How is kinetic energy distributed in an explosion with unequal masses?

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In an explosion that splits an object into two pieces with unequal masses, one piece has 1.5 times the mass of the other, and 7300 J of energy is released. The discussion emphasizes using conservation laws to determine the kinetic energy distribution between the two pieces, with the total kinetic energy after the explosion equating to the released energy. The conservation of momentum is highlighted as a key principle to relate the velocities of the two masses. Participants clarify that the initial kinetic energy is zero in the center of mass frame, and the increase in kinetic energy is equal to the energy released. The problem-solving approach involves expressing the kinetic energy of one mass in terms of the other to find their respective energies.
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Homework Statement


An explosion breaks an object into two pieces, one of which has 1.5 times the mass of the other. If 7300 J were released in the explosion, how much kinetic energy did each piece acquire?

Homework Equations


Po = P

The Attempt at a Solution


I don't even know how to start this...
 
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Use a coordinate system where the center of mass is at rest. Do you know the formula for kinetic energy? There's a conservation law you can use to determine what you should put into that formula. I don't want to tell you any more, because that would make it too easy.
 
Well the only law I know of for Kinetic Energy would be KE=1/2 m(v^2) and that Initial KE = final KE...but I don't know how to use either of those in this situation because I have nothing for velocity, and I'm not even sure if that is relevant.
 
afjunkie said:

Homework Statement


An explosion breaks an object into two pieces, one of which has 1.5 times the mass of the other.

Forget the rest of the question for the moment. Did you draw a before/after picture and label it with everything you know about the two pieces? Does it remind you of anything you've seen before?
 
Yes I did, but it doesn't remind me of anything...all I know about them is that one has a mass of 1.5 m, and the other's mass is m. I guess you could say it looks sort of like a Free Body Diagram?
 
OK, now try imagining time is "in reverse", like you're rewinding a video tape. So the "after" picture comes first, then the "before" picture. What does the video tape look like?

Does that remind you of a type of problem you've solved before?
 
afjunkie said:
Well the only law I know of for Kinetic Energy would be KE=1/2 m(v^2) and that Initial KE = final KE...but I don't know how to use either of those in this situation because I have nothing for velocity, and I'm not even sure if that is relevant.
Energy is not conserved here. The problem statement tells you how much the energy will increase. There is however something else that's conserved here, and that conservation law will tell you the velocity.
 
Did you figure out what other conservation law I was talking about? (It won't tell you the velocities separately. It will just give you one as a function of the other). Did you realize that K_1+K_2=7800?
 
No, i did not realize that. How did you get that?
 
  • #10
Now that I think about it, the specification doesn't say that the 7800 J is the amount of energy that was converted into kinetic energy. Some of it could be heat, or kinetic energy of gas particles that are leftovers from the explosion. But if it is, we can't solve the problem. So I can only assume that the 7800 J is just the energy that was converted to kinetic energy.

In the center of mass frame, the total kinetic energy before the explosion is 0. We interpret the problem specification as saying that the total kinetic energy increases by E=7800 J. That tells us that

E=\frac 1 2 m_1v_1^2+\frac 1 2 m_2v_2^2.

What you have to do is to use a conservation law to eliminate m_2 and v_2 from the equation (express the right-hand side as a function of m_1 and v_1 only).
 
  • #11
Alright, I think I see what you're saying...but the number 7800 seems to have come out of nowhere. Either you mean 7300N because that's what the problem says, or I will need help on how you found that number...
 
  • #12
Oops...you're right. I read the number wrong.
 
  • #13
Ok, no big, just making sure that I didn't miss something.
 

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