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pb23me
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Homework Statement
A .025kg bullet is fired at 400 m/s into a 5kg wooden blockhanging from a string. the block is at rest before the bullet hits. The bullet becomes embedded in the wooden block.
What is the speed of the combined bullet + block?
How high will the block+bullet rise?
Calculate the total kinetic energy before the collision and after the collision.
Homework Equations
[tex]\Delta[/tex]KE=Wnet=1/2mvf2-1/2mvi2
conservation of momentum= Pnetf=Pneti
conservation of energy= [tex]\Delta[/tex]KE=-([tex]\Delta[/tex]PE)
The Attempt at a Solution
To find the speed of the block and bullet together, I used the conservation of momentum equation (.025)(400)=Vf(m1+m2) and got 2 m/s. I took the initial velocitys of both objects just before impact?? To find how high the block and bullet would rise i used the conservation of energy equation; 1/2(2m/s)2= 9.8(yf)... and got .204m for yf.. I set the final velocity equal to zero in this equation. Was that the correct thing to do? Would that really work in reality? Calculating the kinetic energy before the collision you just have the kinetic energy of the bullet so 1/2mv2=2000j Calculating the kinetic energy after the collision; (m1+m2)(2m/s)(.5)=5.025jWhen they ask for kinetic energy after the collision is this immediately after the collision?
When i used the conservation of momentum equation is the final velocity that i get immediately after impact?
How is it that the bullet is no longer putting a force on the wooden block after impact? This is the logic i used to solve the problem. It's what my proffesor told me.
Wnet after collision... Wnet = [tex]\Delta[/tex]KE=-Wg since gravity is the only force acting at this point. Then Wnet=-mgyf... The change in kinetic energy [tex]\Delta[/tex]KEf-[tex]\Delta[/tex]KEi= 5.025j-2000j=1994.975
I should be able to use this number to calculate yf...[tex]\Delta[/tex]KE=Wnet=1994.975=-mgyf...however this doent work...why is that?
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