Notices
Results 1 to 8 of 8

Thread: Elastic collision

  1. #1 Elastic collision 
    Forum Freshman
    Join Date
    Feb 2008
    Posts
    14
    Ok... This one is bugging me...

    (+) - Proton


    Ok, so we need to find the Velocity for W and V as shown in the image...
    u=100m.s<sup>-1</sup>

    Here is what I did:
    if u = 100m.s<sup>-1</sup> with the angles 30° and 60° the collision is elastic because the angle between the two objects is 90°.

    Wx/W = Cos(60)
    Wx = W.Cos(60)

    ---------------
    X-Dir
    ---------------
    ∑<sub>p before</sub> = ∑<sub>p after</sub>
    m<sub>1</sub>V<sub>1</sub>+m<sub>2</sub>V<sub>2</sub> = m<sub>1</sub>V<sub>1</sub>+m<sub>2</sub>V<sub>2</sub>
    (÷m) 100 = W.cos(60) + V.cos(30)
    (1) 200 = W + √(3).V

    ---------------
    Y-Dir
    ---------------
    ∑<sub>p before</sub> = ∑<sub>p after</sub>
    m<sub>1</sub>V<sub>1</sub>+m<sub>2</sub>V<sub>2</sub> = m<sub>1</sub>V<sub>1</sub>+m<sub>2</sub>V<sub>2</sub>
    (÷m) 0 = W.sin(60) + V.sin(30)
    (2) 0 = √(3).W + V

    ---------------
    (2) in (1)
    ---------------
    200=W+√(3).(√(3)W)
    200=2W
    W=100m.s<sup>-1</sup>

    But thats not true. That says that the 1st one stands still and the 2nd one moves. That means that this is 1Dimentional... but its not...

    I'm confused... Can anyone please help...

    Thanks


    Need help with homework?
    http://www.homeworkhelp.co.za/forum
    Reply With Quote  
     

  2.  
     

  3. #2  
    Forum Masters Degree bit4bit's Avatar
    Join Date
    Jul 2007
    Posts
    621
    So you have a proton moving with initial velocity u=100ms<sup>-1</sup>, which collides elastically with a stationary particle, where after collision, the velocity of the particle is w, and the velocity of the proton is v? Is that what you're describing?

    If so, you know that momentum is conserved before and after the collision, both horizontally and vertically. Therefore with u=<u<sub>x</sub>, u<sub>y</sub>>, v=<v<sub>x</sub>,v<sub>y</sub>>, and w=<w<sub>x</sub>,w<sub>y</sub>> being the three velocity vectors, and m<sub>1</sub> and m<sub>2</sub> being the masses of the proton and the other particle respectively, you have:

    for x:

    m<sub>1</sub>u<sub>x</sub>=m<sub>1</sub>v<sub>x</sub> + m<sub>2</sub>w<sub>x</sub>
    100m<sub>1</sub>=m<sub>1</sub>vcos30 + m<sub>2</sub>wcos300
    100m<sub>1</sub>=(1/2)*√3*v*m<sub>1</sub> + (1/2)*w*m<sub>2</sub> (1)

    for y:

    m<sub>1</sub>u<sub>y</sub>= m<sub>1</sub>v<sub>y</sub> + m<sub>2</sub>w<sub>y</sub>
    0=m<sub>1</sub>v<sub>y</sub> + m<sub>2</sub>w<sub>y</sub>
    0=m<sub>1</sub>vsin30 + m<sub>2</sub>wsin300
    0=(1/2)*v*m<sub>1</sub> - (1/2)*√3*w*m<sub>2</sub> (2)

    From equation (1) and (2) you might be able to go ahead and solve this by simultaneous equations, but it's worth asking what other information you have about the scenario (masses velocities etc) first. And you are trying to calculate w right??

    When you are dividing your expressions by "m", then are you assuming that m<sub>1</sub> = m<sub>2</sub>? If not then you cant get rid off the m<sub>1</sub> AND m<sub>2</sub> terms by diving by ONE of them. If you write down the exact question, then we can help you more.


