A uniform wooden rod of length 1.20 m and mass 0.500 kg rests on a horizontal, f

Question

A uniform wooden rod of length 1.20 m and mass 0.500 kg rests on a horizontal, frictionless table. The rod is free to rotate about a pivot at its center.

A bullet of mass 4.20 grams is shot through the rod at a point 0.300 m from one end. The speed of the bullet is 260 m/s before it hits the rod and is 150 m/s after it exits.

(a) What is the magnitude and direction of the angular momentum of the bullet relative to the pivot before it reaches the rod?

(b) What is the magnitude and direction of the angular momentum of the bullet relative to the pivot after it leaves the rod?

(c) What is the angular speed of the stick after the bullet passes through it?

(d) How much kinetic energy is lost in the collision?

Explanation / Answer

A) L1 = m*v*r = 4.2*10^-3*260*0.3 = 0.3276 kg*m^2/S

B) L2 = m*v*r = 4.2*10^-3*150*0.3 = 0.189 kg*m^2/S

C) L1 = L2+I2*w2

I2*w2 = L1-L2 = 0.3276-0.189 = 0.1386

w2 = 0.1386/I2 = 0.1386*12/(0.5*1.2*1.2) = 2.31 rad/s

D) lost in KE = Kf-Ki

Kf = 0.5*m*vf^2 + 0.5*I2*w2^2 = 0.5*4.2*10^-3*150^2 + (0.1386*2.31*0.5) = 47.41 J

Ki = 0.5*m*vi^2 = 0.5*4.2*10^-3*260*260 = 141.96 J

lost in KE= 47.41-141.96 = -94.55 J

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