Conservation of engery of a pulley system

[jwxie]

Homework Statement

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.
(ans: 4.43)

(b) Find the maximum height to which the 3.00 kg object rises.
(ans: 5)

Homework Equations

\Delta K = - \Delta Ug[/text]<br /> mgy - mgyf = - \Delta Ug[/text]&lt;br /&gt; KEi + PEi = KEf + PEf&lt;br /&gt; &lt;br /&gt; &lt;h2&gt;The Attempt at a Solution&lt;/h2&gt;&lt;br /&gt; &lt;br /&gt; 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.&lt;br /&gt; &lt;br /&gt; 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).&lt;br /&gt; &lt;br /&gt; 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 ?&lt;br /&gt; &lt;br /&gt; I have the following data on my hands&lt;br /&gt; &lt;br /&gt; for m2, where m= m2, h = 4&lt;br /&gt; _________________________&lt;br /&gt; | | KE | PE |&lt;br /&gt; | i | 0 | 0 |&lt;br /&gt; | f | 1/2mv^2 | m2h |for m1 where m = m1, h = 4&lt;br /&gt; | | KE | PE |&lt;br /&gt; | i | 1/2mv^2 | mgh |&lt;br /&gt; | f | 0 | 0 |

 

[rl.bhat]

Since the string is unstretchable, if m1 moves a small distance Δx in time Δt, m2 also moves the same distance in same time interval. So vf of m1 is the same as vf of m2.

 

[jwxie]

Hi, thank you.