The last stage of a rocket is traveling at a speed of 9400 m/s . This last stage

Question

The last stage of a rocket is traveling at a speed of 9400 m/s. This last stage is made up of two parts that are clamped together, namely, a rocket case with a mass of 400.0 kg and a payload capsule with a mass of 100.0 kg. When the clamp is released, a compressed spring causes the two parts to separate with a relative speed of 920.0 m/s. Assume that all velocities are along the same line.A) What is the speed of the case after they separate? B) What is the speed of the payload after they seperate? C) What is the total kinetic energy of the two parts before they separate? D) What is the increase in kinetic energy after they separate?

Explanation / Answer

Initial Speed of rocket = 9400 m/s

Let v = speed of rocket case after separation
so, speed of payload =  v + 920 m/s

Mass of Rocket Case = 400 kg
Mass of Capsule = 100 Kg
Mass of Rocket = 400 + 100 = 500 Kg

Momentum is conserved.
Final momentum = Initial momentum
400 Kg* v + 100 * (v + 920) = 500 kg * 9400 m/s
Calulating v,
v = 9216 m/s

(A)
Speed of Case = 9216 m/s
(B)
Speed of Payload = 9216 + 910
Speed of Payload = 10126 m/s
(C)
Initial Kinetic Energy = 0.5 * 500 * 9400^2 J
Initial Kinetic Energy = 2.209 * 10^10 J
Total kinetic energy of the two parts before they separate, = 2.209 * 10^10 J
(D)
Final Kinetic Energy = 0.5 * 400 * 9216^2 + 0.5 * 100 * 10126^2 J
Final Kinetic Energy = 2.2113 * 10^10 J

Increase in kinetic energy after they separate = 2.2113 * 10^10 - 2.209 * 10^10
Increase in kinetic energy after they separate = 2.3 * 10^7 J

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