# This is more a of a theory question of the atwood and energy

1. Oct 26, 2011

### flyingpig

1. The problem statement, all variables and given/known data

I picked this up somewhere and I couldn't get a good answer

Let's say I have m_1 and m_2 hanging by a pulley on an atwood. Let m_1 > m_2 and m_2 on ground and m_1 hanging in the air at a height h

So if I ask you, what is the speed when they meet, what is the set up?

3. The attempt at a solution

Now for some reason, they will always meet at h/2, I don't know why. Why would it be even h/2?? I am not awake right now so I am probably overlooking something

2. Oct 26, 2011

### sandy.bridge

The blocks are connected, and hence they accelerate at the same rate. If they met anywhere other than the midpoint, it would imply that one of the objects is accelerating faster than the other.

3. Oct 26, 2011

### I like Serena

Hi flyingpig!

m_1 and m_2 are connected by a rope.
When m_1 descends by h/2, m_2 ascends by h/2.

4. Oct 26, 2011

### flyingpig

So something like meeting at h/3 never happens?

5. Oct 26, 2011

### I like Serena

No.

If m_1 ascends by h/3,
then m_2 descends by h/3 ending up at (2/3)h.

6. Oct 26, 2011

### sandy.bridge

Look at it this way: if two objects are subject to the same force in opposing direction, and are separated by a distance h, they will always meet in the middle. It is when the forces are are not coordinated and the acceleration of one is not equal and opposite to the other that the objects will meet anywhere else than 0.5h. Hope that helps.

7. Oct 26, 2011

### flyingpig

Yet the masses are different

8. Oct 26, 2011

### I like Serena

If the rope were elastic that would matter, but we're assuming a rope that does not stretch.

9. Oct 28, 2011

### flyingpig

I had to read this again and I think I understand now from your mistake (yeah who would've thought?) from my mistake

I stated that m_2 is on ground level, so m_2 can't descend. But the better way to understand is that if m_1 goes down h/3, m_2 climbs up h/3

Thisis just kinematics then. v0 = 0, but the acceleration has to be the same since they are tied to a rope like you said