1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: 2D inelastic collision

  1. Apr 29, 2014 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    conservation of total momentum

    M1V1+M2V2 = MtotVf

    3. The attempt at a solution

    I tried using the COTM

    as such:

    x: M1V1iCOSΘ1+M2V2SINΘ2=MtotVf

    y: M1V1iSINΘ1+M2V2COSΘ2=MtotVf

    then dividing 1/2:

    COTΘ1 + TANΘ2 = 1

    Θ2 = TAN-1(1-COTΘ1)

    After plugging in the values im getting Θ2 = -10.85°

    What am I doing wrong?
  2. jcsd
  3. Apr 29, 2014 #2
    I think the main issue is in setting both the x and y equations equal to ##M_{tot}V_{f}##, since these are only component equations, so neither will equal the entire momentum. Instead, you'll want to use the components of the total momentum, so ##M_{tot}v_{f,x}## and ##M_{tot}v_{f,y}##
  4. Apr 29, 2014 #3
    I cant see that working because the ending vector is directly along the x-axis, there is no sin or cos involved with that vector.
  5. Apr 29, 2014 #4
    No, it works. The angle from the vertical will just be 0 degrees, so you'll have something like:

    ##P_{x} = P_{tot} \cdot cos(0)## and
    ##P_{y} = P_{tot} \cdot sin(0)##
  6. Apr 29, 2014 #5
    so the equation becomes TANΘ1+COTΘ2=0


    Θ2= -.025°
  7. Apr 29, 2014 #6
    Well, you can't divide the equations like that.
    $$\frac{a + b}{c + d} \neq \frac{a}{c} + \frac{b}{d}$$
  8. Apr 29, 2014 #7
    Ok so before dividing I have

    x: M1V1iCOSθ1+M2V2iSINθ2=MtotVfCOS(0)

    y: M1V1iSINθ1+M2V2iCOSθ2=MtotVfSIN(0)

    look legit?
  9. Apr 30, 2014 #8
    Yes, that looks good. I think the easiest way to solve would be to move the components with ##M_{1}v_{1}## over, and then you could divide the equations.
  10. Apr 30, 2014 #9
    Hey there!

    $$V_{total} = \sqrt{V_{x}^2+V_{y}^2}$$


    $$\theta = tan^{-1}(\frac{V_{y}}{V_{x}})$$

    These seem to be the only two equations that you are forgetting about. As long as you can solve for both component of your resultant velocities, you should be good!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted