Physics/Conservation of momentum
Hi there. How are you? I was shown on TV once how to prove the law of conservation of momentum I.e. momentum before = momentum after. The written proof involved trolleys, steps and explanations and equations. It also used Newton's Third Law of Motion. I wish I still had the darn thing. Would you be able to help me prove it in the same fashion at grade 12 level if its not too much trouble? If possible a hand written proof would do wonders. Its not a homework question. Its just me going back in time to grade 12 and keeping a printed document for my collections. My email address is petergoodwill(at)telkomsa.net if it needs to be hand written. This would be sincerely appreciated.
PS Have a wonderful festive season. God Bless
Please forgive my delay in responding -- it's the only way I can think of, to ensure I am not assisting with academic work, of which homework is just a small part. Also, as I am unable to determine the veracity of what people post, I can not know whether or not a question involves academic work.
Anyway, conservation of momentum in our Universe can be proved in a couple of ways. One way is more complete and involves very simple assumptions, BUT requires graduate level math. The other way -- the way I think you're asking about -- is less complete but uses only algebra. I'll discuss both.
1) Let's say you want to measure the distance between two points, say, the two end points of a piece of wood. In other words, its length. You take a ruler, place the part of the ruler labelled "zero" at one end point, and then look at what's labelled at the other end. Let's say it's 10. You then conclude the wood is ten units long. We reach this conclusion because
10 - 0 = 10
Now let's say we do the same thing, but this time we place the part of the ruler labelled "one" at one end point, and then look at what's now labelled at the other end. I think you'll agree that, in our Universe, it will be 11. We then (a second time) conclude the wood is ten units long because
11 - 1 = 10
"So what?" you have every right to ask. To physicists, this IS a big deal -- it means that, for measuring distance, the zero point doesn't matter. In our Universe, the only important thing is the change in distance between two points -- what one declares to be "zero" is irrelevant. The technical term is that our Universe is "spatially invariant."
Again, "so what?" This is where the complicated math comes in. It turns out that, when you do this math, distance and momentum are "complementary" variables. And this means that, if you assume an interaction is spatially invariant, then you can prove that momentum is conserved.
2) Now let's look at a VERY simple interaction -- a one-dimensional collision between two trolleys on a rail. We'll call them Trolley 1 and Trolley 2, or Tr1 and Tr2. Their respective masses are m1 and m2, the respective forces each experience in the collision are F1 and F2, the respective time each of the two are in contact are t1 and t2, and their respective momentums are P1 and P2 (momentum is abbreviated as 'P').
Newton's Third Law (technically an axiom) states that, whenever one object exerts a force on another object, the latter ALSO exerts a force on the former. This second force is equal in SIZE to the original force, but opposite in direction.
In our example, during the collision, Tr1 exerts F2 on Tr2 for a time t2. Newton's Third Law states that Tr2 also exerts F1 on Tr1 for a time t1.
Now, a little high school physics.
The relationship between force, mass, and acceleration is given by
F = m x a
Acceleration is defined as the change (usually abbreviated as 'delta' or 'D') in velocity over a certain amount of time. In other words
a = Dv/t
From this we see that
F = m x (Dv/t)
F x t = m x Dv
Recall that momentum is defined as mass times velocity, or m x v
So a change in velocity -- a Dv -- will also result in a change in momentum. If mass remains constant, then the CHANGE in momentum will equal the product of force and time.
m x Dv = DP (ie, a change in momentum)
Which leads to
F x t = DP
If the collision of the two trolleys is elastic -- if they simple bounce off each other -- then these two forces last only as long as the two trolleys are in contact with each other. In addition, the mass of both trolleys remains the same.
This leads to
DP1 = F1 x t1
In other words, the CHANGE in momentum for Tr1 is equal to the force that is exerted on it multiplied by the time it experiences the force. Similarly,
DP2 = F2 x t2
Recall Newton's Third Law and we'll see that
F1 = F2 (only in opposite directions)
and an elastic collision means
t1 = t2
DP1 = DP2
Showing that the size of the LOSS is momentum of Tr1 is EXACTLY equal to the GAIN in momentum of Tr2. If the loss experienced by one exactly equals the gain of another -- whether it's momentum, charge, energy, or money -- then we can say that the TOTAL is conserved in the interaction. This means that, in this VERY limited example, momentum is conserved. As I noted at first, it can be shown that momentum is conserved in all interactions.
Here's a couple of URLs that discuss this a little more deeply: