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A 2.00-g bullet hits and becomes embedded in a 500-kg wood block which is hanging from a 1.20-m long string. This causes the block to swing through an arc of 3.50°. What is the speed of the bullet before it hit the block?

Verified Answer :

Height increase h = R(1 - cosΘ) = 1.2(1-cos3.5°) = .00224 m. 

½MblVbl² = Mblgh → 
Vbl = √(2gh) = .2095 m/s 
This is the initial velocity of the block. From momentum considerations, 

MbtVbt = MblVbl → 

Vbt = (Mbl/Mbt)Vbl = (5/.002)*.2095 = 523.8 m/s

this is what i think the answer is im not for sure though... hope it helps...

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