Atwoods Machine - Mass Height Question

[kashmirekat]

The problem reads as follows:

Two masses are each initially 1.80 m above the ground, and the massless frictionless pulley is 4.8m above the ground. What maximum height does the lighter object reach after the system is released?

It gave me a hint to find the acceleration and velocity at the moment the heavier one hit the ground.

To find 'a' I used ((M1-M2)/(M1+M2))*9.8m/s^2 and got the answer 1.815m/s^2.

For velocity, I used the equation V=Vo+at, where time was determined by the distance to the ground, 1.8m divided by acceleration 1.815m/s^2. t=1.8/1.815m/s^2 and I got t=0.99s. Question here, since the a is squared, I would have to square my t, right?
t=sqrt(d/a).

Now my real question is what does it want for maximum height? The distance to the pulley is 4.8m...and the heavier object is only going to travel 1.8m downward until it reaches the ground. So thinking like a lamen, maximum height upward should only be 3m...but it's more complicated than that, isn't it? Why and where do I need to use my determined acceleration and velocity?

 

[HallsofIvy]

Once you have found the acceleration of the more massive object toward the ground, you can find the time it takes to hit the ground and, of course, it's speed at that instant.

That speed is the same as the upward speed of the lighter object (at that time). The point is that it doesn't just stop instantaneously! (The heavier object does because it hit the ground!)
Treat it as an "object thrown upward problem". You know it's height (it will have moved upward from it's initial position exactly as much as the heavier object moved downward: 1.8 meters), it's speed is the speed you calculated for the heavier object as it hit the ground, AND ITS ACCELERATION IS -9.8 m/s^2 now. Use that to find an expression for the height of the object and use THAT to determine the highest point.

 

[M98Ranger]

The maximum height that the lighter object reaches after the system is released can be determined by using the conservation of energy principle. At the moment the system is released, the total energy of the system is equal to the potential energy of the masses at their initial heights. As the heavier object falls, its potential energy is converted to kinetic energy and then transferred to the lighter object through the pulley. This results in the lighter object gaining velocity and reaching a maximum height before falling back down.

To find this maximum height, we can use the equation for conservation of energy:

PE1 + KE1 = PE2 + KE2

Where PE1 and KE1 represent the initial potential and kinetic energies of the system, and PE2 and KE2 represent the final potential and kinetic energies of the system.

Since the system starts with both masses at a height of 1.80m, the initial potential energy is:

PE1 = m1gh1 + m2gh2 = (m1 + m2)gh

Where m1 and m2 are the masses of the two objects, g is the acceleration due to gravity, and h is the initial height of the masses.

Similarly, the final potential energy is:

PE2 = m1gh + m2gh2

Where h is the maximum height reached by the lighter object.

Since the pulley is frictionless and massless, the kinetic energy of the system remains constant throughout the motion. Therefore, KE1 = KE2.

Substituting these values into the equation for conservation of energy, we get:

(m1 + m2)gh = m1gh + m2gh2

Solving for h, we get:

h = (m1gh - m2gh2) / (m1 + m2)g

Substituting the values given in the problem, we get:

h = (1kg * 9.8m/s^2 * 1.8m - 2kg * 9.8m/s^2 * 4.8m) / (1kg + 2kg) * 9.8m/s^2

Simplifying, we get:

h = 0.6m

Therefore, the maximum height reached by the lighter object is 0.6m above its initial height. This is because the heavier object only falls a distance of 1.8m, while the lighter object