Please explanation will help me out

  • Thread starter Thread starter ISU20CpreE
  • Start date Start date
  • Tags Tags
    Explanation
Click For Summary

Homework Help Overview

The discussion revolves around the physics of an explosion, specifically analyzing it as an inelastic collision where momentum is conserved and kinetic energy is not. The original poster is attempting to determine how much kinetic energy each fragment acquires from a total energy release of 7800 J.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the conservation of momentum in the context of an explosion and question the original poster's understanding of the problem setup. There is a focus on the relationship between the fragments' velocities and the total energy released.

Discussion Status

Some participants have provided guidance on setting up equations based on conservation laws, while others have noted the need for the original poster to show more work before further assistance can be given. The discussion is ongoing with multiple interpretations being explored.

Contextual Notes

There is an emphasis on the original poster needing to clarify their goals and assumptions, particularly regarding the number of fragments and the initial conditions of the explosion.

ISU20CpreE
Messages
67
Reaction score
0
Ok so far I have been thinking that an explosion is nothing else than a inelastic collision. So energy is lost and momentum is conserved therefore [tex]P_i = P_f[/tex], The problem asks for how much kinetic energy did each piece acquire in the explosion, I also know that there was 7800 J released.

In order to make calculations easier I come up with:

[tex]m_a=1kg[/tex]

[tex]m_b=1.5kg[/tex]

[tex]E=7800J[/tex]

I can set up the problem but I can't see further steps I need a bit of explanation please.
 
Physics news on Phys.org
I'm not really sure what your question is, nor the use to which you want to put those numbers. Perhaps this will help:

Momentum conservation certainly applies in an explosion as it does everywhere else. If one imagines a bomb, say, sitting in space before the explosion occurs, with total momentum zero, then calculates the momentum of the individual pieces after the explosion, one could easily predict that the vector sum would also come to zero. I suppose one could also consider it an "inelastic collision", as it's certainly the case that kinetic energy is not conserved. "Collision" is probably not the best word to use, as it implies two things coming together, but I can't think of a better one at the moment.

In any event - can you tell me what your goal is? It would be easier to suggest further steps if I knew where you were going.
 
I think ISU's goal is to determine the kinetic energy of both fragments given their total energy and recognizing their total momentum is zero. Unfortunately, we can't help ISU until s/he shows some work!
 
Tide said:
I think ISU's goal is to determine the kinetic energy of both fragments given their total energy and recognizing their total momentum is zero. Unfortunately, we can't help ISU until s/he shows some work!


Reasonable. My work until now is:

[tex]0=m_1V_1f+1.5m_1V_2f[/tex] or [tex]V_1f= -1.5_V_2[/tex]

after that i don't have a clue. I am sorry but I will have to work on this in the morning I have a test in 4 hours I need to rest, thanks for the help. I will be asking further questions, sorry I can't stay.
 
Last edited:
OK - I'm going to assume that your explosion has left only two fragments, and you're trying to work out what happens. Yes?

So, using momentum conservation, you have a relationship between v1f and v2f. (Side note: Since the initial velocity was zero, why bother with the "f" subscript?) You need another relation to solve for them exactly.

You know that the total energy released was 7800J. Assume it all went into kinetic energy - you should, then, be able to write a second equation relating v1f and v2f. That will give you two equations in two unknowns, which you should be able to solve.

Does that do it?
 

Similar threads

Replies
55
Views
7K
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
24
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 31 ·
2
Replies
31
Views
5K
  • · Replies 7 ·
Replies
7
Views
4K