Conservation of Mechanical Energy Question!
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Conservation of Mechanical Energy Question!

[From: ] [author: ] [Date: 11-10-10] [Hit: ]
so y2_before−y2_after = −h.We can rewrite the above equation as:(m1)g(y1_before−y1_after) + (m2)g(y2_before−y2_after) = ½(m1+m2)(v_after)²And substitute to get:(m1)g(h) + (m2)g(−h) = ½(m1+m2)(v_after)²You can solve to that for v_after, and theres your answer for Part 1.For Part 2, I think theyre saying that m2 continues flying upward for a bit (with a slack rope), due to its momentum,......

Now consider the total energy "after":
PE_after = (m1)g(y1_after) + (m2)g(y2_after)
KE_after = ½(m1)(v1_after)² + ½(m2)(v2_after)²

But we can simplify the KE_after, because the two masses are always traveling at the same speed (since they're tied together), at least until the one hits the ground. So instead of using two variables "v1_after" and "v2_after", use just a single variable "v_after" to stand for the speed of both:
KE_after = ½(m1)(v_after)² + ½(m2)(v_after)² = ½(m1+m2)(v_after)²

So, total energy "after" is:
E_after = (m1)g(y1_after) + (m2)g(y2_after) + ½(m1+m2)(v_after)²

Conservation of mechanical energy says the E_before = E_after:
E_before = E_after
(m1)g(y1_before) + (m2)g(y2_before) = (m1)g(y1_after) + (m2)g(y2_after) + ½(m1+m2)(v_after)²

Now we use the fact that m1 fell a known distance (h=1.83 meters), which means that y1_before−y1_after = h. Also, m2 ROSE by that same amount, so y2_before−y2_after = −h. We can rewrite the above equation as:
(m1)g(y1_before−y1_after) + (m2)g(y2_before−y2_after) = ½(m1+m2)(v_after)²
And substitute to get:
(m1)g(h) + (m2)g(−h) = ½(m1+m2)(v_after)²

You can solve to that for "v_after", and there's your answer for Part 1.

For Part 2, I think they're saying that m2 continues flying upward for a bit (with a slack rope), due to its momentum, after m1 hits the ground. In that case, use the conservation of energy just on m2 for this part, since m1 is now out of the picture.
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