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

[strangeeyes]

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 ( $$\mu$$k = 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
$$\mu$$k =0.10
$$\Delta$$d=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= $$\mu$$k(Fn)$$\Delta$$d + 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

 

[gneill]

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.

 

[strangeeyes]

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!

 

[gneill]

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.

 

[rwisz]

, it's the spring...

I would say that your approach to solving this problem is correct. You have correctly identified the initial and final energies of the system and have used the conservation of energy equation to solve for the mass of the wood block. However, as you have mentioned, there may be some additional factors that could affect the final result, such as the loss of energy due to heat and sound. In order to accurately calculate the mass of the wood block, you may need to take these factors into account. Additionally, it may be helpful to consider the conservation of momentum in this situation, as the bullet and wood block are both initially at rest and then move after the collision. Overall, your approach is sound and with some additional considerations, you should be able to arrive at the correct answer.