conservation of momentum vs KE


by Oerg
Tags: conservation, momentum
Oerg
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#1
Aug9-07, 12:41 PM
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1. The problem statement, all variables and given/known data

there is this requirement in my textbook that states:

show an understanding that, whilst momentum is always coneserved in interactions between bodies, some change in kinetic energy usually takes place.

Can anyone explain to me why this is so? If the momentum of a system is conserved before and after an interaction, why is the kinetic energy not usually conserved?


2. Relevant equations
m1u1+m2u2=m2v2+m1v2
K.E.before=K.E.after


3. The attempt at a solution
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Niles
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#2
Aug9-07, 12:55 PM
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I think they are asking for the Q-value, when talking about particle physics; the difference between the mass of a decaying particle and the sum of masses of the daughter particles.
malawi_glenn
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#3
Aug9-07, 12:59 PM
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Total energy is conserved, the kinetic energy can be converted into another kind of energy; potential energy, heat etc.

Oerg
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#4
Aug9-07, 01:02 PM
P: 363

conservation of momentum vs KE


to the second poster: Hmm maybe.. but is there another explanation not involving particle physics?

to the third poster: if kinetic energy has been converted to something else, would not the velocity of the objects be lower than that if kinetic energy is conserved? That would mean a decrease in total momentum too is it not?
malawi_glenn
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#5
Aug9-07, 01:10 PM
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No... the velocity can be distributed different among the objects involved in the collision. Do you own a textbook on newtonian mechanics? I suggest you read a bit in that, or maybe google a bit.

In kinematics, you put up these two conditions:

1. Total linear momentum initial = Total linear momentum final
2. Total energy inital = total energy final

Sometimes you even have a third condtion, total Anlgular momentum, which is also conserved.
Oerg
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#6
Aug9-07, 01:52 PM
P: 363
oh yes thanks for explaining that means although a typical question might ask about 2 objects in interaction, and if they say that kinetic energy is not conserved, that means that we have to include additional objects into the discussion eg. sound waves propagating in air, etc. i understand now thanks a lot. Just that I was assuming that they were only considering the 2 objects before and after and that it didnt make sense if the 2 objects were to lose kinetic energy and still have the same total momentum.
learningphysics
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#7
Aug9-07, 02:19 PM
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Quote Quote by Oerg View Post
oh yes thanks for explaining that means although a typical question might ask about 2 objects in interaction, and if they say that kinetic energy is not conserved, that means that we have to include additional objects into the discussion eg. sound waves propagating in air, etc. i understand now thanks a lot. Just that I was assuming that they were only considering the 2 objects before and after and that it didnt make sense if the 2 objects were to lose kinetic energy and still have the same total momentum.
You don't have to consider additional objects. Even when only considering two objects (no other objects), kinetic energy is not necessarily conserved. Kinetic energy can be converted into heat within the two objects... or some other form of energy within the two objects.

For a closed system: Total energy is conserved and momentum is conserved.

Kinetic energy is not necessarily conserved.
malawi_glenn
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#8
Aug9-07, 02:23 PM
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Also consider a person standing on the ice on skates and trowing an object. Beacause of conservation of linear momentum, the person vill move in the opposite direction of the object she/he trowed. But kinetic energy is a scalar, not a vector, so we can not speak of "kinetic energy in direction y" etc. In this situation, total energy is conserved (chemical energy stored in the persons body is transformed into mechanical energy)
Oerg
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#9
Aug9-07, 09:37 PM
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Then is it true to say that in a perfectly inelastic collision, total momentum is not conserved since all kinetic energy is converted to other forms of energy? If the body possesses no kinetic energy, it wouldnt be moving and the RHS of the moementum equation would be zero compared to the LHS value unless we were to consider air particles that might rocket off during the collision.
lightgrav
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#10
Aug9-07, 10:37 PM
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If the (VECTOR, including direction!)
momentum after a totally INelastic collision is zero,
then the momentum sum (+/- !) before the collision must've been zero.
That is, they initially were travelling in opposite directions, and stopped.

By the way, "totally inelastic" does NOT mean "stops" ...
it means "stick together" , as in "v_1,final = v_2,final " .

You might try conserving momentum with v1=10m/s and v2=0,
(both m=2kg) to see what fraction of original KE is lost.

Then try it with v1=15m/s and v2=5m/s ... "as seen by the bumblebee"
learningphysics
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#11
Aug9-07, 10:40 PM
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Quote Quote by Oerg View Post
Then is it true to say that in a perfectly inelastic collision, total momentum is not conserved since all kinetic energy is converted to other forms of energy? If the body possesses no kinetic energy, it wouldnt be moving and the RHS of the moementum equation would be zero compared to the LHS value unless we were to consider air particles that might rocket off during the collision.
No that's not true.

Total momentum is always conserved in a closed system... so suppose there is a collision between two bodies... and after the collision, neither body has any momentum... the total momentum is 0... that means that the total momentum before the collision was 0 also. That doesn't necessarily mean that the particles weren't moving before the collision... for example two particles of equal mass moving with the same speed in opposite directions, together have a total momentum of 0. Remember that momentum is a vector (magnitude and direction), and it is the total momentum as a vector that is conserved.
Oerg
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#12
Aug9-07, 11:03 PM
P: 363
ok i get it, thanks for all who have posted, i appreciate your help!


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