# Isolated system/conservation of energy

## Homework Statement

Two objects, m1 = 5.00 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 5.00 kg is released from rest, h = 4.00 m above the ground.

Using the isolated system model, determine the speed of the 3.00 kg object just as the 5.00 kg object hits the ground.

## Homework Equations

Kf + Uf = Ki + Ui

## The Attempt at a Solution

So I thought that initially, energy was only in the form of gravitational potential energy for m1 and at the final state there is kinetic energy for m2 and gravitational potential energy for m2.

When I isolate for final velocity of m2, I end up getting:

v = (2*g*h(m1-m2)/m2)^0.5

But I am doing something wrong. Could someone point me in the right direction please?

## Answers and Replies

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Whoop nevermind, caught my mistake -- m1 has a final velocity and both m1 and m2 have this same velocity! (I think the velocity is the same since they're connected by the same string? Not 100% on that though...)

Yes they will have the same velocity because of the string (for about zero time since the ground will slow down one of them while the other will continue to move as it did before)