# Conservation momentum for y- axis

1. Sep 15, 2010

### inky

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

A ball P of mass m kg is dropped from a point A, which is 2 m vertically above a point B on a horizontal floor. After P hits the floor at B, it rebounds and hits another ball Q, of the same mass, which has also been dropped from A. The impact between the two balls is direct and takes place at the mid-point of AB. The coefficient of restitution in each impact is 5/7 . Neglecting air resistance, find the speed of P
(i) immediately after it hits the floor,
(ii) immediately after it collides with Q.

Ans: (i)4.52 m/s (ii) 3.74 m/s

2. Relevant equations

e=(velocities after collision)/(velocities before collision)

Law of conservation of momentum

3. The attempt at a solution

I tried to calculate this problem but some of them said my calculation is wrong because they mentioned that I can't use law of conservation of momentum for last part because of gravity. So how to calculate to get correct answer?Thanks a lot.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

#### Attached Files:

• ###### Impact between P and Q.doc
File size:
61.5 KB
Views:
43
2. Sep 15, 2010

### Delphi51

There is an implied assumption that the duration of the collision is zero. So the impulse applied to the masses F*Δt is zero during the collision and so the change in momentum m*Δv on each mass is zero.

If the time duration of the collision was nonzero, the collision force would have to be specified as a function of time. If it was a nice constant force you could work out the motion during the collision beginning with an F = ma formula for each mass. Looks complicated. The answer would not be 3.74 unless the duration was very short.

3. Sep 16, 2010

### inky

If the time duration of the collision was nonzero, the collision force would have to be specified as a function of time. If it was a nice constant force you could work out the motion during the collision beginning with an F = ma formula for each mass. Looks complicated. The answer would not be 3.74 unless the duration was very short.[/QUOTE]

maP=maQ
(-vP'-0.583)/t={-vQ'-(-4.47)}/t
vQ'-vP'=5.053

But from coefficient of restitution formula
vP'-vQ'=3.61

So I can't find the velocity. I think acceleration is incorrect.We need to consider gravity or not.
Actually this is CIE exam 2002W Question. Please see examiner report.Pls. see attachment.

If I consider v=u-gt for each ball then add , I can get one equation and another from e formula. So I got 0.423 ms-1. This answer is not the examiner' answer.

4. Sep 16, 2010

### inky

Please see examiner report.Pls. see attachment.

#### Attached Files:

• ###### Examiner Report.doc
File size:
69.5 KB
Views:
59
5. Sep 16, 2010

### Delphi51

Sorry, I don't understand your last post. You did get the 3.74 answer given in the latest attachment using the method described in the attachment. Done!

6. Sep 17, 2010

### inky

No. We could not use Law of conservation of momentum for falling objects because of the external force acting on it. I don't want the answer. I want correct method.

7. Sep 17, 2010

### Delphi51

My take on it is that, yes, you can use the law of conservation of momentum during the collision - if you assume the duration of the collision is very short. If it is not very short, then you do not have sufficient information to do the problem and it's answer would depend on the collision force function of time.