# Homework Help: Conserv. of Momentum and Energy Question

1. May 18, 2006

### ksle82

I was able to get the answer to Conservation of Momentum but couln't get the same answer using conservation of energy? Steps are shown below. Where did i go wrong using the conserv. of engery method? Assume frictionless surface and inelastic collision.

Q: A 10 gram bullet with a velocity of 1000 m/sec strikes a 100 gram block of wood initially at rest. What is their combined velocity? Can you work the problem using the principle of Conservation of Momentum? Conservation of Energy?

using Momentum:
m1v1+m2v2= (m1+m2)vf, v2=0, m1=mass of bullet, m2=mass of block
v1=velocity of bullet, v2=velocity of block
vf=combined velocity
so vf= (m1v1)/(m1+m2) = (10*1000)/(110)~91 m/s

Using Cons. of Energy:
KE1 + PE1 = KE2 + PE2, KE=kinetic energy, PE=potential E
(1/2)m1(v1)^2 = (1/2)(m1+v2)vf PE1=PE2=0
-if solved for vf, the aswer would be different from the momentum method

2. May 18, 2006

### nrqed

It is simply that conservation of mechanical energy *does not work* in collisions. The problem is that some energy always goes into heat or to deform the objects (*unless* you are told that the collision is perfectly elastic). So you simply cannot use the equation you used because those effects are not included. In *collisions* you can only use conservation of momentum, in general. And conservation of energy in only certain special cases.

3. May 19, 2006

### maverick280857

In fact if you bring in coefficient of restitution, then energy conservation can be considered to be a special case of momentum conservation when the collision is elastic (except if there are other sources or sinks of energy like springs or heat). If no external force acts on a system, then its momentum is conserved in the direction of zero external force.

4. May 21, 2006

### willydavidjr

There are really factors affecting on conservation of energy because some goes off on this factor. Maybe they are absorbed by friction or something else. There must be an exception on the problem.

5. May 27, 2006

### Dr.Brain

In your question , the collision is not completely elastic , the two boxes after contact stick to each other and act as one mass , so at the instant the box collide , there might be some loss of energy ..... but if no force acts in horizontal direction on the two-box system , yo can always conserve momentum .... for conservation of energy , you also need to take in account loss of energy due to heat/sound and other factors.

6. May 27, 2006

### willydavidjr

But my question Dr. Brain is how can I compute for the ratio of the energy lost from the initial energy?

7. May 27, 2006

### Hootenanny

Staff Emeritus
You know the inital velocities and hence the intial kinetic energies. You can calculate the final velocity using conservation of momentum. Use this velocity (of the bullet and the block - remember to sum their masses) to calculate the final kinetic energy. The difference is the energy lost.

Do you follow?

~H

8. May 29, 2006

### willydavidjr

You mean I need to sum up the two initial kinetic energy and the two final kinetic energy and get their difference?

9. May 29, 2006

### Hootenanny

Staff Emeritus
Spot on.

~H