Conservation momentum for y- axis

In summary: 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.
  • #1
inky
99
1

Homework Statement



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

Homework Equations



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

Law of conservation of momentum

The Attempt at a Solution


Please see attachment.

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.
 

Attachments

  • Impact between P and Q.doc
    61.5 KB · Views: 205
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  • #2
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
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
Please see examiner report.Pls. see attachment.

 

Attachments

  • Examiner Report.doc
    69.5 KB · Views: 215
  • #5
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
Delphi51 said:
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!

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
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.
 

1. What is conservation momentum for the y-axis?

Conservation momentum for the y-axis is a principle in physics that states the total momentum of a system remains constant in the y-direction, unless acted upon by an external force. It is a fundamental law of motion that applies to all objects, from subatomic particles to planets.

2. How is conservation momentum for the y-axis related to Newton's third law?

Conservation momentum for the y-axis is closely related to Newton's third law, which states that for every action, there is an equal and opposite reaction. In the case of conservation momentum, the y-component of the momentum of an object is equal and opposite to the y-component of the momentum of the object it is interacting with.

3. Why is conservation momentum for the y-axis important in physics?

Conservation momentum for the y-axis is important in physics because it allows us to predict and analyze the motion of objects in a system. By understanding how the y-component of momentum is conserved, we can accurately calculate the velocities and trajectories of objects in motion.

4. Can conservation momentum for the y-axis be violated?

No, conservation momentum for the y-axis cannot be violated. This principle is a fundamental law of physics and has been extensively tested and proven to hold true in all known physical systems. Any apparent violation of conservation momentum for the y-axis is likely due to external forces acting on the system.

5. How does conservation momentum for the y-axis apply to real-world situations?

Conservation momentum for the y-axis applies to a wide range of real-world situations, from the motion of objects in space to the collisions of billiard balls on a table. It is a fundamental principle that is used in many fields, including engineering, astronomy, and sports, to understand and predict the motion of objects.

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