinetic energy is dk.e = 94.92-26.21 J = 68.71 J

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