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April 20, 2014

April 20, 2014

Posted by **Jake** on Wednesday, April 18, 2012 at 10:15am.

If m1 starts at a height h=1m above the ground, what is its speed v when it hits the ground?

I've tried a few different ways to get it but I can't seem to get the answer. I've tried using the sum of the torques as well as trying to use energy. I can't seem to figure it out, any help is much appreciated!

- Physics -
**Elena**, Wednesday, April 18, 2012 at 5:47pmIn vector form

m1•¬a = m1•¬g - ¬T1

m2•¬a = m2•¬g - ¬T2

I•¬ε =¬M

Projections on the vertical axis:

m1•a = m1•g - T1,

m2•a = m2•g – T2,

(mR^2/2)•(a/R) = (T1-T2)•R.

Solving this system for acceleration a, we obtain

a= (m1-m2)•g/(m1 +m2+m/2) =(3-2)/(3+2+10/2) = 0.1 m/s^2.

Since

s =v^2/2•a.

v =sqrt(2•a•s) = sqrt(2•0.1•1)= 0.45 m/s.

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