Explosion energy dissipation and Ek

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Homework Help Overview

The discussion revolves around the energy dissipation during an explosion and its relation to kinetic energy (Ek). Participants explore the conservation of momentum and energy in a system involving two masses before and after an explosive event.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of kinetic energy and the conservation of momentum, questioning how to relate momentum changes to energy dissipation. There are attempts to clarify the roles of different masses and velocities in the system.

Discussion Status

The conversation has progressed with participants providing guidance on the conservation laws applicable to the problem. There is a recognition of the need to calculate initial and final momenta and kinetic energies, with some participants confirming their calculations and seeking validation.

Contextual Notes

Participants are navigating through potential misunderstandings regarding mass values and the implications of momentum being a vector quantity. There is an emphasis on ensuring the correct total mass is used in calculations.

Maggie
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Homework Statement
A 9.6 kg mass slides on a frictionless one- dimensional track with a speed of 14.6 m/s. Suddenly it explodes so that afterwards a section of mass 0.2 kg is moving in the opposite direction with speed 2.4 m/s. Find the energy (in J) that released in the explosion.
Relevant Equations
Ek=0.5mv^2
Hello, I think the only energy involved is kinetic energy. So I subtract to get the difference between two stages to find the energy cost by explosion.
(0.5*9.6*14.6^2)-(0.5*0.2*2.4^2)=1022.59J
However the system said that this answer is wrong, So I wondered where did I make a mistake?
 
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Maggie said:
I think the only energy involved is kinetic energy
No. Something exploded.
What conservation laws do you know?
 
Last edited:
haruspex said:
No. Something exploded.
What conservation laws do you know!
I remembered energu conservation, momentum conservation, and probably charge (not relevant)... should I use momentum in this question?
 
Maggie said:
I remembered energu conservation, momentum conservation, and probably charge (not relevant)... should I use momentum in this question?
Yes.
 
haruspex said:
Yes.
but how? I calculated the initial p is 140.16 and final p is 0.48, the difference is 139.68, how can I transfer this answer into energy?
 
Maggie said:
but how? I calculated the initial p is 140.16 and final p is 0.48, the difference is 139.68, how can I transfer this answer into energy?
The mass has two parts and some explosive in between. To answer the question you need to assume the explosive matter has negligible mass. The system consisting of the two mass parts and the explosive is 'closed', i.e. no external forces acting on it (horizontally, that is). What do you know about the momentum of such a system?
 
haruspex said:
The mass has two parts and some explosive in between. To answer the question you need to assume the explosive matter has negligible mass. The system consisting of the two mass parts and the explosive is 'closed', i.e. no external forces acting on it (horizontally, that is). What do you know about the momentum of such a system?
I only know that the momentum should be conserved since no external force applied, so the difference between pi and pf should be the momentum of the explosion? I'm confused now...
 
Maggie said:
I only know that the momentum should be conserved since no external force applied, so the difference between pi and pf should be the momentum of the explosion? I'm confused now...
The explosion only shifts momentum between the two parts of the mass. Their total momentum won't change.
Find expressions for total momentum before and after.
 
Last edited:
haruspex said:
The explosion only shifts momentum between the two parts of the mass. Their total momentum won't change.
Find expressions for total momentum before and after.
is the momentum calculated by mv? then: momentum of the mass before= explosion momentum+momentum of the mass after? and initial momentum is 140.16 and final is 0.48, the difference is 139.68
 
  • #10
Maggie said:
is the momentum calculated by mv? then: momentum of the mass before= explosion momentum+momentum of the mass after? and initial momentum is 140.16 and final is 0.48, the difference is 139.68
The total momentum of the two masses before the explosion is equal to the total momentum of the two masses after the explosion. That's conservation of momentum.
 
  • #11
jbriggs444 said:
The total momentum of the two masses before the explosion is equal to the total momentum of the two masses after the explosion. That's conservation of momentum.
Do you mean that the momentum of the two masses before should be calculated by v(m1+m2)? which is 9.8*14.6?? But I don't understand why the initial momentum should include both masses...
 
  • #12
Maggie said:
Do you mean that the momentum of the two masses before should be calculated by v(m1+m2)? which is 9.8*14.6??
Yes, the momentum of the two pieces before should be calculated by ##mv## where m is the combined mass of the two pieces (9.6 kg) and v is the velocity of the combined pieces (14.6 m/s).
Maggie said:
But I don't understand why the initial momentum should include both masses...
It should include both pieces.
 
Last edited:
  • #13
Maggie said:
I don't understand why the initial momentum should include both masses...
We are discussing the momentum of the system consisting of the whole mass, whether as one piece or two. Initially the two pieces have the same velocity, so the momentum of the system is (m1+m2)v. After the explosion they have different velocities but the same total momentum.

