Bullet hits wood block, slides, hits ideal spring: cons. of energy/momentum

AI Thread Summary
A bullet embeds itself in a wooden block, which then slides and compresses a spring, prompting a discussion on energy and momentum conservation. The initial kinetic energy of the bullet is not fully transferred to the block due to energy losses from heat and sound. The user is attempting to derive the mass of the block using conservation equations but struggles with algebraic manipulation. They have calculated a mass of 1.3 kg for the block but are advised to verify this by substituting back into the quadratic equation. The discussion emphasizes the importance of correctly accounting for energy losses and checking calculations for accuracy.
strangeeyes
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Homework Statement



A 50g bullet traveling horizontally at 200 m/s embeds itself in a much more massive wooden block initially at rest on a horizontal surface ( \muk = 0.10). The block then slides 1.2 m toward an ideal spring and collides with it. The block compresses the spring (k=600.0 N/m) a maximum of 20.0 cm. Calculate the mass of the block of wood.

let (b) represent the bullet and (w) the wood block:
Mb=0.05 kg
Vb=200 m/s
Vw=0 m/s
\muk =0.10
\Deltad=1.2 m(+ 0.2 m)?
k=600 N/m
x=0.2m
Mw=?
Vw+b'=?

My thinking is that the initial energy of the bullet does not equal the final energy of the system because the bullet would lose energy when it hits the block (through heat, sound etc).

to find an equation to substitute for Vw+b:
MwVw+MbVb=(Mw+Mb)Vw+b
Mw0 + 0.05(200)=(Mw+ 0.05)Vw+b
Vw+b= 10/(Mw + 0.05)

now to put that into the conservation of energy equation:
Et=Et'
Ek=Eth+Ee
0.5Mw+b(Vw+b)2= \muk(Fn)\Deltad + 0.5kx2
0.5(0.05+Mw)(10/Mw+0.05)2=0.1[9.8(0.05+Mw)]1.2+0.5(600)0.22

can someone tell me if I'm on the right path? i have no clue how to complete the left side algebraically, so it'd be nice to know
if it is even a necessary effort...

also i don't know if the displacement should include the 20 cm the block travels after hitting the spring because at that point it's not just friction acting against the block
 
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It looks like you're on the right track. Your final expression will resolve into a quadratic equation which you'll have to solve to find the mass. I'd suggest moving everything to one side of the "=" and expanding it out, multiplying out all the arithmetic operations you can to reduce the confusion.
 
thanks, i think I've figured it out. i got Mw=1.3 kg

also, i doubt you remember this, but you were helping me with an explosion question i was having trouble with

https://www.physicsforums.com/showthread.php?t=456273

i got an answer but it's wrong. your help would be greatly greatly appreciated; this problem is driving me mad!
 
strangeeyes said:
thanks, i think I've figured it out. i got Mw=1.3 kg
You might want to check that answer (plug the value back into the quadratic and see if it produces zero). This particular quadratic has two real roots; one positive and one negative.

also, i doubt you remember this, but you were helping me with an explosion question i was having trouble with

https://www.physicsforums.com/showthread.php?t=456273

i got an answer but it's wrong. your help would be greatly greatly appreciated; this problem is driving me mad!

I'll have a look.
 
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