A spherical satellite of approximately uniform density with radius 5.4 m and mas

Question

A spherical satellite of approximately uniform density with radius 5.4 m and mass 270 kg is originally moving with velocity (2600, 0, 0) m/s, and is originally rotating with an angular speed 2 rad/s, in the direction shown in the diagram. A small piece of space junk of mass 6.1 kg is initially moving toward the satellite with velocity (-2200, 0, 0) m/s. The space junk hits the edge of the satellite as shown in the figure below, and moves off with a new velocity (-1300, 480, 0) m/s. Both before and after the collision, the rotation of the space junk is negligible. (a) Just after the collision, what are the components of the center-of-mass velocity of the satellite (vx and vy) and its rotational speed co? (For vx, enter your answer to at least four significant figures.)

Explanation / Answer

from conservation of momentum

momentum before collision = momentum after collision

M*v1 + m*v2 = Mv + m*v3

(270*2600)i - (6.1*2200)i = (270*v) + (6.1*(-1300i + 480j))

702000 i - 13420 i = 270v - 7930i + 2928j

v = 2579.7 i - 10.84 j

vx = 2579.7 m/s

vy = -10.84 j

v = sqrt(vx^2+vy^2) = 2579

+++++

from conservation angular momentum

Li = Lf

I*w1 + m*v2*R = I*w + m*v3x

(2/5)*M*R^2*w1 + m*v2*R = (2/5)*M*R^2*w + m*v3x*R)

-((2/5)*270*5.4^2*2)+(6.1*2200*5.4) = ((2/5)*270*5.4^2*w)+(6.1*1300*5.4)

w = 7.41 rad/s

++++++++++++++++++++++++

total energy before collision

Ei = (2/5)*M*R^2*w1^2 + (1/2)*M*v1^2 + (1/2)*m*v2^2

Ei = ((2/5)*270*5.4^2*2^2) + ((1/2)*270*2600^2)+((1/2)*6.1*2200^2)

Ei = 927374597.12 J

Ef = (2/5)*M*R^2*w^2 + (1/2)*M*v^2 + (1/2)*m*v3^2

Ef = ((2/5)*270*5.4^2*7.41^2) + ((1/2)*270*2579^2)+((1/2)*6.1*(1300^2+480^2))

Ef = 903947675.98 J

internal energy = Ei - Ef = 23426921.14 J

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