Two objects, m1 = 4.50 kg and m2 = 3.00 kg, are connected by a light string passing over a light frictionless pulley as shown in the figure below. The object of mass 4.50 kg is released from rest, h = 3.00 m above the ground.
(a) Using the isolated system model, determine the speed of the 3.00 kg object just as the 4.50 kg object hits the ground.
(b) Find the maximum height to which the 3.00 kg object rises.
[tex]\Delta K[/tex] = - [tex]\Delta Ug[/text]
mgy - mgyf = - [tex]\Delta Ug[/text]
KEi + PEi = KEf + PEf
The Attempt at a Solution
I know I can solve problem a differently. Suppose I solve each mass separately using law of conservation of energy, I can find Vf for mass 2 separately.
Since the system is connected by the same uniform massless string, so the accerlation must the same. But I did not assume their Vf will be the same (the impact of m1 as it hits the ground, and at that instant the Vf of m2 raising to 4 meter).
What I want to know is, how do you prove that the final velocity of m1 and m2 at that same instant is the same using the isolated system model KEf + PEf = KEi + PEi ?
I have the following data on my hands
for m2, where m= m2, h = 4
| | KE | PE |
| i | 0 | 0 |
| f | 1/2mv^2 | m2h |
for m1 where m = m1, h = 4
| | KE | PE |
| i | 1/2mv^2 | mgh |
| f | 0 | 0 |
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