Question

Two manned satellites approaching one another at a relative speed of 0.450 m/s intend to dock. The first has a mass of 3.00 ✕ 10³ kg, and the second a mass of 7.50 ✕ 10³ 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

Answer 1

Given,

Masses, m1 = 3000 kg, m2 = 7500 kg

Velocity, v2 = 0.45 m/s

Using law of conservation of momentum,

m2 x v2 + m1 x 0 = (m1+m2) v

7500 x 0.45 + 0 = 10500 v

v = 0.32 m/s

B)

KEi = 0.5 x 7500 x (0.45^2) = 759.38 J

KEf = 0.5 x 10500 x (0.32^) = 537.6 J

so KE loss = 221.78 J

C)

0 - 3000 x 0.45 = 10500 v

v = -0.129 m/s

KEi = 0.5 x 3000 x (-0.45^2) = 303.75 J

KEf = 0.5 x 10500 x (-0.129^2) = 87.365 J

So the KE loss is 216.38 J

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