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

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Homework Help Overview

The problem involves a bullet embedding itself in a wooden block, which then slides and compresses a spring. The context includes concepts of momentum and energy conservation, specifically in a collision scenario and subsequent motion involving friction and spring mechanics.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to apply conservation of momentum and energy principles to find the mass of the wooden block. They express uncertainty about energy loss during the collision and the role of friction.
  • Some participants suggest algebraic manipulation of the equations to simplify the problem and check the validity of the results.
  • Questions arise regarding the inclusion of the spring compression in the displacement calculation and whether the derived mass value is accurate.

Discussion Status

Participants are exploring various interpretations of the problem, with some guidance provided on algebraic approaches. There is no explicit consensus on the correctness of the mass value derived by the original poster, and further verification is suggested.

Contextual Notes

The original poster mentions a previous related problem, indicating ongoing challenges with similar concepts. There is an acknowledgment of potential errors in earlier calculations, which may influence the current discussion.

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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