Help with Conservation of Energy

[moodyflower]

I need help setting up this problem. I need help setting up the equation, knowing what cancels and what doesn't and what values to use.

Two objects are connected by a light string passing over a light, frictionless pulley as in Figure P5.63 (see attached file). The 5 kg object is released from rest at a point 4 m above the floor. (a) determine the speed of each object when the two pass each other. (b) determine the speed of each object at the moment the 5 kg object hits the floor, (c) how much hgiher does the 3 kg object travel after the 5 kg object hits the floor?

Any help is appreciated. Thanks.

 

[Sirus]

Please show us what you have done so far so we can help you. Think about the energy transfers (potential-kinetic and vice versa) that occur here.

 

[moodyflower]

I just need help setting up the equation. I am not sure what cancels out of the conservation equation and what doesn't. I know the initial kinetic energies cancel out because it's not moving yet, but other than that I am not sure what to do.

 

[arildno]

The easiest way is Sirus' energy conservation idea:
1) Regard the two weights as point masses, and compute their total potential energy at initial time t=0. Since at t=0, neither mass is moving, you have therefore found the total MECHANICAL energy of the system.
Hint: It is easiest to use the floor as the reference level for the potential energy.
2) Since the string length remains constant, this clearly implies that the two objects' speed must be equal to each other at all times in the descent (but not their velocities; they are the negative of each other)
Use conservation of energy plus 1)+2) to answer a)+b).

In order to answer c), assume that the speed of the non-colliding object does not change during the collision period.
Use your results from b) and conservation of the still-moving object's mechanical energy to determine its maximal height.

 

[moodyflower]

questions

Can you tell me if this equation that I came up with for part (a) is correct? m2gh2 initial= 1/2 (v)squared(m1+m2)+ mgh final.