    Reply With Quote  
     

  4. #3  
    Forum Freshman
    Join Date
    Feb 2008
    Posts
    14
    Sorry, didnt see you edited your post after the food...

    Well it was kinda explained in class. So I mostly have it in my head.

    Its basicaly
    u is 100m.s-1
    Then the 30° and 60° added up give 90 which makes it a perfectly elastic collision.
    All the protons have the same mas
    Because the chance a proton has to actually collide with an air molecule is so small we say that it is frictionless so we can use conservation of momentum.

    Think basic. Nothing big.

    So now i need to work out the x and y speed for V and W...

    But I keep getting W=100... tried like 3 different ways already and now im just stuck...

    Thanks
    Need help with homework?
    http://www.homeworkhelp.co.za/forum
    Reply With Quote  
     

  5. #4 Re: Elastic collision 
    Suspended
    Join Date
    Apr 2007
    Location
    Pennsylvania
    Posts
    8,795
    Quote Originally Posted by Dark_Fire
    (÷m) 0 = W.sin(60) + V.sin(30)


    ---------------
    One of those has to be negative.
    Reply With Quote  
     

  6. #5  
    Forum Masters Degree bit4bit's Avatar
    Join Date
    Jul 2007
    Posts
    621
    OK, if both particles are protons, then they will have the same mass and the collision will be elastic as in generally the case for sub-atomic particles.


    So from my working above take (1) and (2) and if m<sub>1</sub>=m<sub>2</sub>, then we can just call this 'm'. Divide (1) and (2) by 'm' to obtain:

    (1) / m ----> 100 = (1/2)*√3*v + (1/2)*w ------(3)
    (2) / m ----> 0 = (1/2)*v - (1/2)*√3*w ----------(4)

    Now we can solve the equations simultaneously by making the 'v' terms equal. To do this mulitply (3) by (1/2), and (4) by (1/2)*√3, to obtain:

    (3)*(1/2) --------> 50 = (1/4)*√3*v + (1/4)*w -----(5)
    (4)*(1/2)*√3 ---> 0 = (1/4)*√3*v - (3/4)*w --------(6)

    Then subtract the equations:

    (5) - (6) ----> 50 = (1/4)*w + (3/4)*w = w( 1/4 + 3/4 ) ----> w = 50 ms<sup>-1</sup>

    You can then put this back into one of the equations to get a value for v, and then put both back into the original to find the mass.

    Alternatively, the mass of a proton = 1.673×10<sup>-27</sup> kg, so you could use that too.
    Reply With Quote  
     

  7. #6  
    Forum Freshman
    Join Date
    Feb 2008
    Posts
    14
    Wow. It actually worked.

    I don't quite understand why you multiplied with 1/2 and 1/2 root3...

    Ill look at it again in the morning thou. But currently Im still confused. But good thing is your magic touch works. Thanks a million.

    I just like to know why instead of just use it. I like to learn. Im addicted to knowledge...
    Need help with homework?
    http://www.homeworkhelp.co.za/forum
    Reply With Quote  
     

  8. #7  
    Forum Masters Degree bit4bit's Avatar
    Join Date
    Jul 2007
    Posts
    621
    Glad to help

    The reason I multiplied one by (1/2) and one by (√3/2) is to make all the terms in the equation with a 'v' in them, equal....so that it's then possible to get rid of the 'v' terms when we subtract the equations, and are just left with something in terms of w, which is easily solved. In general you can cross-multiply co-efficients from the y-terms in each equation leaving you with equal terms to subtract. Have you done simultaneous equations? They almost always come in helpful with these momentum questions.
    Reply With Quote  
     

  9. #8  
    Forum Freshman
    Join Date
    Feb 2008
    Posts
    14
    Jup. We have done them.

    And I can't thank you enough. Its funny How you can sometimes do the smallest thing wrong and still being in that mind set you'll keep doing it wrong as far as you go. :P i now realize How stupid I was. Hehe.

    Thanks again.
    Have a wonderfull week.
    Need help with homework?
    http://www.homeworkhelp.co.za/forum
    Reply With Quote  
     

Bookmarks
Bookmarks
Posting Permissions
  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •