A thin metal rod of mass 1.8 kg and length 0.9 m is at rest in outer space, near

Question

A thin metal rod of mass 1.8 kg and length 0.9 m is at rest in outer space, near a space station (see figure below). A tiny meteorite with mass 0.08 kg traveling at a high speed of 210 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°.

(b) Afterward, what is the angular velocity

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

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

(0.08*210*cos26)i + (0.08*210*sin26) j = (1.8*Vcm) - (0.08*60*cos82)i + (0.08*60*sin82)j

Vcm = 8.8 i + 1.45 j

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

(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

-0.08*210*cos26 = 0.08*60*cos82 + ((1/12)*1.8*0.9^2*w)

w = -130 k

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

Ki = (1/2)*m*vi^2 = (1/2)*0.08*210^2 = 1764 J

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

Kf = ((1/2)*0.08*60^2) + ((1/2)*1.8*(8.8^2+1.45^2)) + ((1/12)*1.8*0.9^2*130^2)

Kf = 2269 J

increase = Kf - Ki = 2269 - 1764 = 505 J

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