A thin metal rod of mass 1.5 kg and length 0.7 m is at rest in outer space, near

Question

A thin metal rod of mass 1.5 kg and length 0.7 m is at rest in outer space, near a space station (see figure below). A tiny meteorite with mass 0.09 kg traveling at a high speed of 245 m/s strikes the rod a distance 0.2 m from the center and bounces off with speed 60 m/s as shown in the diagram. The magnitudes of the initial and final angles to the x axis of the small mass's velocity are theta_i = 26 degree and theta_f = 82 degree. (a) Afterward, what is the velocity of the center of the rod? (Express your answer in vector form.) V_CM = m/s (b) Afterward, what is the angular velocity omega of the rod? (Express your answer in vector form.) omega = rad/s (c) What is the increase in internal energy of the objects?

Explanation / Answer

A)

linear momentum before collision = linear momentum after collision

(m*vi*costhetai)i + (m*vi*sinthetai)i = (M*vcm) - (m*vf*costhetaf)i + (m*vfIsinthetaf)j

(0.09*245*cos26)i + (0.09*245*sin26) j = (1.5*Vcm) - (0.09*60*cos82)i + (0.09*60*sin82)j

Vcm = 13.71 i + 2.88 j

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(b)

angular momentum before collision = angular momentum after collision

-m*vi*d*sinthetai = m*vf*sinthetaf - I*w

I = 1/12*M*L^2

-m*vi*d*costhetai k = m*vf*costhetaf k - (1/12)*M*L^2*w k

-0.09*245*cos26 = 0.09*60*cos82 - ((1/12)*1.5*0.7^2*w)

w = 335.8 k

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Ki = (1/2)*m*vi^2 = (1/2)*0.09*245^2 = 2701.12 J

Kf = (1/2)*m*vf^2 + (1/2)*M*vcm^2 + (1/2)*I*w^2

Kf = ((1/2)*0.09*60^2) + ((1/2)*1.5*(13.71^2+2.88^2)) + ((1/12)*1.5*0.7^2*335.8^2)

Kf = 7215.84 J

increase = Kf - Ki = 7215.84 - 2701.12 = 4514.72 J

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