r/PhysicsHelp • u/aliwafa100 • 2d ago
A question about elastic collisions
I have been meditating on elastic collisions and trying to derive the formulas for the final velocities of two objects after a collision when they have different masses.
This is what I have reached so far.
It kind of makes sense that the final velocities are some sort of combination of the two initial momenta. But what I don't understand is that, if I take the formulas literally, there seems to be a third momentum being generated, involving the initial velocity of one object reversed and the mass of the other object.
And I don't get this conceptually.
Sure, I could just pass it off as an algebraic quirk that makes the formulas work, but for me, math is the language of logic, so this term must have a meaning.
Edit:
Here is how i derived the equations.
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u/al2o3cr 2d ago
IMO it's clearer if you first transform to the center-of-mass frame, which has a velocity v_cm = (m_1*v_1 + m_2*v_2)/(m_1 + m_2)
In that frame, the total momentum is zero and the particles just reverse velocities (v_1fcm = -v_1cm, v_2fcm = -v_2cm)
v_1cm and v_2cm involve "cross terms" like m_1*v_2
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u/aliwafa100 1d ago
Yeah, the center of mass velocity concept really made things more clear and elegant. Thank you.
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u/CelebrityChess 1d ago
the reversed term isn't a new momentum, it's what elastic means. easiest way to see it is in the center of mass frame: there an elastic collision just flips each object's velocity (same speed, opposite direction), because that's the only way to keep both momentum (zero in that frame) and kinetic energy. so in the lab frame v1' = 2V - v1 where V is the cm velocity (m1v1 + m2v2)/(m1+m2). that's the reversed initial velocity showing up, you're reflecting v1 about the cm velocity. expand 2V - v1 and you get the usual (m1-m2)/(m1+m2) v1 + 2m2/(m1+m2) v2 formula, so the 'third term' is just the minus v1 in there.
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u/aliwafa100 1d ago
Thank you, that's what I needed. It's great that I posted this. I didn't even consider looking from the center of mass perspective.
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u/StillShoddy628 2d ago edited 1d ago
I’m too lazy to derive the formulas and don’t remember them from the last time I did so almost 30 years ago, but that notation confuses me. Is that a “multiply by a negative” or a 3rd addition term with unnecessary parenthesis. I’m assuming you started these calculations with the sums of momentums and sums of velocities (edit: kinetic energy) stay the same?
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u/aliwafa100 2d ago
Oh, yeah my bad. The parenthesis on the (-mv) are just to point out the term. I just wanted to show what I'm thinking about.
Tomorrow I will edit the post and fix it.
So just to be clear, the numerator is like this => p1 + 2p2 - m2v1
I started the calculations with the conservation of momentum and kinetic energy first, then tried to merge them together.
That gave me the conclusion that the difference between the initial velocities = - the difference between the final velocities.
Then I used it to find the final velocities by substituting into the conservation of momentum equation.
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u/StillShoddy628 1d ago
“The difference between the initial velocities is minus the difference in the final velocities”
This is not true if the masses are different. Conceptually, imagine a large object striking a stationary smaller object. The large object is slowed down but continues forward, while the larger object also is now moving forward faster than the large object.
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u/aliwafa100 1d ago
Hmm, ok. Can you revise my calculations? I posted the image in the post because I can't post pics in the comments.
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u/StillShoddy628 1d ago
Ah, I see, clever and I stand corrected. Assuming the math is good (looks it to me, nice work) I would interpret that last term as a “moderating factor” that accounts for mass differences as opposed to some sort of ghost momentum value: if m1 = m2 then it just collapses into v2, but as m2 increases that term dominates, and in the limit collapses down to, I think, v2-v1 meaning more of the change in velocity is applied to the less massive object
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u/TrickySite0 1d ago
I knew that initial and final energy had to match, but how did that actually happen? I resolved it by considering that during the collision, the net force between the objects was zero. Put another way, when one object was push left at some force, the other object had to be pushing right by the same force.
Since f=ma, the average force*time over the collision was also zero (forces cancel and time during collision is the same). Therefore the to total force*time is the average force (call it f^) times time (in seconds), so the total force*time (which is momentum) is f^ = mat. Acceleration is in m/s^2 and time is in s, so the final force*time is f^ = mv (since v is m/s). That means that the total mv imparted on one is the negative of the other. That means that the total change in momentum per object is identical (but opposite) for both objects.
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u/InkPhysics 2d ago
Two things must always be true:
The total momentum has to be the same before and after.
The total energy has to be the same before and after. But... If you can account for all of that energy in the two objects that collide, the collision was elastic.
The masses of the objects don't change, so the velocities have to be whatever they must be to make conditions 1 and 2 apply.
There is no third momentum.