# Conservation of momentum/mass when dropping an object on a spring.

1. Dec 5, 2011

### JustinLiang

1. The problem statement, all variables and given/known data
A spring of negligible mass and force constant k=400 is hung vertically and a 0.2kg pan is suspended from its lower end. A butcher drops a 2.2kg steak onto the pan from a height of 0.4m. The steak makes a totally inelastic collision with the pan and sets the system into vertical SHM. What is the speed of the pan and steak immediately after the collision?

2. Relevant equations
PE=mgh
KE=mv^2/2
mv=mv
E(total energy in spring)=kA^2/2

3. The attempt at a solution
First I found out the energy produced from dropping the steak 0.4m:
PE=mgh
=2.2kg(9.8)(0.4m)
=8.624J

Then I used this to find the speed of the steak before the collision:
PE=KE
8.624J=(2.2kg)(v^2)/2
v=2.8m/s

When the steak goes into the pan by conservation of momentum we have:
mv=mv
(2.2kg)(2.8m/s)=(2.4kg)(v)
v=2.56666666m/s

My problem: However if I plug 2.566666m/s back into the KE equation I do not get 8.624J, instead I get 7.095J which means that the energy is not conserved? Do you guys know why? If I have 7.095J as the energy and I plug it into the E(total energy in spring) equation I do not get the right amplitude, instead plugging in 8.624J will get me the right answer.

HERE IS A SIMPLER PROBLEM
I found the same problem when solving for the final velocity of a block m=0.992kg when a bullet with v=280m/s and m=0.008kg strikes it and embeds itself into it.

By conservation of momentum:
(0.008)(280)=(1kg)(v)
v=2.24m/s

Where initially the KE in the bullet initially is
KE=0.008(280)^2/2=313.6J

But the KE after the collission is
KE=1kg(2.25m/s)^2/2=2.5088J

So clearly the energy is not conserved??? I am confused.

Last edited: Dec 5, 2011
2. Dec 5, 2011

### BruceW

You have done it correctly, and it is the correct answer. So you are right that energy is not conserved during the collision. That's why the collision is called inelastic.

3. Dec 5, 2011

### JHamm

Any energy 'lost' has just been converted in to heat and sound during the collision :)

4. Dec 5, 2011

### JustinLiang

If that's the case shouldn't 7J get me the correct amplitude? The answer suggests I use 8.6J, but this energy is not conserved?

5. Dec 5, 2011

### BruceW

7J should get you the right amplitude. Maybe the answer book is incorrect?

6. Dec 5, 2011

### BruceW

hmm, actually maybe the lost KE is turned into PE in the spring (so that although the collision is fully inelastic, energy is conserved).

7. Dec 5, 2011

### JustinLiang

Yeah it is totally inelastic so all the energy would be conserved. But how do you explain the differences in the energy before collision 8.624J and after collision 7.905J (I initially typed out the wrong values).

8. Dec 5, 2011

### JustinLiang

I just tried a different calculation:

If all the energy is conserved then:
PE=KE(with both masses)
8.624J=(2.4kg)(v^2)/2
v=2.68m/s

This answer tells us all the PE becomes KE in the combined mass. However the answer key has a v=2.6m/s so I don't think this is right. Maybe the answer key is wrong...

9. Dec 5, 2011

### BruceW

I'm guessing the answer key is wrong. It does happen on occasion.

10. Dec 5, 2011

Okay thanks!