The last stage of a rocket is traveling at a speed of 9200 m

The last stage of a rocket is traveling at a speed of 9200 m/s. This last stage is made up of two parts that are clamped together, namely, a rocket case with a mass of 260.0 kg and a payload capsule with a mass of 280.0 kg. When the clamp is released, a compressed spring causes the two parts to separate with a relative speed of 800.0 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?

Solution

Here ,

mass of rocket case , mr = 260 Kg

mass of payload , mp = 280 Kg

relative velocity = 800 m/s

let the final speed of payload is vp

final speed of rocket is vr

Using conseravtion of momentum

260 * vr + 280 * vp = (260 + 280) * 9200 ---(1)

for the relative velocity

vp - vr = 800 ---(2)

solving for vp and vr

vp = 9585.2 m/s

vr = 8785.2 m/s

the speed of the case after they seperate is 8785.2 m/s

the speed of the payload after they seperate is 9585.2 m/s

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total initial kinetic energy = 0.5 * mr * vr^2 + 0.5 * mp * vp^2

total initial kinetic energy = 0.5 * 280 * 9200^2 + 0.5 * 260 * 9200^2

total initial kinetic energy = 2.285 *10^10 J

the total initial kinetic energy was 2.285 *10^10 J

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increase in kinetic energy = final kinetic energy - initial kinetic energy

increase in kinetic energy = 0.5 * 260 * 8785.2^2 + 0.5 * 280 * 9585.2^2 - 2.285 *10^10

increase in kinetic energy = 4.601 *10^7 J

the increase in kinetic energy after seperation is 4.601 *10^7 J