# Kinetic Energy Problem

1. Oct 28, 2008

### Bones

1. The problem statement, all variables and given/known data

An internal explosion breaks an object, initially at rest, into two pieces, one of which has 1.6 times the mass of the other. If 25000 J is released in the explosion, how much kinetic energy does each piece acquire?

2. Relevant equations

3. The attempt at a solution
I am not sure where to begin.

2. Oct 28, 2008

### LowlyPion

Here is a similar problem:

3. Oct 28, 2008

### Bones

Ok so m1=1.6m2
How do I figure out v1?

Last edited: Oct 28, 2008
4. Oct 29, 2008

### Bones

Would v1 be 1/1.6v2?

5. Oct 29, 2008

### LowlyPion

It depends on which piece is the heavier.

What must be true in this problem?

6. Oct 29, 2008

### Bones

The heavier piece will have less velocity than the lighter piece.

7. Oct 29, 2008

### Bones

25000 J = 2.6/1.6 * 1/2 * m2v22
So 25000J*.81=20250J for the lighter piece and 25000J-20250J=4750J for the heavier piece.
Is that correct??

8. Oct 29, 2008

### LowlyPion

Not quite.

They want the KE of the pieces. And the KE is mv2/2 not mv2

9. Oct 29, 2008

### Bones

I am not sure what I am doing...that was just kinda of a guess.

Last edited: Oct 29, 2008
10. Oct 29, 2008

### LowlyPion

So what does mv2/2 equal in your equation?

Isn't that the KE of m2?

Isn't that 1 part of what they asked?

11. Oct 29, 2008

### Bones

So 25000 J=2.6/1.6*1/2mv2v2^2 which is the same thing as 25000 J=2.6/1.6*mv2v2^2/2

25000J=0.8125m2v2

Am I on the right track?

Last edited: Oct 29, 2008
12. Oct 29, 2008

### LowlyPion

Yes but 25000 J = 2.6/1.6 * 1/2*m*v2 = 2.6/1.6 * KE2

Isn't the question "Find KE1 and KE2" ?

13. Oct 29, 2008

### Bones

Yes.
So 1.625*KE2=25000
KE2=15385J the lighter piece
KE1=9615J the heavier piece
?

14. Oct 29, 2008

### LowlyPion

That's what it looks like.

15. Oct 29, 2008

### Bones

Great! Thanks ;)