# Homework Help: Using the energy principle

1. May 31, 2014

### Maxo

1. The problem statement, all variables and given/known data
A wooden block with mass 1.25 kg is hit by a bullet with mass 5.00 g and the speed vi according to the figure. The bullet sticks in the block and together they go up to the maximum height of 22.0 cm. Calculate the initial speed of the bullet.

2. Relevant equations

3. The attempt at a solution
hi = 0 so it can be removed from the equation. Before the collision the objects must be treated as individual, afterwards they can be treated as one single object. vf = 0 when the objects reach their maximum height so it can also be removed from the equation. So with m1=bullet and m2=block we only have left:

(m1+m2)*g*hf = 0.5*m1*vi2

Solving for vi=âˆš(2/m1*(m1+m2)*g*hf)

This gives me vi = 32.9 m/s. Which apparently is incorrect. What did I do wrong?

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Last edited: May 31, 2014
2. May 31, 2014

3. May 31, 2014

### Simon Bridge

Kinetic energy is not conserved in the collision.

shortcut: KE=p^2/2m

4. May 31, 2014

### Maxo

Done. Do you see it now?

Do you mean that the energy principle can't be used then? Or how should it be written instead?

That's good. I know the linear momentum is preserved. But if we use Pf=P0 then we need an expression for vf to solve for vi.

5. May 31, 2014

### ehild

There are two stages of the problem. The first one is the inelastic collision between the bullet and block. Using conservation of momentum, determine the initial velocity of the block with the bullet in it. Energy is not conserved during this process.

The block with the embedded bullet rises to maximum height h. At this stage, energy is conserved.

ehild

6. May 31, 2014

### Simon Bridge

If you mean that vf is the velocity of the block+bullet after the collision - you do not need it.

To avoid subscripts - write u=vi, as the initial velocity of the bullet.
(saves typing)
define:
m=mass of bullet
M=mass of the block
h=height m+M end up at.
g=acceleration of gravity

Before the collision: write the initial momentum of the bullet in terms of m and u.

Right after the collision:
What is the momentum of the block+bullet in terms of m and u?
Therefore: using "EK=p^2/2m", what is the kinetic energy of the block and bullet, in terms of m, M, and u?
Put that expression equal to (m+M)gh and solve for u.

(or do it in two stages...)

7. May 31, 2014

### BiGyElLoWhAt

break it up into chunks

1) The point where the block is at it's max height.
2) The instant after the collision
3) The instant before the collision

How can you relate these periods of time together? And what do you know about them? (What is conserved in each case?)

8. May 31, 2014

### Maxo

p0 = m*u

pf = (m+M)*vf = m*u
Since momentum is preserved from the moment bullet is fired (p0) until it hits (pf).

Can you please write "KE0" or "KEf" to show whether it is the initial or the final KE(EK)? I assume here it's initial (from the moment that the bullet hits)??

KE0 = (m^2*u^2) / (2m + 2M)
So far so good?

You mean, that it is equal to "PEf"? The final potential energy.

And the reason for that is that from the moment the bullet hits the box, there are no non-conservative forces doing work (Wnc=0), and PE0 = 0, and KEf = 0. Right?

What are the forces doing Wnc acting on the bullet/box before the bullet hits the box? Since the KE is not the same before and after bullet hits, there must be some nc force? What is it?

Last edited: May 31, 2014
9. May 31, 2014

### BiGyElLoWhAt

yes
You have a perfectly inelastic collision, meaning the 2 objects merge together to form 1 new object. This means that energy is definitely NOT conserved throughout the collision.

However, after the collision, you have 1 object that is moving from a height $h=0$ to a point $h=h_f$
This is just translational motion, no collision, and therefore energy IS conserved over the path of the motion. When the object is at $h=0$ you have $E_{total_0} = KE_0 + PE_0 = KE_0 + 0$ as $mgh = 0$.
At the highest point the object reaches, you have $E_{total_f} = KE_f +PE_f = 0 + PE_f$
Through conservation of energy, (with no work being done on the system) we know that $E_{total_0} = E_{total_f}$ and therefore $KE_0 + 0 = 0 + PE_f$ â†’ $KE_0 = PE_f$

make sense?

10. May 31, 2014

### Maxo

Please see my post again, I edited it and added some questions. I would very much appreciate if they could be answered/clarified.

Why is that so obvious? How can we know that energy is not conserved? What is the non conservative force acting on it so that energy is not conserved? Friction?

11. May 31, 2014

### Simon Bridge

well done: pf=p0=mu so we can drop the extra letter and put p=mu.

The general equation for kinetic energy that I wrote is no different for initial or final kinetic energy. Those labels are part of the interpretation needed to fit the equation to the situation - that's your job.

You need to get used to seeing formulae and equations in general forms.

Well, before the bullet hits would not get you anywhere so - yes, that would be a good inference.

... so far so good.

That is correct.

Thats right, the only thing KE can turn into here is PE because the only thing the masses interact with is a slope. Only gravity is there to remove kinetic energy from the system.

Nothing - but while the bullet is penetrating the box, it is deforming the box, making noise and heat. Some of the bullet's initial kinetic energy has to go into producing those things.

[edit: I thought of a better way of putting it...]

before collision:
p0=mu (only bullet moving)
KE0=mu^2/2 (ditto)
PE0=0 (bottom of slope)

during collision:
some KE turns into heat and sound.

after collision
p1=p0=mu (conservation of momentum)
KE1=p1^2/2(M+m) (â‰ KE0; inelastic collision)
PE1=0 (bottom of slope)

on slope: KE is exchanged for PE because of work against gravity.

after slope
p2=0 (at rest)
KE2=0 (at rest)
PE2=(M+m)gh (up slope)

But you have to make sure you track everything ... the final kinetic energy is KE2=0.
All you need after that list is the link between KE1 and PE2.

You could notice that you don't have to write anything for the terms that are zero, which allows you to drop the subscripts. I hate subscripts.

Aside:
What percentage of the initial kinetic energy turned into sound and heat in the collision? (in terms of M and m.)

Was linear momentum conserved on the slope?

Last edited: May 31, 2014
12. Jun 1, 2014

### BiGyElLoWhAt

Another way that deformation can be looked at is as lots and lots of force interactions, between pieces of the block and the bullet. Sections of the block are being compacted, bent, and moved out of place, and since we can't really account for all of these interactions, we say that energy is not conserved; in all actuality, it is, but you would have to keep track of EVERYTHING: the heating, and the sonic waves emitted, and exactly how the block was deformed.

This is simply too much, as we can't realistically know that much about our system. Fortunately, momentum doesn't care that there was deformation.