You are given the velocity of one piece. Find the velocity of the other.
 
  • #14
haruspex said:
We are discussing the momentum of the system consisting of the whole mass, whether as one piece or two. Initially the two pieces have the same velocity, so the momentum of the system is (m1+m2)v. After the explosion they have different velocities but the same total momentum.

You are given the velocity of one piece. Find the velocity of the other.
Now I got it clearer, so the initial momentum is (9.6+0.2)*(14.6)=143.08, and the final momentum of m2 is 0.48, so the difference is 142.6, then 142.6/9.6=14.854m/s, this is the final v of m1. Then how can I deal with this final v?
 
  • #15
Maggie said:
Now I got it clearer, so the initial momentum is (9.6+0.2)*(14.6)=143.08, and the final momentum of m2 is 0.48, so the difference is 142.6, then 142.6/9.6=14.854m/s, this is the final v of m1. Then how can I deal with this final v?
Getting closer, but momentum is a vector. The direction matters. If we say the original velocity is +14.6 m/s then the initial momentum is +143.08 kgm/s (always state the units), and after the explosion the 0.2kg piece has velocity -2.4m/s and a momentum -0.48 km/s.
What does that give you for the momentum of the larger piece after the explosion?
 
  • #16
haruspex said:
Getting closer, but momentum is a vector. The direction matters. If we say the original velocity is +14.6 m/s then the initial momentum is +143.08 kgm/s (always state the units), and after the explosion the 0.2kg piece has velocity -2.4m/s and a momentum -0.48 km/s.
What does that give you for the momentum of the larger piece after the explosion?
ohhh I just knew that energy is scalar but forgot about momentum... okay. So the final momentum of m1 would be (143.08-(-)0.48)/9.6=+14.954m/s, right?
 
  • #17
Maggie said:
ohhh I just knew that energy is scalar but forgot about momentum... okay. So the final momentum of m1 would be (143.08-(-)0.48)/9.6=+14.954m/s, right?
Almost, but I just noticed you are using 9.8kg for the total mass and 9.6, 0.2 for the two pieces. The total mass is 9.6kg.
 
  • #18
haruspex said:
Almost, but I just noticed you are using 9.8kg for the total mass and 9.6, 0.2 for the two pieces. The total mass is 9.6kg.
do you mean that the m1+m2=9.6kg? So actually m1 is only 9.4kg? oh I misunderstood all the time... so the initial momentum is 9.6*14.6=+140.16kgm/s, then the final would be (140.16+0.48)/9.4=14.962m/s?
 
  • #19
Maggie said:
do you mean that the m1+m2=9.6kg? So actually m1 is only 9.4kg? oh I misunderstood all the time... so the initial momentum is 9.6*14.6=+140.16kgm/s, then the final would be (140.16+0.48)/9.4=14.962m/s?
Yes.
 
  • #20
haruspex said:
Yes.
okay now I understand, but once I got the final velocity of m1, what should I do next? using EK equation?
 
  • #21
Maggie said:
okay now I understand, but once I got the final velocity of m1, what should I do next? using EK equation?
Yes, you can now work out all the KE before and after the explosion.
 
  • #22
haruspex said:
Yes, you can now work out all the KE before and after the explosion.
okkkkay thanks. but just to confirm my calculation. the initial Ek will be 0.5*9.6*(14.6)^2=1023.168J, and the final Ek for m1 is 0.5*9.4*14.962^2=1052.149J, since energy is scalar, the direction doesn't matter, the final Ek for m2 is 0.576J, the total final Ek is 1052.725J. And the difference is 29.56J, is this the final answer?
 
  • #23
Maggie said:
okkkkay thanks. but just to confirm my calculation. the initial Ek will be 0.5*9.6*(14.6)^2=1023.168J, and the final Ek for m1 is 0.5*9.4*14.962^2=1052.149J, since energy is scalar, the direction doesn't matter, the final Ek for m2 is 0.576J, the total final Ek is 1052.725J. And the difference is 29.56J, is this the final answer?
I make it more like 29.5J, but yes.
 
  • #24
haruspex said:
I make it more like 29.5J, but yes.
thank you! so tricky this question
 
  • #25
Maggie said:
thank you! so tricky this question
If you ignore the numbers and just do it algebraically it is ##\frac 12m_2(u+v)^2\frac{m_1+m_2}{m_1}##, where u is the initial speed and v is the fragment's final speed (assuming they are in opposite directions).
 
  • #26
haruspex said:
If you ignore the numbers and just do it algebraically it is ##\frac 12m_2(u+v)^2\frac{m_1+m_2}{m_1}##, where u is the initial speed and v is the fragment's final speed (assuming they are in opposite directions).
Yes I got the same answer, but this equation is more abstract to me, I like that momentum method better:)
 

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