Explosion Question did I do this correctly?

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In summary, the question was about determining the speed and direction of motion of two chunks of 4.0 kg mass each after an internal explosion occurs in a body of mass 8.0kg traveling at 2.0 m/s. The conversation discusses different methods to solve this problem, with the final answer being that the chunks will have final velocities of v1' = 0 m/s and v2' = 4 m/s due to the law of conservation of energy and momentum.
  • #1
siemieniuk
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Hey all... I got an answer to this, but I'm not sure if I did it correctly.

q: A body of mass 8.0kg is traveling at 2.0 m/s under the influence of no external agency. At a certain instant an internal explosion occurs, splitting the body into two chunks of 4.0 kg mass each; 16 joules of translational kinetic energy are imparted to the two chunk system by the explosion. Neither chunk leaves the line of the original motion. Determine the speed and direction of motion of each of the chunks after the explosion.

a: I assumed the original block was just 2 4kg blocks with the same kinetic energy traveling in the same direction. (to the right)

Then, I subtracted 16J from the 8J of kinetic energy that the left most piece had because that's the way the 16J would be directed from the explosion.

When I put that into the calculation, I got 2.0 m/s (to the left) for the left block, and when I put 24 J (to the right) into the kinetic energy equation, I got the right block traveling 3.46 m/s (to the right).

Did I do this correctly??

Steve
 
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  • #2
siemieniuk said:
a: I assumed the original block was just 2 4kg blocks with the same kinetic energy traveling in the same direction. (to the right)

Then, I subtracted 16J from the 8J of kinetic energy that the left most piece had because that's the way the 16J would be directed from the explosion.

When I put that into the calculation, I got 2.0 m/s (to the left) for the left block, and when I put 24 J (to the right) into the kinetic energy equation, I got the right block traveling 3.46 m/s (to the right).

Did I do this correctly??
Sorry, but I don't understand what you did here at all. What you need to do is use the idea of conservation of momentum:

[tex]\mathbf{p}_{tot} = \mathbf{p}'_{tot}[/tex]

where, if x is an initial quantity, then x' is the value after the explosion.

[tex]\mathbf{p}_{tot} = \mathbf{p}'_{tot}[/tex]

[tex]8.0kg \times 2.0m/s [forward] = 4.0kg(v'_1 + v'_2)[/tex]

[tex]4.0 m/s [forward] = (v'_1 + v'_2)\ \dots \ (1)[/tex]

Note: [itex]v'_1[/itex] and [itex]v'_2[/itex] are the final velocities of chunk 1 and chunk 2, respectively (although it doesn't really matter which chunk we call chunk 1 or chunk 2).

Also, from the law of conservation of energy:

[tex]\frac{1/2}(4.0kg)[(v'_1)^2 + (v'_2)^2] = 16 J[/tex]

[tex](v'_1)^2 + (v'_2)^2 = 8m^2/s^2\ \dots \ (2)[/tex]

Now, solve for your velocities using equations (1) and (2). I would suspect you might get two possible answers, one that [itex]v'_1 = v'_2 = 2.0 m/s[/itex] meaning that the explosion has no effect, or another solution which is the one you'd be looking for. Note, you can and should use the law of conservation of momentum because there are no external forces acting on the system, I believe. If this is wrong, let me know.
 
  • #3
yea AKG i tried it and got 2m/s, therefore it had no effect, and they both had the same speeds.Btw how did you get 3.46m/s? Maybe you used another method...so either that or 2.82m/s of the sqrt of 8..
 
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  • #4
AKG said:
Also, from the law of conservation of energy:

[tex]\frac{1/2}(4.0kg)[(v'_1)^2 + (v'_2)^2] = 16 J[/tex]

... If this is wrong, let me know.

I think that should be :

[tex]\frac{1}{2}(4.0kg)[(v'_1)^2 + (v'_2)^2] = 16J + \frac {1}{2}(8.0kg)(2.0m/s)^2[/tex]

The way you solved it, the explosion adds no energy to the system, so the final velocities will be the same as the initial velocities.

However, you are actually getting an extra 16J from some form of chemical energy.

Answer v1' = 0 m/s and v2' = 4 m/s.
 
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  • #5
Gokul43201 said:
I think that should be :

[tex]\frac{1}{2}(4.0kg)[(v'_1)^2 + (v'_2)^2] = 16J + \frac {1}{2}(8.0kg)(2.0m/s)^2[/tex]

The way you solved it, the explosion adds no energy to the system, so the final velocities will be the same as the initial velocities.

However, you are actually getting an extra 16J from some form of chemical energy.

Answer v1' = 0 m/s and v2' = 4 m/s.
Yeah, that makes more sense.
 
  • #6
thanks a million for the help
 

1. How can I prevent an explosion?

Preventing an explosion involves understanding the factors that can lead to an explosion, such as flammable materials, oxygen, and a source of ignition. Proper storage and handling of materials, as well as implementing safety protocols and procedures, can help prevent explosions.

2. What causes an explosion?

An explosion occurs when a rapid increase in pressure and release of energy happens in a confined space. This can be caused by a chemical reaction, combustion, or a sudden release of compressed gas.

3. What should I do if an explosion occurs?

If an explosion occurs, it is important to stay calm and follow emergency procedures. This may include evacuating the area, seeking medical attention if necessary, and reporting the incident to the proper authorities.

4. What are the potential dangers of an explosion?

Explosions can cause significant damage to property and can also result in injuries or fatalities. They can also release toxic gases and particles into the air, which can be harmful to human health.

5. Can explosions be predicted?

While it is not always possible to predict an explosion, there are warning signs that can indicate a potential risk, such as unusual odors, leaks, or changes in temperature or pressure. Regular maintenance and monitoring can also help identify potential hazards before they lead to an explosion.

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