A space vehicle is traveling at 4300Kmph relative to earth when the exhausted motor (mass 4m) is disengaged and sent back ward with a velocity 82Kmph relative to the command module (mass m) . What is the speed of the command module relative to earth just after separation

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Solution

Let the total mass of the particle is M and its initial velocity is U. After the disintegration the velocity of the mass of the rocket motor is m (say) and let the mass of the command module be (M-m). Let the final velocity of the command module be V and the velocity of the rocket motor be v.

Then according to the law of conservation of linear momentum

MU = (M-m)V + mv

Also we are given that M= 5, m=4 , (M-m)=1 , U= 4300km/h, (V-v)= 82km/h. then the equation becomes

(5×4300) = V + 4(V-82)

then V= 4365.6

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