How Much Kinetic Energy Does the Heavier Piece Acquire in an Explosion?

[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?

 

[Doc Al]

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.)

 

[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?

 

[maverick280857]

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?

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.

 

[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?

 

[maverick280857]

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?

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.

 

[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?

 

[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 that's because this is an ill-posed problem. You should still apply the correct equations though.

 

[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?

 

[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

 

[MD2000]

any help guys?

 

[Doc Al]

Wouldnt the second be the sum?

Written seperately..

8200 J = 1/2m1v1^2 + 1/2m2v2^2

This is correct.

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

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

 

[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 acquired 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:
$$8200=E_{1}+E_{2}=2.4E_{2}$$

 

[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

 

[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:
$$E_{1}=1.4E_{2}$$

 

[MD2000]

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

 

[arildno]

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

 

[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

 

[maverick280857]

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

 

[thiotimoline]

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?
QUOTE]

The problem is, is the velocity of 2 objects same or different? If it is the same then it is easier because the ratio of the 2 energies is related to the ratio of the 2 masses. Somehow I feel this question is incomplete in the first place.

If we assume both objects have equal final velocity, then:

0.5m(v^2) + 0.5(1.4m)(v^2) = 8200

 

[Doc Al]

If we assume both objects have equal final velocity, then:

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

 

[quantumdude]

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.