Law of Conservation of Mechanical Energy Exercise

[Kernul]

Homework Statement

A mass ##m_1 = 5.0 kg## is hanging from the end of a thread, of negligible mass, that slides on a pulley, of negligible mass too and without friction. At the other end of the thread, at the same height of ##m_1##, there is another hanging mass ##m_2 = 3.5 kg##. Using the Law of Conservation of Mechanical Energy, determine the velocity of ##m_1## when it goes down, starting still, of a height of ##2.5 m##.

Homework Equations

##E = K + U##
##K_f + U_f = K_0 + U_0 \Rightarrow E = const##

The Attempt at a Solution

So, since we have two masses that move at the same time but have different masses, we can start writing these:
##K_0 = \frac{1}{2} * m_1 * v_0^2 + \frac{1}{2} * m_2 * v_0^2##
But since we know it starts still, ##v_0 = 0## and so ##K_0 = 0##.
##U_0 = 0## because it hasn't moved yet.
##K_f = \frac{1}{2} * m_1 * v_f^2 + \frac{1}{2} * m_2 * v_f^2##
with ##v_f## being the velocity we are searching for.
##U_f = m_2 * g * h - m_1 * g * h##
So, using the second relevant equation we have:
##K_f = - U_f##
##\frac{1}{2} * m_1 * v_f^2 + \frac{1}{2} * m_2 * v_f^2 = m_1 * g * h - m_2 * g * h##
##\frac{v_f^2}{2}(m_1 + m_2) = g * h(m_1 - m_2)##
##v_f =\sqrt{ \frac{2 * g * h(m_1 - m_2)}{(m_1 + m_2)}} = 3 \frac{m}{s}##
Is this method I used to find the velocity with the law good? Or I could have done better and faster?

 

[BvU]

This is about it, can't do much better than ##\Delta K + \Delta U = 0##

 

[Kernul]

Okay! Thank you for replying! :D

 

[BvU]

You're welcome.