A thin metal rod of mass 1.7 kg and length 0.8 m is at rest in outer space, near

Question

A thin metal rod of mass 1.7 kg and length 0.8 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 165 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

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

Explanation / Answer

(a) The change in momentum of the meteorite in the y direction divided by mass of rod equals the y velocity and the change in momentum of the meteorite in the x direction divided by the mass of the rod equals the x velocity.

Change in momentum y-dirn = mvfsinf - mvisini

= 0.07*60sin82°- 0.07*165sin26° = 4.159-5.063 = -0.904

Change in momentum x-dirn = -mvfcosf - mvicosi

= - 0.07*60cos82°- 0.07*165cos26° = -0.584-10.381 = -10.965

The velocity of the center of the rod y-dirn = -0.904/1.7 = -0.531m/s

The velocity of the center of the rod x-dirn = -10.965/1.7 = -6.45 m/s

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