Physics Help Forum momentum- really urgent

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 Aug 3rd 2008, 08:07 AM #1 Junior Member   Join Date: Jul 2008 Posts: 4 momentum- really urgent state the principle conservation of linear momentum and discuss , in the case of a single body, how thi s principle is consistent with the first law of newton? what is meant by perfect elastic collision? the below diagram shows the initial ppostion of two bodies A n B respectively of mass (a) and mass ( b) . A is released from rest and it undergoes elastic collision with B. 1) if M(a) = M (b) and the vertical displacement for B after collision is L/2, show that the speed of A is zero instantly after the collision with B and find the value of cos (alpha). 2) if M(b) > M(a), describe the motion of A subsequent to the collision with B. 3) if M(a) > M ( b), what is the ratio of angular speed of A after collision with b/ angular speed of A before collision with B
Aug 4th 2008, 07:06 PM   #2
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 Originally Posted by qweiop90 state the principle conservation of linear momentum and discuss , in the case of a single body, how thi s principle is consistent with the first law of newton? what is meant by perfect elastic collision?
In fact, the rule of conservation of linear momentum states that without applying external forces on a system, the total momentum before and after should be the same.

Some books would introduce two terms, " perfectly elastic collision" and " elastic collision", frankly , they are actually the same. They both refer to the collision with no loss in kinetic energy in the whold system before and after .

But for the question you asked, maybe you should check the link of picture first as I cannot see the image.

 Aug 5th 2008, 10:21 AM #3 Junior Member   Join Date: Jul 2008 Posts: 4 by the way how to attach the pic?
Aug 5th 2008, 05:28 PM   #4

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 Originally Posted by qweiop90 by the way how to attach the pic?
Go to the page where you are posting a reply/creating a new thread. Go down below the "Submit Reply" and "Preview Post" buttons. The second box down in the "Additional Options" pane is for "Managing Attachments." You can click on that and a menu will come up.

Alternately if the file is on the web you can insert a weblink as an image if you click the button on the toolbar above (the one that starts off "B I U" etc.) The sixth button from the right looks like a little landscape. That will insert an image from a web page.

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Aug 6th 2008, 12:31 AM   #5
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huh ok. here is the pic
Attached Images
 phy.bmp (576.1 KB, 4 views)

Aug 7th 2008, 10:41 PM   #6
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 Originally Posted by qweiop90 the below diagram shows the initial ppostion of two bodies A n B respectively of mass (a) and mass ( b) . A is released from rest and it undergoes elastic collision with B. 1) if M(a) = M (b) and the vertical displacement for B after collision is L/2, show that the speed of A is zero instantly after the collision with B and find the value of cos (alpha). 2) if M(b) > M(a), describe the motion of A subsequent to the collision with B. 3) if M(a) > M ( b), what is the ratio of angular speed of A after collision with b/ angular speed of A before collision with B
Sorry if I have answered it wrongly.

Finding $\displaystyle cos \alpha$, I guess it is 2.5/3 which is 5/6 after simplification.
As after collision, the vertical displacement of B is L/2. Constructing a simple triangle, we would see the hypotenue which always equals to 3L and the adjacent side equatls( 3L-0.5L)=2.5L
Sorry that I don't know the way to prove velocity is zero.

For question 2, due to conservation of momentum, if the momentum of b is greater than that of a, then momentum of a after collision should be negative keeping momentum conserved. Then a should swing backwards.

For question 3, is there something related to mgh= mv^2/2 ?Or maybe related to SHM. I 'm not quite sure.Sorry.

By the way, does the notation M(a)>M(b) means the momentum of a is greater than that of b after collision?

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