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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 3.50Ā°. What is the speed of the bullet before it hit the block?

Respuesta :

Height increase h = R*(1 - cosĪ˜) = 1.2*(1-cos3.5Ā°) = .00224 m.Ā 

Ā½Mbl*VblĀ² = Mbl*g*h ā†’Ā 
Vbl = āˆš(2gh) = .2095 m/sĀ 
This is the initial velocity of the block. From momentum considerations,Ā 

Mbt*Vbt = Mbl*Vbl ā†’Ā 

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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