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 wooden block after a collision. A 10.0g bullet traveling at an initial speed of 450 m/s strikes a 1.0 kg block, causing the block to rise 1.0 cm. The key equations involved are the conservation of momentum and energy, specifically mgh + 1/2mv^2 = mgy + 1/2mv^2. The bullet's speed upon exiting the block is determined by analyzing the energy lost during the collision and the work done against gravitational potential energy.

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Homework Statement


1. Homework Statement [/b]
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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Okay think about you started out with this much energy then you had to spend that energy on other things then how much energy do you have left?
 
I think its this equation: mgh +1/2mv^2=mgy +1/2mv^2, using my kinetic energy as my velocity
should be:(I THINK) v=radical of 2*g*h

Is this correct?
 
okay so, you start out with kinetic energy of the bullet then you hit the wood of certain mass, that slows down the bullet, but it slows down by raising the wood and bullet up some height. then after that the bullet comes out of the wood at lower kinetic energy thus comes out slower then ititiallly started. Can you form an equation from that info?
 
Is it mv(initial)=mv(final)?
 
Antineutron said:
slows down the bullet, but it slows down by raising the wood and bullet up some height.

No. Read again, this part. It must have done some work? What did the bullet do work on? What kind of collision is it when this occurred?
 
Last edited:

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