Ball collision when one is dropped and the other thrown up

Plugging values into random equations is as far from good physics as it can possibly get. No wonder you get nonsensical results.

Before using the equations, think about the physics of the problem. You have observed, correctly, that the time of motion for the two balls is equal. Can you find this time from what you know about any of the balls? Can you then find the velocities of the balls?

 

For the top ball,
v²=2a(H-h)

Taking on magnitude for bottom ball,

1/4(2a(H-h)) =u²-2a(H)

2aH-2ah=4u²- 8aH

10aH-2ah=4u²

10a -2ah/H=4u²/H

 

Consider ball A. What do you NOT know about it that would prevent you from solving for its velocity at h and the time it takes to get there?

 

The mindset, as I said, is to think less about the repertoire of the equations you have in front of you, and to think more about the real physics. In the end, you need only a couple of equations, everything else can be derived from them.

For ball A, falling down from rest, everything depends solely on the original height H and height h it is at any given time. Once this is clear, you can look at your equations. From 2.15 you obtain the time. From 2.11 you obtain its velocity. Both will still depend on the unknown h.

Then consider ball B. For this ball, everything depends on its initial velocity and the time elapsed. Since we already know the time, and also know the final velocity, we obtain its initial velocity. Check what equation could be used for this. The result will again depend on the unknown h.

So we can express this height h as the distance traveled by ball B (what equation?), and that will give you an equation, from which h can be found.

 

This is the concept of determinism. Every aspect of motion of a mechanical system depends solely on (or is determined by) its initial state (positions + velocities of all its constituent parts) and time elapsed.

 

The initial conditions for A are its initial height H and its initial velocity, which is zero. So everything depends just on H and the time elapsed. Everything, including h at any given time. But that equally means that at any given h, time can be found from h and H. So we can exclude the time and express any aspect of A's motion as dependent on h and H.

 

Well, I can't say I analyze this really. I analyzed this quite a while ago and I just KNOW now that, for example, I can trade one kinematic variable of a system for another (t for h in this case). Another thing I immediately take note of is that A has ALL the initial conditions given, while for B we know only the initial height. So it makes sense to start with A. Yes, all of this takes a little bit of practice.

There is also another method, which is less physical and more mathematical. It usually is more roundabout, but sometimes, when you are stuck, you might consider it. Every uniformly accelerated 1D motion is fully described by two equations: [tex]x = x_0 + v_0t + \frac {at^2} 2 \\ v = v_0 + at [/tex]So you just express your problem in terms of these equations - for every part of the system. Physics mostly ends here, and mathematics begin: you just solve the resultant system of equations. For this problem, for example, you have two balls, so you have four equations: [tex]
x_A = x_{A0} + v_{A0}t + \frac {at^2} 2 \\ v_A = v_{A0} + at
\\ x_B = x_{B0} + v_{B0}t + \frac {at^2} 2 \\ v_B = v_{B0} + at
[/tex]Plugging here all the known parameters, we end up with [tex]
h = H - \frac {gt^2} 2 \\ v_A = - gt
\\ h = v_{B0}t - \frac {gt^2} 2 \\ v_B = v_{B0} - gt
[/tex] We further know that ## v_A = 2v_B ##, so we just add this to our system, and end up with five equations and five unknowns: ## h, \ t, \ v_A, \ v_{B0}, \ v_B ##, so we should be able to solve this system purely algebraically without any more physical considerations. Well, not completely, as we have quadratic equations there, so we may have to choose from two different roots, and that typically needs some physical consideration.