# Homework Help: Conservation of Momentum/Energy! AHH!

1. Jan 16, 2010

### mc8569

1. The problem statement, all variables and given/known data
http://i48.tinypic.com/2cfb3py.png
^^Diagram of situation
There is a projectile of mass m and initial velocity v fired into a block of width L and mass M at rest on a frictionless surface.

a) Find the speed of block M as the bullet emerges from the back side with a velocity of v/3

b)If the block is fixed (can't slide) and the block emerges with v/2 from back of block, determine the loss of kinetic energy.

c)If the frictional force exerted by the block on the bullet is constant, in terms of L, what is the minimum width W should a fixed block have in order to stop the bullet?

2. Relevant equations
Conservation of Momentum
Conservation of Energy

3. The attempt at a solution
a) I am pretty sure this is right:
Pi = Pf
mv = Mv(f) + m(v/3)
v(f) = (2mv)/(3M)

b) I am pretty sure this is right as well:
KEi = KEf + Wnc
*nc = non conservative
Wnc = - (.5mv(i)^2 - .5mv(f)^2)
Wnc = .5m(v/2)^2 - .5mv^2
Wnc = -(3/8)mv^2

c) I don't know where to go with this. I want to say that I should begin using b) and say that Wnc = Wf (work done by friction of block) and from there solve for the frictional force. This frictional force will be equal to the only force working on the block, and therefore use this friction to solve for W by doing:
W = fdcos@ = .5mv^2 <--and using v for initial velocity.

But the problem is I don't think you can use conservation of energy because this isn't an elastic equation (or inelastic? I always get the two mixed up, regardless you know what I mean) and therefore you can't use "Work = fdcos@ = .5mv^2" because the initial velocity v is not the actual initial velocity of kinetic energy that moves the block since when it hits the block some energy is transferred into heat or sound or whatever it may be - and therefore I feel I can't use v as the initial velocity to solve for the force. (To be concise, what I mean is in b) they ask for the TOTAL kinetic energy lost, not the energy lost due to friction while inside the block. If it were to ask for energy lost due to friction while inside the block I feel I cannot answer that question either.)

So I know now other method in solving for the force, and it is only through solving for this force vector that I can see anyway of finding the minimum width W.

Therefore, since I can't use

2. Jan 16, 2010

### Matterwave

I think you're overthinking the question. Yes, energy is lost to sound, the deformation of the material, the deformation of the projectile, heat, etc. But in this question, all that energy is lost due to the frictional force. The frictional force, in this case, is all encompassing. I.e. it disperses kinetic energy of the projectile into all those forms of energy losses that you're thinking of.

3. Jan 16, 2010

### mc8569

So are you saying I should use what I found in b), divide it by L thereby solving for the frictional force f. Using this f use work = fdcos@ = KEi and solve for this new d accordingly?

Therefore, is this right:
From b) solve for frictional force by dividing that found value by L:
*can't use a) to solve for frictional force because a) is a situation where block is not fixed, so frictional force is not the same as in situation c), but b) is same situation - block is fixed
fdcos@ = (3/8)mv^2
fL = -(3/8)mv^2
f = -(3/8)mv^2/L

and then the setup to solve for minimum distance W to stop bullet using the found frictional force from b) should be:
KEi - fdcos@ = 0
.5mv^2 = fd
d = W = (mv^2)/(2f)
W = (4/3)L

Please let me know if this is correct.

Last edited: Jan 17, 2010
4. Jan 17, 2010

### glueball8

b) work is equal to change in kinetic energy, I think there's just a computational error.

(1/2)^2-1^2 ???

For c) you know the loss in energy. What is the formula for work? Then which is what in this situation?

5. Jan 17, 2010

### mc8569

glueball, i don't understand what you're talking about. could you please explain to me? I'm very confused..

6. Jan 17, 2010

### glueball8

"b) I am pretty sure this is right as well:
KEi = KEf + Wnc
*nc = non conservative
Wnc = - (.5mv(i)^2 - .5mv(f)^2)
Wnc = .5m(v/2)^2 - .5mv^2
Wnc = -(3/8)mv^2
"

how did you get Wnc = -(3/8)mv^2 ???

