Calculating the collusion and loss of energy?

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In summary: So I'm going the right track?In summary, the conversation discusses a problem involving the momentum of a full shell and its two parts after an explosion. The first step is to use the conservation of momentum equation and the second step is to find the sum of the two masses after the explosion. The conversation also clarifies that energy is not conserved in this problem and that additional energy is released as part of the explosion. To find the kinetic energy, one needs to subtract the kinetic energy of the shell from the sum of the kinetic energies of the two parts. The conversation also mentions that explosions tend to increase kinetic energy and suggests finding the energy released by the explosion by subtracting the kinetic energy of the shell from the sum of the kinetic energies
  • #1
Matriculator
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Homework Statement


Homework Equations


m1v1i+m2v2i=m1v1f+m2v2f

The Attempt at a Solution


I tried doing MV=m1v1f+m2v2f

I know that the sum of the momentum of the two parts is equal to the the momentum of the the full shell. Since I have two unknowns, I used the conservation of energy. I set used KE=KE1+KE2. The kinetic of the full shell equals to the sum of the kinetic energy of the two parts. Since I have two unknowns here too, I used a system of equations solve for v. Am I on the right path? I found two "ms" didn't add up to 5kg. Thank you in advance.
 

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  • #2
It would be helpful if you stated the problem - preferably in its original version. If it's not in English, someone may be able to translate it accurately. Supplementally, give your own translation as best you can.
 
  • #3
Matriculator said:

Homework Statement





Homework Equations


m1v1i+m2v2i=m1v1f+m2v2f


The Attempt at a Solution


I tried doing MV=m1v1f+m2v2f

I know that the sum of the momentum of the two parts is equal to the the momentum of the the full shell. Since I have two unknowns, I used the conservation of energy.
Don't use conservation of energy. Energy isn't conserved in this problem.

Yes, you will come back to that later when you find the difference in kinetic energies such that you can calculate the energy released. But that comes later.

So go back to your first equation regarding conservation of momentum. That's one of the two equations you will need.

The second equation that you'll need is actually quite simple. What is the sum of the two masses after the explosion? [Hint: it's the same as the shell's mass before the explosion.] Put that in equation form. :wink:

I set used KE=KE1+KE2. The kinetic of the full shell equals to the sum of the kinetic energy of the two parts.
That's not true for this problem. Additional energy is released as part of the explosion.

But what you can do is find the kinetic energy of the shell before the explosion and compare that to the sum of kinetic energies of the two parts.

I found two "ms" didn't add up to 5kg. Thank you in advance.
According to your attachment, the two masses should add up to 4.5 kg. (not 5 kg).
 
  • #4
collinsmark said:
Don't use conservation of energy. Energy isn't conserved in this problem.

Yes, you will come back to that later when you find the difference in kinetic energies such that you can calculate the energy released. But that comes later.

O-oh.. ohhh. Wow I feel slow. Thank you very. So it's 410m1+215m2=1440 and m1+m2=5. I understand this part now(although I'm 85% sure on the former's signage, I'm still loose on signage in physics).

After this to find kinetic energy, I'm thinking of subtracting the kinetic energy of the full shell to the sum of the kinetic energies of the two parts using the new found masses, is that correct? Basically ΔKE=KEShell-(KE1+KE2).
 
  • #5
Matriculator said:
O-oh.. ohhh. Wow I feel slow. Thank you very. So it's 410m1+215m2=1440
That looks right to me :smile:

(Except for the lack of units. But perhaps I'm being nit-picky.)

and m1+m2=5.
I think you mean 4.5 kg.

I understand this part now(although I'm 85% sure on the former's signage, I'm still loose on signage in physics).
Everything is going in the same direction for this problem, so the signs look good to me the way you have them.

After this to find kinetic energy, I'm thinking of subtracting the kinetic energy of the full shell to the sum of the kinetic energies of the two parts using the new found masses, is that correct? Basically ΔKE=KEShell-(KE1+KE2).

Almost right. You can usually expect the energy of the explosion to cause the kinetic energy of the subsequent pieces to be greater overall, compared to the original kinetic energy. And that holds true for this problem too. So what does that tell you about what you need to subtract from the other?
 
  • #6
collinsmark said:
That looks right to me :smile:

(Except for the lack of units. But perhaps I'm being nit-picky.)

Almost right. You can usually expect the energy of the explosion to cause the kinetic energy of the subsequent pieces to be greater overall, compared to the original kinetic energy. And that holds true for this problem too. So what does that tell you about what you need to subtract from the other?

Oh sorry about the unit thing. So I have it backwards? It should be the sum of the potential energy of the two parts minus the kinetic energy of the shell. Could I think of it as the absolute value of the change in kinetic energy, although there's no such convention? Thank you again.
 
  • #7
Matriculator said:
Oh sorry about the unit thing. So I have it backwards? It should be the sum of the potential energy of the two parts minus the kinetic energy of the shell. Could I think of it as the absolute value of the change in kinetic energy, although there's no such convention? Thank you again.

I don't think potential energy plays a role in this problem.

What I meant in my last post is that explosions tend to increase the kinetic energy of the things around them. So if you want to find the energy released by the explosion, and assuming all of the energy released by the explosion went into increasing the kinetic energy of the two pieces (in addition to what they already had before the explosion, back when the two pieces were a single shell), then... :wink:
 
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  • #8
collinsmark said:
I don't think potential energy plays a role in this problem.

What I meant in my last post is that explosions tend to increase the kinetic energy of the things around them. So if you want to find the energy released by the explosion, and assuming all of the energy released by the explosion went into increasing the kinetic energy of the two pieces (in addition to what they already had before the explosion, back when the two pieces were a single shell), then... :wink:

Sorry I miswrote that, I meant Kinetic energy. As in ƩKEParts-KEShell. Thank you for everything by the way.
 
  • #9
Matriculator said:
I meant Kinetic energy. As in ƩKEParts-KEShell.
That looks like the correct idea to me. :approve:
 

Related to Calculating the collusion and loss of energy?

1. How do you calculate the collusion and loss of energy?

The collision and loss of energy can be calculated using the formula: E = 1/2mv2, where E is the energy, m is the mass of the colliding objects, and v is the velocity at which they collide. This formula takes into account the kinetic energy of the objects before and after the collision.

2. What factors affect the collusion and loss of energy?

The factors that affect the collision and loss of energy include the masses of the objects, the velocity at which they collide, and the types of materials involved. Inelastic collisions, where some energy is lost in the form of heat or sound, also result in a greater loss of energy compared to elastic collisions.

3. How does the angle of collision affect the collusion and loss of energy?

The angle of collision can affect the collusion and loss of energy in two ways. First, it can affect the direction and magnitude of the resulting forces, which can impact the amount of energy transferred. Second, it can also affect the coefficient of restitution, which is a measure of how much energy is lost in a collision. A glancing collision, for example, may result in a lower coefficient of restitution and a greater loss of energy.

4. Can the collusion and loss of energy be calculated for all types of collisions?

While the formula for calculating the collision and loss of energy is applicable to most types of collisions, it may not accurately account for all factors in certain scenarios. For example, in collisions involving deformable objects, additional factors such as the elasticity and deformation of the objects may need to be considered in the calculation.

5. How is the collision and loss of energy used in practical applications?

The calculation of collision and loss of energy is used in various practical applications, such as designing safety features for cars and other vehicles. It is also used in analyzing the efficiency of energy transfer in mechanical systems, such as in the design of roller coasters or other machines. In sports, understanding the collision and loss of energy can help improve equipment and techniques to reduce the risk of injury.

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