The last stage of a rocket is traveling at a speed of 8200 m/s . This last stage
ID: 1428547 • Letter: T
Question
The last stage of a rocket is traveling at a speed of 8200 m/s. This last stage is made up of two parts that are clamped together, namely, a rocket case with a mass of 380.0 kg and a payload capsule with a mass of 290.0 kg. When the clamp is released, a compressed spring causes the two parts to separate with a relative speed of 950.0 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
according to momentum conservation
(m1+m2)*v = m1*v1 + m2*v2
where v2 = v1+950
=> 670*8200 = 380*v1 + 290*(v1+950)
=> v1 = 7788.80 m/s ....speed of case after separation
=> v2 = 8738.80 m/s ...speed of payload after separation
total KE before separation = (m1+m2)*v^2/2 = 670*(8200)^2/2 = 2.2525*10^10 J
total KE after separation = [m1*v1^2 + m2*v2^2 ]/2 = [380*(7788.80)^2 + 290*(8738.80)^2]/2 = 2.2599*10^10 J
SO INCREASE IN KE = (2.2599 - 2.2525 )*10^10 = 7.4*10^7 J
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