Conservation of Energy: Solving for Initial and Final Energy

[Neon32]

Homework Statement

2. Homework Equations [/B]

Inital energy=Final energy
K.Ei+P.Ei=K.Ef+P.Ef

The Attempt at a Solution

 

[gneill]

Describe the state of m3 (position, velocity) at the instant m5 hits the floor. What do you expect to happen to m3 after that instant?

 

[Neon32]

Describe the state of m3 (position, velocity) at the instant m5 hits the floor. What do you expect to happen to m3 after that instant?

The position of m3 when m5 hits the ground is 4 m high above the ground and m3 will come to rest after that.

 

[gneill]

The position of m3 when m5 hits the ground is 4 m high above the ground and m3 will come to rest after that.

It's the "m3 will come to rest after that" part that you need to look into. How fast and what direction is m3 moving at that instant? Can you think of a way to use energy conservation to find out how high m3 will go?

 

[Neon32]

It's the "m3 will come to rest after that" part that you need to look into. How fast and what direction is m3 moving at that instant? Can you think of a way to use energy conservation to find out how high m3 will go?

m3 is moving up and with velocity 4.43m/s at the instant m5 hits the ground

 

[gneill]

m3 is moving up and with velocity 4.43m/s at the instant m5 hits the ground

So it starts this phase of its journey at a height of 4m and moving upward with velocity 4.43 m/s. Thus it has some initial PE and KE. What happens to the PE and KE from there?

 

[Neon32]

So it starts this phase of its journey at a height of 4m and moving upward with velocity 4.43 m/s. Thus it has some initial PE and KE. What happens to the PE and KE from there?

So you mean in the second case, the m3 will start from height 4m and velocity 4.43m and go some distance above 4m?

This is how I understood it:
Inititaly the m3 is at height 4m and inital velocity 4.43m.
so Initial energy= m3gh+1/2 m V² (1)

Then it will move some distance above 4m until it reaches its maximum distance and comes to rest.
so Final energy of the whole system(both bodies)=m3gh(new highet)+ 0 (2)

So from (1) and (2)
m3gh+1/2 m V²=m3gh(new height)

 

[gneill]

That looks good. What do you find for the new height?

 

[Neon32]

That looks good. What do you find for the new height?

height approximatley equal 5.001 m.

 

[gneill]

height approximatley equal 5.001 m.

Remember to round to the correct number of significant figures. But otherwise, that looks great!

 

[Neon32]

Remember to round to the correct number of significant figures. But otherwise, that looks great!

Thanks for taking the time to help me. Appreciated :).