How high does m2 rise above its initial position before momentarily coming to rest?.?

Question

The two masses in the Atwood's machine have an initial speed of v = 0.17 m/s, and m2 is moving upward.
 m1=3.0 kg    m2=4.6 kg

gintable · Accepted Answer

Use kinematics to find distance, should we know acceleration:
vf^2 = v^2 - 2*a*d, negative sign exists because m2 is slowing down.

solve for d, knowing that since it comes to rest, vf = 0:
d = v^2/(2*a)

And to find a, use the standard result of Atwood's machine:
http://hyperphysics.phy-astr.gsu.edu/hbase/Atwd.html

a = g*(m2 - m1)/(m1 + m2)

Thus:
d = (m1 + m2)*v^2/(2*g*(m2 - m1))

Data:
m1:=3 kg; m2:=4.6 kg; v:=0.17 m/s; g:=9.8 N/kg;

Result:
d = 7 millimeters

How high does m2 rise above its initial position before momentarily coming to rest?.?

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