A spherical satellite of approximately uniform density with radius 6.3 m and mas

Question

A spherical satellite of approximately uniform density with radius 6.3 m and mass 260 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 ? (For vx, enter your answer to at least four significant figures.)

(b) Calculate the rise in the internal energy of the satellite and space junk combined.
J

Explanation / Answer

linear momentum before collision = linear momentum after collision

(m*v2) = (M*vcm) + (m*v3)

m*(v2x i + v2yj + v2zk) = (M*(vxi + vyj + vzj)) + (m*v3xi +v3yj + v3zk)

(vxi + vyj + vzj) = (m/M)*(( v2x-v3x)i + (v2y-v3y)j + (v2z-v3z)k)

(vxi + vyj + vzj) = (6.1/260)*((-2200+1300)i + (0-480)j + (0-0)k )

vxi + vyj + vzj = -21.1 i - 11.26 j + 0k

Vx = -21.1 m/s

vy = -11.26 m/s

vz = 0

angular momentum before collision = angular momentum after collision

m*v2*r - I*w1 = m*v3x + I*w

I = (2/5)*M*r^2

(6.1*2200*6.3) - ((2/5)*260*6.3^2*2) = (6.1*1300*6.3) + ((2/5)*260*6.3^2*w)

w = 6.4 rad/s

==============================

(b)

Ki = (1/2)*m*v2^2 + (1/2)*M*v1^2 + ((1/2)*(2/5)*M*r^2*w1^2

Ki = ((1/2)*6.1*2200^2) + ((1/2)*260*2600^2) + ((1/2)*(2/5)*260*6.3^2*2^2)

Ki = 893570255.5 J

Kf = (1/2)*m*v3^2 + (1/2)*M*v^2 + ((1/2)*(2/5)*M*r^2*w^2

Kf = ((1/2)*6.1*(1300^2+480^2)) + ((1/2)*260*(21.1^2+11.26^2)) + ((1/2)*(2/5)*260*6.3^2*6.4^2)

Kf = 6016116.22 J

increase = Ki - Kf =   893570255.5 - 6016116.22 = 887554139.28 J

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