# Homework Help: Kinetic Energy Problem

1. Jun 17, 2006

### MD2000

An explosion breaks an object into two pieces, one of which has 1.4 times the mass of the other. If 8200 J were released in the explosion, how much kinetic energy did the heavier piece acquire?

What I was initially using was:
Momentumi = Momentumf..
0 = m1v1-1.4m2v2 (x-comp)
v1 = 1.4v2

The thing is I'm just not sure what to do next..am I on the right path?

2. Jun 17, 2006

### Staff: Mentor

You are on the right path. Now consider the total KE of the two pieces. (Assume that all the energy released in the explosion went into the KE of the pieces.)

3. Jun 17, 2006

### MD2000

hmm..so how does the following look?

8200 = 1/2 m1v1^2 + 1/2(1.4m1)(1.4v1)^2

The thing is we would still have two unknowns..I dunno..theres soemthing I'm obviously not doing right..or can we use KEi = KEf?

4. Jun 17, 2006

### maverick280857

You can't use $KE_{i} = KE_{f}$ here becuse this is essentially an inelastic collision (energy is not conserved) but what you are doing is $Q = KE_{f} - KE_{i}$ where Q is the explosion energy. Looks like you are missing something, the equations are correct.

5. Jun 17, 2006

### MD2000

Shouldn't Q just equal to KEf since there was no energy initially? What I don't understand is how to differentiate between the KE of the two different pieces. Won't the KEf - KEi be of the entire system?

6. Jun 17, 2006

### maverick280857

In your case $KE_{i} = 0$, but I have written it to make sure you don't use conservation of kinetic energy which is not even applicable here.

$KE$ refers to the kinetic energy of both masses (before the explosion, the kinetic energy of the combined mass is zero). The system consists of the combined unexploded mass and hence (after the explosion has taken place) two two resulting masses which are born at the instant the explosion takes place.

7. Jun 17, 2006

### MD2000

What I still am a little confused about is whether the two pieces will have different final KEs.

Q = KEf in this case..

8200J = 1/2m1v1^2 (block 1)
8200J = 1/2 (1.4m1)(1.4v1)^2 (block 2)

Am I looking at this right?

Or would it be

8200 = 1/2 (m1 + 1.4m1)(v1 + 1.4v1) ^ 2

Regardless..wouldn't we still be left with two unknowns?

8. Jun 17, 2006

### maverick280857

Q has to equal the sum of the kinetic energies of the two masses. You apply it to the system, not to the chunk.

You would still have two unknowns, but thats because this is an ill-posed problem. You should still apply the correct equations though.

9. Jun 17, 2006

### MD2000

Wouldnt the second be the sum?

Written seperately..

8200 J = 1/2m1v1^2 + 1/2m2v2^2
8200 J = 1/2m1v1^2 + 1/2(1.4m1)(1.4v2)^2

Wouldn't that be the sum of the two?

10. Jun 17, 2006

### MD2000

So according to that last eq I posted..

2190.17 = 1/2m1v1^2

So 8200 - 2190.17 = 6009.83 for the larger block? I dunno..I think I jus made that all up..lol

11. Jun 18, 2006

### MD2000

any help guys?

12. Jun 18, 2006

### Staff: Mentor

This is correct.

This is not. (You misapplied your result from the momentum equation.)

Better yet is to find the relationship between the two KEs:

KE1 = 1/2m1v1^2
KE2 = 1/2m2v2^2 = 1/2(1.4m1)(v1/1.4)^2 = (1/1.4) KE1

13. Jun 18, 2006

### arildno

Okay, let's sort this out:
Just PRIOR to the explosion, the object contains 8200J of CHEMICAL energy $E_{i.c}$that is going to be converted into KINETIC energy through the explosion. Thus, we have:
$$E_{i.c}=\frac{1}{2}(m_{1}\vec{v}_{1}^{2}+m_{2}\vec{v}_{2}^{2})$$
Furthermore, we have conservation of momrentum:
$$m_{1}\vec{v_{1}}=-m_{2}\vec{v}_{2}\to\vec{v}_{1}=-\frac{m_{2}}{m_{1}}\vec{v}_{2}$$

The RATIO of their aquired energies are therefore:
$$\frac{E_{1}}{E_{2}}=\frac{m_{1}}{m_{2}}\frac{\vec{v}_{1}^{2}}{\vec{v}_{2}^{2}}=\frac{m_{2}}{m_{1}}=1.4$$

Thus, we get:
$$8200=E_{1}+E_{2}=2.4E_{2}$$

14. Jun 18, 2006

### MD2000

Hmm..I'm a little confused as to why we do the ratio..and arildno..how did you get that final equation? Shouldn't it be 1.4E2 = 8200 which is 5857.14?

Or is it 2.4E2 = 8200 which is 3416.67..not exactly sure how u got the 2.4 tho

I really appreciate your help guys..srry for being an idiot..lol

15. Jun 18, 2006

### arildno

First off, we have by definition:
$$E_{1}=\frac{1}{2}m_{1}\vec{v}_{1}^{2},E_{2}=\frac{1}{2}m_{2}\vec{v}_{2}^{2}$$

Thus, we may rewrite our original equation as:
$$E_{i.c}=E_{1}+E_{2}$$
Get that?
As for "why doing the ratio?" remember that if we know:
$$\frac{E_{1}}{E_{2}}=1.4$$
then we must have:
$$E_{1}=1.4E_{2}$$

16. Jun 18, 2006

### MD2000

Actually Im srry I got that part..I wasnt sure about the 2.4E2 tho?

17. Jun 18, 2006

### arildno

Look at my edit, and figure that last bit out for yourself.

18. Jun 18, 2006

### MD2000

Ohhh..you know what I completely forgot the coefficient 1..which would make it 1.4 + 1..I got it now..

so..
2.4E2 = 8200
E2 = 3416.67

19. Jun 18, 2006

### maverick280857

EDIT: Oh I just saw Arildno said just what I came to say here :rofl:

20. Jul 10, 2006

### thiotimoline

21. Jul 10, 2006

### Staff: Mentor

Then you've missed the point of the problem, which is to realize that momentum is conserved in an explosion.

22. Jul 10, 2006

### Tom Mattson

Staff Emeritus
You cannot assume that the velocities are the same. As arildno pointed out you have to conserve both KE and momentum. That's how you get 2 equations for the 2 unknowns.