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bjac
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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?
[tex]a=\frac{mg}{2M+m}[/tex]
[tex]v^2=2aH[/tex]
[tex]v=\frac{D}{t}[/tex]
[tex]v^2=\frac{2mgH}{2M+m}[/tex]
[tex]\frac{D^2}{t^2}=\frac{2mgh}{2M+m}[/tex]
[tex]g=\frac{(2m+m)D^2}{2mHt^2}[/tex]
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?
[tex]a=\frac{mg}{2M+m}[/tex]
[tex]v^2=2aH[/tex]
[tex]v=\frac{D}{t}[/tex]
[tex]v^2=\frac{2mgH}{2M+m}[/tex]
[tex]\frac{D^2}{t^2}=\frac{2mgh}{2M+m}[/tex]
[tex]g=\frac{(2m+m)D^2}{2mHt^2}[/tex]
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