Suppose that two bodies collide in a closed system. If I denote their masses and velocities 1 and 2, and their velocities before and after the collision b and a, then the law of conservation of momentum says that the following equation must hold true:
m1bv1b + m2bv2b = m1av1a + m2av2a
But the motion of the bodies means they have kinetic energy, and energy is also preserved, so the following equation must also be true (assuming that the whole thing takes place in vacuum, so no loss of energy to friction):
0.5m1bv1b² + 0.5m2bv2b² = 0.5m1av1a² + 0.5m2av2a²
Those two equations have the exact same terms, but once linearly and once quadratically. How can they both hold true at the same time? I’m sure it is possible to find some combination of the respective ms and vs where both equations are fulfilled. But they must be true in every case, and I don’t see how the maths work out for that, considering that quadratic terms grow very differently from linear ones.
Kinetic energy is not conserved, in general, if the colliding bodies have any internal structure.
Mathematically there are two equations with two unknowns: the final speed of each body. (More if you vectorize the equations, but still sufficiently constrained.) If nothing else, the speed of each body can remain unchanged.
Elaborating on what @Pleonast said. There’s one obvious solution to the equations above - the absolute values of the velocities for each object before the collision equal the absolute values for the velocities after the collision. That’s the elastic solution, in which kinetic energy is conserved. For inelastic solutions, energy is conserved, but kinetic energy is not; the missing kinetic energy is converted to something else (almost certainly internal heat).
Yeah, there are many other forms of energy, besides kinetic energy. In some kinds of collisions, such as billiard balls, these other forms of energy will have little relevance, and so you’ll get the elastic collision condition, or something very close to it. But if you collide, say, two blobs of Play-doh together, then almost all of the energy will go into other forms.
And as @Andy_L alluded to, whenever the question is “Where did the energy go?”, the answer is almost always “heat”.
A good example is Newton’s Cradle - the requirement to conserve both energy and momentum explains why n incoming balls results in the ejection of n balls at the same velocity, rather than (say) just the end ball at a higher velocity.
I’d add that, even ignoring other types of energy, kinetic energy is easy to “hide” because it is dependent only on speed, not velocity. An object with a joule of kinetic energy might be very noticeable if all that energy is due to movement of all parts of it in a single direction, but if you divide up that joule of kinetic energy between a whole bunch of particles and make sure they’re all moving in different directions, the energy is barely noticeable (which is basically what is happening in inelastic collisions – the energy is still there, just moved from a configuration where it is conspicuous, to one where it is not).
Contrast this with momentum. Because it depends on the direction of the velocity, as well as its magnitude, it’s impossible to divide it up in a way where it is inconspicuous. If you have an object with 1 kg*m/s of momentum, and you try to split it between a bunch of particles, you end up with a bunch of particles all moving in the same direction that the original object was, which is just as hard to ignore as the original object.
The difference in the ability to “hide” energy vs momentum, at least in tabletop experiments, is why momentum seems to be “more conserved” than energy.
Billiards is an interesting case to think about because though billiard ball collisions look very elastic, it’s obvious that they can’t be - because we hear the “click” between the balls, meaning that air has been made to vibrate, using energy (that will dissipate into heat).
Another interesting case: imagine a hard, but hollow and liquid-filled billiard ball. Now you’ll still hear the click when the balls strike, but the post-hit behavior will be very different, even on a completely frictionless table, because of internal friction (just after the strike, the outside of each ball will have reversed the direction of its rolling, but the liquid inside will be rotating the ‘wrong’ way, and internal friction will stop both balls giving zero momentum for both and making them both warmer)
Just to expand on what Pleonast already said, assuming a 1-D problem (as you the OP originally did), if we could only say that momentum is conserved, then that would be only one equation constraining the two unknowns of the final speeds…and hence we would be able to come up with infinitely-many possible solutions (i.e., pairs of final speeds) that would satisfy this constraint. Each of these solutions would have a different fraction of the original kinetic energy conserved and which one is chosen depends on what fraction of the original kinetic energy is in fact conserved. This could range from a completely-inelastic collision, which would be if the two particles stick together (conserving the least amount of kinetic energy possible given the constraint of conservation of momentum) to a completely-elastic collision, which would be if the two particles move apart at the fastest possible speed that they can while still conserving momentum and not ending up with more final kinetic energy than initial kinetic energy. [It is even possible to have a collision where the final kinetic energy is greater than the initial kinetic energy because there is some other source of energy that gets converted to kinetic energy. A firecracker exploding is one example.]
