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A uniform thin rod of length 0.60 m and mass 3.5 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a 3.0 g bullet traveling in the rotation plane is fired into one end of the rod. As viewed from above, the bullet's path makes angle θ = 60° with the rod. If the bullet lodges in the rod and the angular velocity of the rod is 14 rad/s immediately after the collision, what is the bullet's speed just before impact? (Answer is in m/s)

Verified Answer :

The moment of inertia for the rod is 
mrL^2/12 
After collision 
I=mrl^2/12+mbr^2 
I=0.073 
The momentum of the rod after collision is 
Iw 
=1.022 

The momentum of the bullet that is radial 
to the rod is 
vr=vsin(60) 
vr/r=w 

so the radial momentum of the bullet is 
3vr/(0.251000) 

vr0.012 

applying conservation of momentum 
vr*0.012=1.022 
vr=85 m/s 

and 
v=85/sin(60) 
v=98 m,/s 

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