Atwood's Machine Simulation: Determining Height with Given Masses

[cajunchrisbu]

help! atwood machine

The two masses in the Atwood's machine shown in the figure below are initially at rest at the same height. After they are released, the large mass, m2, falls through a height h and hits the floor, and the small mass, m1, rises through a height h.

In this Atwood's machine, the mass m2 remains at rest once it hits the floor, but the mass m1 continues moving upward. How much higher does m1 go after m2 has landed? Give your answer for the case h = 5.8 m, m1 = 3.7 kg, and m2 = 3.9 kg.


http://phga.pearsoncmg.com/phga2/modules/unproctoredTest.Print

(the answer is not 5.8 or 0 meters)

 

[Astronuc]

The picture is not showing. One may wish to save it as a jpg and host it at imageshack.

Basically the masses accelerate and when m2 stops, m1 is still traveling with some velocity (speed). One has to determine that speed, which then starts decreasing under the force of gravity on m1 only.

These notes might help.
http://hyperphysics.phy-astr.gsu.edu/hbase/atwd.html

 

[cajunchrisbu]

got it thanks

oops...my mistake (probably from lack of sleep or carelessness)

basicly i forgot to find the velocity of the large mass and that velocity is also for the small mass. so the small mass has kinetic energy until it is over come by gravity at which point (the instant its velocity is 0) it has a potential energy of (v=mgh). after which if freefalls until the string pulls it tight and it stops so basicly.....

Ke=-u
(1/2)m(v^2)=(-)mgh
solve for h as the height of the short freefall and the answer is

tada-------------->0.15m

geez i feel like an idiot...haha but i got it :rofl:

 

[aew7r]

I am working on the same problem, but I am still confused. How did you find the velocity?