By the way, the technical statement of when momentum of a system is conserved is when no external forces act on the system, which is usually true to a very good approximation during the short time in which collisions occur. (Even if it is not, you can often divide the problem up into the collision, which you consider to be essentially instantenous, and for which momentum is conserved, and the rest of the time when you worry about the action of the external forces.) An example of such a problem would be a ballistic pendulum where you shoot a bullet at a block of wood that is hung as a pendulum and then use the extent of the bullet-block’s subsequent rise to determine the speed of the bullet: You use conservation of momentum (for the bullet-block system) for the collision but then conservation of energy for the subsequent pendulum motion.
The technical statement of when mechanical (kinetic + potential) energy is conserved in a system is when no non-conservative forces act on a system. A non-conservative force is one where the work done by the force depends on the path and not just on the initial and final positions (and hence cannot be described by a potential energy function). The more intuitive idea is that a non-conservative force is basically any force that converts kinetic energy to some other form of energy like thermal energy (“heat”)…or vice versa, as in the case of the exploding firecracker.
Billiards may not be very interesting based on college physics because the balls hold kinetic energy in terms of spinning. Depending on where and how (height and angle) the cue hits the ball, it spins differently.
I’d like to come back to the question of “how can the maths work out?” There’s a nice geometric way of looking at this for 1-D collisions. We know that the momentum is fixed, so before and after we have
m_1 v_1 + m_2 v_2 = p
If we imagine a plane with v_1 on the horizontal axis and v_2 on the vertical axis, then the set of all points satisfying this equation (i.e., satisfying momentum conservation) is a straight line; different amounts of momentum p will correspond to different lines. Similarly, the equation for energy conservation is
\frac12 m_1 v_1^2 + \frac12 m_2 v_2^2 = E
which defines an ellipse. By changing the amount of energy E, the size of the ellipse changes.
The velocities that correspond to a given amount of energy and a given amount of momentum are the intersections of this line and this ellipse. But a line and an ellipse can intersect at two, one, or zero points.
The case where they intersect at two points is the usual case: one intersection point corresponds to the initial velocities, and the other to the final velocities.
The case where they intersect at zero points tells you that it’s impossible to choose your velocities such that they have the particular amounts of momentum p and energy E that you are considering.
And the case where the line and the ellipse intersect at exactly one point (it can be shown) corresponds to the case where v_1 = v_2, i.e., the objects can’t collide at all.
Cool. And the non-elastic case makes the equality an inequality - (E (post collision) is less than E (prior to collision)), which mean that in any case where you can have a collision, you have a variety of possible collisions, depending on how non-elastic the collision is.
Actually, as I pointed out a great many years ago on this board, although the solution that n balls incoming results in n balls outgoing is consistent with conservation of momentum and energy, it’s not the only solution that is – You have five balls whose movement must be accounted for and conservation of energy and momentum only give you two equations to determine those motions. You still have too many unrestricted variables to say that you’ve uniquely determined the solution. You could say that if there were only two balls involved. In fact, there are other factors determining the number of balls moving and the speeds they move at after the collision.
I was about to comment on that, too. A Newton’s cradle is actually dependent on the sound waves moving through the balls, and reflecting off of the ends of the line.
This thread reminds me of something I have long wondered about. It is often claimed that it is easier to hit a home run off a 100 MPH fastball, than an 80 MPH curve, because the former holds more energy. Now it does of course, but this ignores momentum. You have to reverse the momentum. Clearly, a lot of the change in momentum will have to be absorbed by the earth. But even so, the batter has to impart that impulse using his strength. The fact that a faster pitcher ball thrown against a fixed target will come back further than will a slower one somehow still fails to convince me. A batter is not a fixed target.
It doesn’t have to be a fixed target, though. The bat weighs more than the ball, so you could have a bat hanging freely in the air (perhaps released by a batter at the last second, or perhaps hung carefully by a thread) and aim a pitch right at the bat, and the ball will come back toward the pitcher faster if it is thrown faster.
If that feels counterintuitive, consider not a bat floating in the air but a 1000 lb block. It isn’t connected to the earth, but it’s so heavy that it acts basically like a completely fixed target. Here, a faster pitch certainly bounces back faster.
What if we lower the weight from 1000 lb to 100 lbs? Or 50 lb? Or 2 lb? When does the “faster backward is faster pitch” effect stop happening? It stops when we lower the mass of the target (bat or block) down to the mass of the ball since that’s the point where the ball doesn’t bounce back at all anyway. And if we keep lowering the mass of the target, the ball “bounces” (continues) forward toward to backstop, and its speed will be higher it the pitch was faster.
That last line should be interesting: whether the ball bounces back to the pitcher after striking the freely floating target or whether it continues in the pitched direction depends only on which item is heavier: ball or target. But in both cases, it’s speed in the post-collision direction (whichever one) will be made higher by pitching the ball faster.