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 0.07 kg traveling at a high speed of 150 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. Afterward, what is the velocity of the center of the rod? (Express your answer in vector form.) V_CM rightarrow = m/s Afterward, what is the angular velocity omega rightarrow of the rod? (Express your answer in vector form.) omega rightarrow = rad/s What is the increase in internal energy of the objects? J

Explanation / Answer

(A) Applying momentum conservation for the system,

0.07 x 150 (cos26i + sin26j) + 1.5x 0 = 1.5v + (0.07 x 60 (- cos82i + sin82j))

9.437i + 4.603j = 1.5v - 0.584i + 4.159 j

v = 6.681i + 0.296j

Vcm = <6.681, 0.296, 0 > m/s

(B ) Applying angular momentum conservation about the centre of mass.

(-0.07 x 150 x 0.2 x cos26) + 0 = (0.07 x 60 x 0.2 x cos82) + I w

for rod: I = m L^2 / 12 = 0.06125 kg m^2

w = - 32.72 rad/s

w = <0, 0, - 32.72 > rad/s

(C) Ki = 0.07 x 150^2 /2 = 787.5 J

Kf = (0.07 x 60^2 / 2) + (1.5 x (6.681^2 + 0.296^2) / 2) + (0.06125 x 32.72^2 / 2)

= 126 + 33.54 + 32.79 = 192.33 J

Increase in internal energy = Ki - Kf= 593.2 J

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