Stages of a Rocket The last stage of a rocket is traveling at a speed of 7200 m/

Question

Stages of a Rocket

The last stage of a rocket is traveling at a speed of 7200 m/s. This last stage is made up of two parts that are clamped together, namely, a rocket case with a mass of 370.0 kg and a payload capsule with a mass of 270.0 kg. When the clamp is released, a compressed spring causes the two parts to separate with a relative speed of 940.0 m/s. Assume that all velocities are along the same line.

What is the speed of the case after they separate?

What is the speed of the payload after they separate?

What is the total kinetic energy of the two parts before they separate?

What is the increase in kinetic energy after they separate?

Explanation / Answer

Let mc be the mass of the rocket case and mp the mass of the payload. At first they are traveling together with
velocity v. After the clamp is released mc has velocity vc and mp has velocity vp. Conservation of momentum yields

(mc+mp)v = mc*vc + mp*vp

A. After the clamp is released the payload, having the lesser mass, will be traveling at the greater speed. We write vp = vc + vrel, where vrel is the relative velocity. When this expression is substituted into the conservation of momentum condition, the result is

(mc +mp)v = mc*vc +mp*vc +mp*vrel

vc = [(370+270)*7200 - (270*940)]/(370+270) = 6803.43 m/s

B. vp = vc + vrel = 6803.43 + 940 = 7743.43 m/s

C. The total kinetic energy before the clamp is released is

Ki = 0.5*(mc + mp)v^2 = 0.5(270 + 370)*7200^2 = 1.65888*10^10 J

D. The total kinetic energy after the clamp is released is

Kf = 0.5*mc*vc^2 + 0.5*mp*vp^2 = 0.5*370*6803.43^2 + 0.5*270*7743.43^2 = 1.66577*10^10 J

So it increase by

deltaK = (166577 - 1.65888)*10^10 = 0.00689*10^10 = 6.89*10^7 J

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