Find the speed of the bullet as it emerges from the block

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SUMMARY

The discussion focuses on calculating the speed of a bullet as it exits a wood block using principles of linear momentum and conservation of energy. A 10.0g bullet traveling at an initial speed of 450 m/s impacts a 1.0 kg block, causing the block to rise 1.0 cm. The relevant equations include linear momentum conservation (mbullet*vi = mbullet*vf + mblock*vblock) and energy conservation (mgh + 1/2mv^2 = mgy + 1/2mv^2). The key to solving the problem lies in first determining the velocity of the block after the bullet passes through.

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Homework Statement


A 10.0g bullet is shot through a 1.0 kg wood block suspended on a string 2.0m long. The center of mass of the block rises a distance of 1.0cm. find the speed of the bullet as it emerges from the block if its initial speed is 450m/s.

Homework Equations


I think this has to do with linear momentum, p(i)=p(f)(mv=mv). Either that or conservation of energy: mgh +1/2mv^2=mgy +1/2mv^2

The Attempt at a Solution


Im confused as to how to go about solving this problem. for the initial mv, I plugged in the values for the bullets mass and used 450 m/s as the initial velocity. for the final mv, i plugged in the the mass value for the block and bullet, and used that to divide my initial mv, which was good for linear momentum but still not good enough, i don't think its right. so when i use the conservation of energy,i know my h=2.0m and y=.01m

Am I on the right track. Which equation should i use? Thanks
 
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The idea is to use:

mbullet*vi = mbullet*vf + mblock*vblock

where vf is the velocity of the bullet right after it goes through, and vblock is the velocity of the block right after it goes through.

so you need to find vf. but before you can do that you need vblock... do you see a way you can find the velocity of the block right after the bullet goes through?
 

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