The last stage of a rocket is traveling at a speed of 7250 m/s . This last stage

Question

The last stage of a rocket is traveling at a speed of 7250 m/s. This last stage is made up of two parts that are clamped together, namely, a rocket case with a mass of 330.00 kg and a payload capsule with a mass of 135.00 kg. When the clamp is released, a compressed spring causes the two parts to separate with a relative speed of 910 m/s. What are the speeds of the two parts after they have separated? Assume that all velocities are along the same line.a) What is the speed of the payload?

b)What is the speed of the rocket case?

c)Find the total kinetic energy of the two parts before and after they separate; account for any difference. The kinetic energy before they separate.

d)The kinetic energy after they separate.

Explanation / Answer

Consider moving with the rocket so it appears at rest. When the spring does it's thing, it will impart equal and opposite momentums to each section of the rocket.

150*u = 290*v .... direction of v is opposite that of u

And we also know that: u + v = 910 m/s

135*u = 330*(910 - u)
465u = 330*910 = 263900
u = 645.80
v = 255.2

Speed of payload = 7250 + 645.80 = 7895.8 m/s
Speed of case = 7250 - 255.2 = 6994.8 m/s

kinetic energy before = (1/2)(465)(7250)^2 = 1.22*10^10 J

kinetic energy capsule (after) = (1/2)(135)(7895.8)^2 = 4208196891 J
kinetic energy case (after) = (1/2)(330)(6994.8 ^2) = 8072992462 J
Sum of both = 1.22*10^10 J

Difference = 3864795571 J

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