Kinetic Energy of many-particle systems

  • #1
Okay, so I'm at conundrum with my physics text book and I cannot seem to understand a few concepts. Allow me to elaborate.

1. Why is the total momentum of a firecracker sliding on ice the same before and after it explodes?
Shouldn't the explosion cause the momentum to increase since the explosion is really like force? I just don't see how this is possible.

2. Why does term Ʃ (m.sub i)(vector velocity of CM) dot (velocity of i relative to cm) = 0 in the total Kinetic energy for a system??

Some mathematical proofs and diagrams would be great!! I have provided a picture to show what I am talking about. I would really appreciate some explanations of this and this section. I am getting a little bit frustrated :( thank you.

P.S. This is for my University Physics course.
 

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  • #2
Simon Bridge
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Okay, so I'm at conundrum with my physics text book and I cannot seem to understand a few concepts. Allow me to elaborate.

1. Why is the total momentum of a firecracker sliding on ice the same before and after it explodes?
Shouldn't the explosion cause the momentum to increase since the explosion is really like force? I just don't see how this is possible.
You only get a change in momentum if the object is acted on by an external force. The explosive forces are all internal - every action has an equal and opposite reaction: so the gain in momentum of one bit has an equal and opposite change in momentum elsewhere on another part of the cracker.

Note: the passage starts "we have seen how..." ... having trouble understanding this part suggests you have not seen how - and you should go back over the previous chapter. It's because it's the momentum of the center of mass that counts.

2. Why does term Ʃ (m.sub i)(vector velocity of CM) dot (velocity of i relative to cm) = 0 in the total Kinetic energy for a system??

$$\sum m_i \vec{v}_{com} \cdot\vec{v}_{rel,i} = 0$$ ... because $$\sum m_i\vec{v}_{rel,i}=0$$ ... because that is the total momentum with respect to the center of mass ... which comes from the previous work on the center of mass and momentum. Recall above: it's the momentum of the center f mass that counts.
 
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  • #3
I understood completely. I was able to answer all the problems regarding momentum before this section. It was the firecracker that was throwing me off. I was thinking about it and I guess I did not grasp it because I did not think of the exploding particles as a whole system - I thought of them as individual systems. It did occur to me that perhaps momentum was conserved because of Newton's third law, like you have just said. Also, I found a proof and it matches what you are saying.
Thank you, Simon Bridge.
 
  • #4
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Have a look..
 

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  • #5
ehild
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1. There are forces acting on all parts of the firecracker, but these are internal forces. One part acts on the other and the other part acts on the first one with equal and opposite force.

Assume two parts:

Newton Second Law says that m1a1=F1, m2a2=F2. According to the Third Law, F2=-F1. If you add the two equations m1a1+m2a2=F1+F2=0

As a=dv/dt,

m1dv1/dt+m2dv2/dt=d/dt(m1v1+m2dv2)=0, the total momentum m1v1+m2v2 stays unchanged.
-----


2. As for the KE of a system of point masses, lets do it again for two particles.

The velocities are Vi=VCM+vi , vi is the relative velocity with respect to the CM.

The velocity of the CM is VCM= (m1V1+m2V2)/(m1+m2)

Writing it with the relative velocities,

(m1+m2)VCM=m1(VCM +v1)+m2(VCM+v2)=VCM(m1+m2)+(m1v1+m2v2)

The term (m1v1+m2v2) has to be zero.



ehild
 
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  • #6
Borek
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My bet is you are mistaking firecracker for a rocket. Rocket expels all the combustion products in one direction, firecracker explodes in all directions.

Do you see it now?
 
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  • #7
Thanks everyone I understand now. I appreciate everyone's help! :-)
 

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