The last stage of a rocket is traveling at a speed of 9600 m/s . This last stage

Question

The last stage of a rocket is traveling at a speed of 9600 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 250.0 kg. When the clamp is released, a compressed spring causes the two parts to separate with a relative speed of 850.0 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 separate?

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

here ,

let the final speed rocket case is vr

final speed of payload is vp

Using conservation of momentum

9600 * (250 + 380) = 250 * vr + 380 * vp ----(1)

for the relative speed

vp - vr = 850

vp - vr = 850 ----(2)

solving 1 and 2

vp = 9937.3 m/s

vr = 9087.3 m/s

a) the speed of case after seperate is 9937.3 m/s

b)

the speed of the payload after they seperate is 9087.3 m/s

c)

total kinetic energy before they seperate = 0.5 * (250 + 380) * 9600^2

total kinetic energy before they seperate = 2.903 *10^10 J

the total kinetic energy before they seperate is 2.903 *10^10 J

d)

increase in kinetic energy = 0.5 * 250 * 9087^2 + 0.5 * 380 * 9936.3^2 - 2.903 *10^10

increase in kinetic energy = 50407086.1 J

the increase in kinetic energy is 50407086.1 J

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