In school, I usually had good grades in physics, but there is one thing I never fully understood in classic mechanics (thankfully, I never faced such a question in exams - all we had to do was applying the formulas in specific calculations, which I did) - how can both momentum and kinetic energy be preserved in an elastic collission at the same time?
Consider two bodies of masses m[sub]1[/sub] and m[sub]2[/sub] colliding at velocities of v[sub]1[/sub] and v[sub]2[/sub]. The principle of conservation of momentum demands that
But how can term 1 and 2 be constant at the same time? I mean, (1) is a linear function while (2) is a quadratic one. One would assume that with any change of velocities, kinetic energy (term 2) would grow (or decrease) faster than momentum (term 1). If you place the two functions in the same coordinate system, you’ll see that they intersect at two points (at most), which, I suppose, shows that (1) and (2) cannot be constant for every possible set of values. Yet apparently they are.
I previewed the post itself, but apparently didn’t care about the thread title. Of course I am asking about conflicts between conservation of energy and conservation of momentum.
I’m a little confused by your comments about “plotting the two functions in a coordinate system”; if you’re doing it the way I think you’re doing it, it’s actually a nice way of seeing the two solutions (corresponding to “before” and “after”).
Suppose you plot all the points with m[sub]1[/sub]v[sub]1[/sub][sup]2[/sup] + m[sub]2[/sub]v[sub]2[/sub][sup]2[/sup] = 2 E, with v_1 along the x=axis and v_2 along the y-axis; this curve will be an ellipse, and it represents all the configurations of the system with a fixed energy E. Similarly, the set of all points with m[sub]1[/sub]v[sub]1[/sub] + m[sub]2[/sub]v[sub]2[/sub] = P (the set of all points with a given momentum) will be a straight line in the plane. These two curves will, as you mention, intersect at two points; one of these points will correspond to the values of v[sub]1[/sub] and v[sub]2[/sub] before the collision, and the other point will correspond to the values of v[sub]1[/sub] and v[sub]2[/sub] after the collision.
As a simple example, suppose m[sub]1[/sub] = m[sub]2[/sub] = 1, E = 1, and P = 0. (Pretend I’m using MKS units if you need to attach more physical meaning to this.) If you graph the two curves corresponding to this, you find that the two points where the curves intersect are (v[sub]1[/sub], v[sub]2[/sub]) = (1,-1) and (v[sub]1[/sub], v[sub]2[/sub]) = (-1,1). So if the first configuration is “before” and the second one is “after”, this would correspond to two equal masses coming together with equal velocity, colliding, and then heading back out after the collision with their velocities exactly reversed.
Finally: if you’re concerned about the solutions jumping between the two points discontinuously, and therefore that the energy and momentum can’t be conserved at all times, you’re right: during the course of the collision, a certain amount of energy will be stored as an elastic deformation of the masses. So during the collsion, you don’t have 1/2 (m[sub]1[/sub]v[sub]1[/sub][sup]2[/sup] + m[sub]2[/sub]v[sub]2[/sub][sup]2[/sup]) = E, but rather 1/2 (m[sub]1[/sub]v[sub]1[/sub][sup]2[/sup] + m[sub]2[/sub]v[sub]2[/sub][sup]2[/sup]) + E[sub]elastic[/sub] = E.
You have to remember that in (2) there is another term “E” that is a collection of all the energy that isn’t in the Kinetic energy of the two bodies, i.e. heat, sound, deformation etc. In a perfectly elastic collision that term is assumed to be zero. In non-elastic equations, this E term changes to keep the total energy constant regardless of what two velocities you select.
If the collision is elastic, all the energy remains kinetic. Think of a simple case where the two objects are mounted on frictionless rails so they can only interact along one dimension. There are two kinetic energies and two momenta, or four numbers, that exist before the collision. There are also the same four numbers (though probably each having new values) after the collision. So, there are four degrees of freedom for the new numbers, and four constraints posed by the old ones. Clearly, one way of solving the problem that would have to work would be adjusting each of the four final numbers to minimize the errors in your equations; you could make the errors vanishingly small by adjusting and adjusting again and again. Of course, you could also find an exact answer algebraically, but for me the insight into why that would work is more obscure.
The other answers to your OP seem overly complex to me. Since m[sub]1[/sub] & m[sub]2[/sub] don’t change and v[sub]1[/sub] & v[sub]2[/sub] are known quantities (before the collision), your unknowns are v[sub]1[/sub] & v[sub]2[/sub] after the collision. For that calculation, you have two equations and two unknowns. Very simple and straightforward.
It’s a little unclear what, exactly, the OP is concerned about. However, I suspect it’s what MikeS touched on at the end of his post. Let me take a guess and reword the OP:
The answer is that kinetic energy does not remain the same during the collision, but total energy does remain the same. During the collision, some of the energy is stored as potential energy, as elastic deformation in whatever is colliding. In a perfect collision, all of the potential energy is changed back into kinetic energy by the end of the collision, and the particles continue on their merry way. In collisions that are not perfect, some of the energy is converted into permanent plastic deformation or heat, and doesn’t make it back into kinetic energy. In either case, the total energy remains the same at all times.
It’s a little clearer to me now what the OP is asking. I think the problem is confusion resulting from failing to fully form the equations at play here. They are:
The only true variables are v[sub]1after[/sub] & v[sub]2after[/sub]. For any given set of contants (all the other values other than v[sub]1after[/sub] & v[sub]2after[/sub]), the result for v[sub]1after[/sub] & v[sub]2after[/sub] will be unique, as the OP finds when graphed - they have one point of intersection. For different constants, the lines move around, but still have a single point of intersection.
“Conservation of momentum” just means the center of mass of the system doesn’t change velocity. If the center of mass was originally stationary (e.g. two masses each moving at the same speed, towards each other), then the center of mass will be stationary after the collision. That could mean the two masses stick together and form a stationary lump, in which case kinetic energy after collision is zero (inelastic collision), and momentum is conserved. Or the two masses could fly away from each other at a milliion miles per hour and still the momentum would be conserved.
As for the formulas, perhaps you are neglecting the fact that v[sub]1[/sub] and v[sub]2[/sub] can be negative values. (Well, they’re really vectors.) So if v[sub]1[/sub] is positive and v[sub]2[/sub] is negative, (m[sub]1[/sub]v[sub]1[/sub]+m[sub]2[/sub]v[sub]2[/sub]) can be as small as you want, and (1/2)((m[sub]1[/sub]v[sub]1[/sub][sup]2[/sup]+m[sub]2[/sub]v[sub]2[/sub][sup]2[/sup]) can be as large as you want.
Thanks for all the replies, and thanks to the mod for fixing the title.
Seems I phrased the question poorly; I was wondering how it is possible to get two values, v[sub]1[/sub] and v[sub]s[/sub], which fulfil the condition that both momentum (which is a linear function) and kinetic energy (which is a quadratic function) remain constant. It’s not a problem of finding these values when masses and velocity before collision are given - that’s just a matter of solving the equations; I’m wondering how it can be that there is a solution for any given collision. It does seem to have something to do with the fact that the total momentum and total kinetic energy is distributed among two bodies, so there are two post-collision velocities which can be “adjusted” by the laws of physics so everything works out fine.
I’m not having problems with handling these equations mathematically, I’m having problems with imagining theoretically how it all fits together, but this seems to have been covered by your input. Thanks.