The last stage of a rocket is traveling at a speed of 8650 m/s. This last stage

Question

The last stage of a rocket is traveling at a speed of 8650 m/s. This last stage is made up of two parts that are clamped together, namely, a rocket case with a mass of 290.00 kg and a payload capsule with a mass of 150.00 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 befroe and after they separate; account for any difference. The kinetic energy before they separate.?

d.The kinetic energy after they separate.?

Explanation / Answer

v =8650 m/s , m1 =290 kg , m2 =150 kg

relative speed u1 +u2 = 910 m/s ...(1)

From conservation of momentum

m1u1+m2u2 = (m1+m2) v

(290*u1) +(150*u2) = (290+150)*8650 ... (2)

from (1) and (2)

u1 =26210. 7 m/s

u2 = -25300.7 m/s

a) speed of payload u2 = 25300.7 m/s

b) speed of rocket case u1 = 26210.7 m/s

(c) K1 = (1/2)(m1+m2)v^2

K1 = 0.5*(290+150)(8650*8650)

K1=1.6461*10^10 J

(d) K2 = (1/2)m1u1^2 +(1/2)m2u2^2

K2 = (0.5*290*(26210.7^2))+(0.5*150*(25300.7)^2)

K2 =14.76*10^10 J

K2 - K1 = 14.76*10^10 - 1.6461*10^10 J

K2 -K1 = 13.12*10^10 J

The additional energy is derived from the potential energy of the spring before the clamp is released.

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