Deriving an Expression for g in Terms of M, m, D, H, and t

[bjac]

There are two blocks of equal mass M that hang from a massless, frictionless pulley. There is a small square block of mass m sitting on top of one of the blocks. When the block is released, it accelerates downward for a distance H until the block of mass m sitting on tip gets caught by a ring. The block of mass M continues to fall at constant speed for a distance D.

Derive an expression in terms of g using M, m, D, H, and t, where t is the time that M takes to move at constant speed through the distance D.

I have made drawing describing the situation:

http://img691.imageshack.us/img691/3714/38860857.png

I actually have this solution, but I don't understand it. Could someone explain how it was derived?

a=\frac{mg}{2M+m}

v^2=2aH

v=\frac{D}{t}

v^2=\frac{2mgH}{2M+m}

\frac{D^2}{t^2}=\frac{2mgh}{2M+m}

g=\frac{(2m+m)D^2}{2mHt^2}

 

[ahmadmz]

What part are you having difficulty with? What did you try?

 

[fatra2]

Hi there,

By looking at your problem, and making the calculations myself, I found this error in your solution:

a = \frac{mg}{M + m} without the 2M at the bottom

 

[ahmadmz]

Hi there,

By looking at your problem, and making the calculations myself, I found this error in your solution:

a = \frac{mg}{M + m} without the 2M at the bottom

Are you sure? I did it too and I'm getting 2M.

 

[fatra2]

From where do you find the second M?

Let's look at the forces acting on the (M+m) system. You have the tension in the rope, and the weight of both M+m.

\sum F = (M+m)a
T - F_M - F_m = (M+m)a
\cancel{Mg} - \cancel{Mg} - mg = (M+m)a
a = -\frac{mg}{M+m}

 

[ahmadmz]

What i did was :

(M+m)g - T = (M+m)a for the right side
T - Mg = Ma => T = M(g+a) for the left
Then

(M+m)g - M(g+a) = (M+m)a
mg-Ma = Ma + ma
mg = 2Ma + ma
a = mg / (2M+m)