Linear impulse and linear momentum

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krnhseya
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Homework Statement


given mass of bullet and box and given initial and final velocity of projectile, find kinetic friction coefficient


Homework Equations



G1 + integral of sum force respect to time = G2

The Attempt at a Solution



I setup like this...

mv(projectile) + mv(block, which is 0) = mv(projectile) + mv(block)

I was trying to use the equation under 2 but i don't know how that can help me anything.
 
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krnhseya said:
G1 + integral of sum force respect to time = G2
drop this one

mv(projectile) + mv(block, which is 0) = mv(projectile) + mv(block)

I was trying to use the equation under 2 but i don't know how that can help me anything.

I can't tell what it's supposed to mean, but the other equation will help. Use that to find the speed of the block after the collision.

Then the block slides to a stop as friction does the (negative) work to take away the kinetic energy.
 
yes I've found the velocity of the block.
i don't know how to go from there...
do i use work energy equation to finish it?
1/2mv^2(for block) + 1/2mv^2(for projectile) - work done by friction = 1/2mv^2(for block) + 1/2mv^2(projectile)?
 
The work done by friction will just be on the block, forget the bullet after the collision, it's no longer your concern. So the KE' of the block is taken away by the work done by friction. KE = W
 
i forgot to mention what cancells out from the equation above.
well anyways, i think my brain is dead...
-(coefficient of kinetic)mgd = 1/2mv^2(velocity that i got from the above).
i got coefficient, which is 0.3 something but it's negative...
do i take the absolute value of it since it depends on how i draw it?
thank you very much.
 
the negative cancels out because the change in KE is negative (lost)