1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Momentum and angles problem

  1. Dec 3, 2004 #1
    I had made an error on my previous post so i will take it back to the beginning. I can't get this one for some reason.

    First look at the diagram: http://bosco.iwarp.com/diagram.jpg

    OK here's the problem:

    Two identical balls (mass m) undergo a collision. Initially one ball is stationary, the other has kinetic energy of 8J. The collision is partially inelastic with 2 J energy converted to heat. What is the maximum deflection angle (alpha or beta) at which one of the balls is observed?

    I have come up with the following relationships using conservation of momentum and energy:

    1. Vo = V1 cos a + V2 cos B

    2. 0 = V1 sin a - V2 sin B

    and 12 = m (V1f^2 + V2f^2) which i believe can be written

    3. 0.75Vo = V1^2+V2^2

    by rearranging initial kinetic energy of ball one to solve for m.

    How do i go about comparing alpha and beta? Are my equations correct
    Help please! thanks
     
    Last edited: Dec 3, 2004
  2. jcsd
  3. Dec 3, 2004 #2

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    1. and 2. are fine. 3. should have Vo^2:

    This is not a trivial problem to solve. Use 1 and 2. to find [itex]cos\beta \text{ in terms of } sin\alpha \text{ and } cos\alpha[/itex]

    Then look at the ranges of values that [itex]cos\beta[/itex] can have. There should be a [itex]\sqrt{.75}[/itex] term in there somewhere.

    AM
     
  4. Dec 4, 2004 #3
    Well i have attempted this another way and have got the following:

    [itex]2 = mv_{1}v_{2} (cos \alpha }+ \beta)[/itex]

    now i do not know how to compare [itex]\alpha[/itex] and [itex]\beta[/itex]

    any help
     
  5. Dec 4, 2004 #4
    ??????????
     
  6. Dec 4, 2004 #5

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    It has to have [itex]\beta[/itex] in terms of [itex]\alpha[/itex] ONLY. You can do this with these equations: determine v2f in terms of v1f using 2. and substitute into 1. to find v1f in terms of [itex]\alpha \text{ and } \beta[/itex]. Substitute also into 3. to find v1f. [itex](v_0^2 = 16/m)[/itex]. Then combine the two to find [itex]beta[/itex] in terms of [itex]\alpha[/itex] (the m falls out).

    AM
     
  7. Dec 5, 2004 #6
    I dont know why this is not working out but after trying what you said i got:

    [tex] 3sin^2\beta (cos^2\alpha + 2cos\alpha + cos^2\beta) = 4(sin^2\beta+sin^2\alpha)[/itex]

    It should probably be simpler shouldnt it?

    I really want to get this one done but it just isnt working out
    Thanks for all your help on this one
     
    Last edited: Dec 5, 2004
  8. Dec 5, 2004 #7
    ???????????
     
  9. Dec 6, 2004 #8

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    I told you it was non-trivial. Have a look at the solution for an elastic collision at: http://rustam.uwp.edu/201/L12/lec12_w.html (scroll down to the collision in 2 dimensions). For elastic collisions the two angles add to 90 degrees. The question is asking how the sum of the two angles in an inelastic collision where 25% of the energy is lost compares to 90 degrees.

    AM
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Momentum and angles problem
  1. Momentum with an angle (Replies: 8)

  2. Momentum and Angles (Replies: 3)

Loading...