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A simple maybe 8th or 7th grader question, but I can't really get over it. Elastic collision: momentum is conseved and so is the kinetic energy whereas in an inelastic collision, momentum is conserved but not the energy. Although I don't get it why is momentum of a closed and isolated system is supposed to be constant/conserved just accounting to the fact that system is closed (no mass enters or leave it) and isolated (no net external force) but, ok, let me accept this for a minute and think about how is it possible that momentum is conserved but the kinetic energy of system changes as happening in an inelastic collision.

I mean since both quanties depend on velocities of object initially and finally, both of them should act alike. If one changes, other does too. If one doesn't, other doesn't as well.

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marked as duplicate by stafusa, Chris, Kyle Kanos, Emilio Pisanty, Community Feb 12 '18 at 15:57

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Energy is conserved. It just transforms from one form to the other. $\endgroup$ – LRDPRDX Feb 11 '18 at 11:20
  • $\begingroup$ But we are and shall keep talking about the conservation of KINETIC energy not Energy in general. Only kinetic energy is concerned here @BogdanSikach $\endgroup$ – Anonymous Feb 11 '18 at 11:30
  • $\begingroup$ Possible duplicate of Difference between momentum and kinetic energy $\endgroup$ – stafusa Feb 11 '18 at 11:33
  • $\begingroup$ I just read the question and its answers from your link, @stafusa it didn't help me and Idk how can I be more clear with my query so I am prolly not going to edit it, kindly reconsider your flagging. $\endgroup$ – Anonymous Feb 11 '18 at 11:41
  • $\begingroup$ @Anonymous, From the equations that define both quantities it's straightforward that it's not true that, as you put, "If one changes, other does too.". A numerical example from the linked answer: momentum $= mv$, kinetic energy $=mv^2/2$, right? OK, so increase the mass $100$ times and divide the velocity by $100$. Same momentum, right? $100mv/100=mv$ But what does it do to the kinetic energy? $[100m(v/100)^2]/2=[mv^2/100]/2$. So by increasing the mass $100$ times, for the same momentum, we've cut the kinetic energy by a factor of $100$. $\endgroup$ – stafusa Feb 11 '18 at 11:48
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There is general conservation law - conservation of total energy and momentum. Imagine that two balls of the same mass move towards each other with the same velocity (in opposite directions) and then collide absolutely inelastic so at the end we have one entire body at rest. So momentum is conserved (and equal to zero). It is not in a contradiction with the fact that kinetic energy is not conserved because there are other forms of energy here.

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  • $\begingroup$ It supposed to be a comment . Cannot delete it now. $\endgroup$ – LRDPRDX Feb 11 '18 at 11:54
  • $\begingroup$ ''So momentum is conserved (and equal to zero)'' But system's momentum doesn't become zero just because of the fact that one body after collision stops dead? What about the monentum of other one? $\endgroup$ – Anonymous Feb 11 '18 at 11:56
  • $\begingroup$ I do not understand. Both of the balls stopped. Total momentum of the system is zero. $\endgroup$ – LRDPRDX Feb 11 '18 at 12:04
  • $\begingroup$ I got. My bad. I will fix the formulation. $\endgroup$ – LRDPRDX Feb 11 '18 at 12:06
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This is not a complete answer to your question, but it may give you some insight… Consider a head-on collision, observed in the laboratory frame of reference, between two spheres of equal mass. There will be another frame, known as the centre of mass (CM) frame, in which the same collision is seen as the spheres approaching each other with equal and opposite velocities.

For example suppose the spheres have lab-frame initial velocities in the x-direction of $u_1$ and $u_2$. The CM frame is the frame moving in the x-direction with velocity $\frac{u_{1}+u_{2}}{2}$. In this frame the spheres will have initial velocities $\frac{(u_{1}-u_{2})}{2}$ and $\frac{(u_{2}-u_{1})}{2}$ [Galilean transform].

Now, whether the collision is elastic or inelastic, it is a matter of symmetry that in the CM frame the spheres' velocities after the collision will be equal and opposite. Call them $v_{CM}$ and $-v_{CM}$. If the collision is inelastic these velocities will be of smaller magnitude than the initial velocities in the CM frame. In the laboratory frame these velocities will be $v_{CM} +\frac{u_{1}+u_{2}}{2}$ and $-v_{CM} +\frac{u_{1}+u_{2}}{2}$. So the vector sum of 'final' velocities in the lab frame is $u_{1}+u_{2}$, that is momentum is conserved!

Note that we've achieved this result by using little more than symmetry and the Galilean velocity transform between reference frames. Clearly things aren't quite as simple for bodies of different mass, but the message is the same: conservation of momentum can be regarded as a consequence of a spatial symmetry.

We note that the argument depended on the vector nature of momentum (or of velocity). The same argument could not be used for the scalar, kinetic energy.

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  • $\begingroup$ I went thru some similar stuff as yours in a book and I have a question. When we say a reference frame is fixed w.r.t to a moving point/particle do we mean that frame is attached to the particle i.e. it moves with particle OR frame is fixed and particle is moving w.r.t this frame? $\endgroup$ – Anonymous Feb 11 '18 at 18:54
  • $\begingroup$ We'd mean that the frame of reference is moving with the same velocity as the particle. Note that the frame of reference I'm using isn't fixed to a particle but moves with the centre of mass of the two particles. [It is therefore an inertial frame: it suffers no acceleration even when the particles collide.] $\endgroup$ – Philip Wood Feb 11 '18 at 19:22
  • $\begingroup$ Nah I was talking about a general case when I said particle but ok yeah I got it. Can't vote your post tho cuz it requires minimum 15 reputation but ty :) $\endgroup$ – Anonymous Feb 11 '18 at 19:43
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Since both quanties depend on velocities of object initially and finally, both of them should act alike

Kinetic Energy has a v squared term in it so you simply consider the magnitude.
Momentum has only a v term in it so you consider both the direction and magnitude.

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