>>16701438
Cons. of energy. The init. dist. between the two balls is H = r_1 + r_2 + d and the final when they collide is h = r_1 + r_2, so the init. energy of the state is E = -Gm_1 m_2 / H and the final energy of the state is E = -Gm_1 m_2 / h + Kinetic_Energy, meaning
>Kinetic_Energy = Gm_1 m_2 / h - Gm_1 m_2 / H

Now understand this is a 2 body problem; by ignoring the center of mass, we reduce the degrees of freedom from 2 objects to 1. Look up the 2 body problem right now if you don't know it, and don't continue reading until you've looked it up.

Instead of (m_1, x_1) and (m_2, x_2), we work with (\mu = reduced mass, r = x_2 - x_1). The center of mass R ain't moving, so the total energy is
>E = 1/2 \mu (r')^2 + U(r) = Kin. + Pot.
From
> Kinetic_Energy = 1/2 \mu (r'_final)^2 = Gm_1 m_2(1/h - 1/H)
we find
>r'_final = \sqrt( 2G[m_1 + m_2][1/h - 1/H] )

x_1 and x_2 are both linear func. of r, whose form is even easier to guess than to derive (you should already know what they are by now). Since the constant R in the linear eq ain't moving (all we got is the slope), it reduces down to || x_1' || = \mu/m_1 r' and || x_2' || = \mu/m_2 r', which is guessable and intuitive considering momentum.

So
>v_1 = || x_1' || = \sqrt( 2G\mu[m_2/m_1][1/h - 1/H] ).
This is CLASSICALLY EXACT (not general rel.), so no approximation needed. Cons. of energy allows us to bypass a lot of complex eq. of motion when our concern is at certain nice coord. points in (x, v) cause of hamiltonian and lagrangian. Finding the corresponding (x,t) for these same points are sometimes harder since that's the area of newton laws. You can say the same the opposite way around. It's good to recognize when cons. of energy and cons. of momentum should be used to save time.

Here's a great follow up question: How do we reduce v = \sqrt( 2G\mu[m_2/m_1][1/h - 1/H] ) to \sqrt( 2gd )? Approx g = GM_2/h, \mu = m_1, and 1/h - 1/H = d/h and voila.
Derive these and fill in the blanks as practice.