A thin metal rod of mass 1.8 kg and length 0.7 m is at rest in outer space, near

Question

A thin metal rod of mass 1.8 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 175 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 i = 26° and f = 82°.

(a) Afterward, what is the velocity of the center of the rod? (Express your answer in vector form.)

(b) Afterward, what is the angular velocity of the rod? (Express your answer in vector form.)

(c) What is the increase in internal energy of the objects?

A thin metal rod of mass 1.8 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 175 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 -260 and f= 82°. (a) Afterward, what is the velocity of the center of the rod? (Express your answer in vector form.) VCM-1 (6.44.0.67.0 m/s (b) Afterward, what is the angular velocity of the rod? (Express your answer in vector form.) -| 0.0,-31.55) | rad/s (c) What is the increase in internal energy of the objects? 1062.13

Explanation / Answer

c)

initial total kinetic energy is given as

KEi = (0.5) m vi2 = (0.5) (0.07) (175)2 = 1071.875 J

final total kinetic energy is given as

KEf = (0.5) m vf2 + (0.5) M V2 + (0.5) I w2 = (0.5) (0.07) (60)2 + (0.5) (1.8) (sqrt(6.44^2 + 0.67^2))2 + (0.5) ((1.8) (0.7)2/12) (31.55)2 = 200.3

increase in energy = 1071.875 - 200.3 = 871.6

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