Then you know the loss in energy (which is Kinetic energy) is equal to the W done on it. And since its linear, you can use W = F_{frictional}*L. There is no angle.

7. Jan 17, 2010

### mc8569

Er, my part b) I am 99.99% sure I am correct.
Initial velocity (initial kinetic energy) before entering/collision with the block (before it loses any kinetic energy) is v. Its final velocity when it exits the block (v/2) consists of its final kinetic energy. Therefore,
KEi - Wnc = KEf
*You can logically see that this makes sense because the KEi > KEf, therefore that is why it is set up that way, leading to Wnc = KEi - KEf.
*But also, if you think about it, say it is all lost due to friction. You say there is no angle, but there certainly is. Friction is ALWAYS at a 180 degree angle, because friction always acts directly opposite. So, pretend Wnc is work by friction, you have:
Wnc = Wf = fdcos@ = KEi - KEf
*cos@ = -1
-fd = KEi - KEf
fd = KEf - KEi
Again, think logically that KEf < KEi, and therefore it makes sense that the value you obtain from this lastly solved equation is negative, because of course work done by friction is ALWAYS negative.
That is how I got Wnc = -(3/8)mv^2

And for c) I was saying, if you said that all of the non conservative force was enacted by friction within the block, and you don't consider any energy lost due to sound or other collisional losses, then:
Wnc = Wf = -(3/8)mv^2
Wf = fdcos@ = -fd = -(3/8)mv^2
Given this equation, I would see to solve for the frictional force f by dividing by distance d (or rather, distance L, which is what you have to plug in) thereby getting:
fd = fL = (3/8)mv^2
f = (3/8)mv^2 / L
With this force, you re-setup the equation of work = Fdcos@ <-in this case, F meaning force in general - it just happens to be that the only force acting on the block would be the frictional force by the bullet.
With Work = Fdcos@ you set it equal to
Fdcos@ = .5mv^2
fWcos@ = .5mv^2 <--(f plugged in for F, and W plugged in for d, W is what we must solve for in terms of L)
because all the work done on the object, will give it its final kinetic energy right?
So using these two energy equations, you solve for the unknown W:
W = .5mv^2 / f
W = .5mv^2 / ((3/8)mv^2/L)
W = (4/3)L

Last edited: Jan 18, 2010
8. Jan 18, 2010

### mc8569

9. Jan 18, 2010

### mc8569

Can someone pleaseeee help me! I need help by tomorrow... which is when this problem is due! I would appreciate it a lot.. =(

10. Jan 18, 2010

### ideasrule

Suppose the frictional force is F. Acceleration would then be F/m. What distance would be required to slow down the bullet to v/3? To 0?

11. Jan 18, 2010

### mc8569

Omg.. no one is really giving me much help.
The case for when the bullet leaves the block at v/3 is when the block is not fixed, and moves due to the frictional force of the bullet.
In c) the block is stated to be fixed, and must use v/2 from part b). The main problem I have is, can we say that energy is conserved? That it is an elastic collision? It obviously cannot be an elastic collision because if you solve part a) using conservation of energy, you do not get the same value for the final speed of the block as you do using conservation of momentum.

I guess, I am asking more of a conceptual question, that's why my comments consist so much of words rather than equations. Do you think that in this problem, the problem makers did not consider the error in making the problem? Because I know that part a) is right and that you must use conservation of momentum, not conservation of energy. But if you can't use conservation of energy, then that is only because energy is not conserved. If energy is not conserved, you cannot solve for part c) because in part b), all of the kinetic energy is not lost due to only the frictional force when it is going through the block - some is also lost due to sound and bullet compression and whatnot. Should I consider this problem as having a mistake in concept and answer part c) as if all the kinetic energy is lost only while it is in the block (none lost due to the collision).

12. Jan 19, 2010