# Homework Help: Conservation of mechanical energy

1. Oct 20, 2008

### nvictor

1. The problem statement, all variables and given/known data

I have two blocks attached by a string (massless) over a pulley (frictionless, massless). Block 2 (which weighs less than block one) is released and the two blocks meet momentarily at the same height. I have the find the speed at which the meet at that moment. The blocks are separated by a distance of 1m.

I have drawn a figure in paint (sorry for this pic): http://i36.tinypic.com/2076at1.png

2. Relevant equations

Emechf = Emechi
Emech = Ug + K

3. The attempt at a solution

Emechf = Emechi
<=> (m_1gh + m_2gh) + (1/2m_1v^2 + 1/2m_2v^2) = m_1gH (where H = 1m, and h is the height at which they meet)

But then I have this h which is unkown and I can't figure out.

Thanks a lot in advance.

2. Oct 20, 2008

### Staff: Mentor

If the lower (and lighter) block starts out at height "h", the higher (heavier) block starts out at height "h + 1".

At what height will they meet? Hint: One block lowers by the same distance that the other rises.

3. Oct 20, 2008

### nvictor

Thanks a lot, I see it clearly now. I really appreciate you help.