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 problem -- The collision of two particles with different masses

  1. Apr 30, 2015 #1
    1. The problem statement, all variables and given/known data
    So I have this practice problem with the solution, but I don't understand how:

    "A particle of mass m1 and speed v1 in the +x direction collides with another particle of mass m2. Mass m2 is at rest before the collision occurs, thus v2 = 0. After the collision, the particles have velocities v'1 and v'2 . in the xy plane with directions θ1 and θ2 with the x axis as shown below. There are no external forces acting on the system. Express all of your answers in terms of m1, m2, v1, θ1 and θ2.

    Q:What is the ratio of the speeds v'2/v'1 ?"

    The answer goes like this:

    "Conservation of momentum in the y direction: m1v'1 sin θ1 = m2v'2sin θ2
    and by solving for ratio we get:

    m1sinθ1/m2sinθ2"

    Why did we choose the conservation in the y not the x direction?
    Can this problem be solved any other way?


    2. Relevant equations

    -

    3. The attempt at a solution

    When I first attempted the problem, I thought of solving it like this:
    v1'=v1'(sqrt[(cos(θ1)+sin(θ1))^2])
    v2'=v2'(sqrt[(cos(θ2)+sin(θ2))^2])
    And since momentum is conserved,
    m1v1=m1v1'(sqrt[(cos(θ1)+sin(θ1))^2])+m2v2'(sqrt[(cos(θ2)+sin(θ2))^2])

    But this feels wrong, and I don't know what's with it.
    Please feel free to correct me and give me more info about momentum and collision if needed, I understand it but I don't feel like having a full grasp of the concept.
     
  2. jcsd
  3. Apr 30, 2015 #2
    I don't understand how you got these equations?
     
  4. Apr 30, 2015 #3

    gneill

    User Avatar

    Staff: Mentor

    With no external forces acting the momentum of a system is conserved. Thus the momentum of the system after the collision must be identical to the momentum before the collision.

    Momentum is a vector quantity (magnitude and direction). It is conserved separately in both x and y directions and in total. That means if the initial momentum had no y-component (as in this problem), the final momentum must also have no y-component when you sum up the contributions from all the moving parts. Similarly, the final momenta in the x-direction must sum to what it was before the collision.

    The given solution for determining the ratio of the speeds V2'/V1' took advantage of the fact that the total momentum in the y-direction happens to be zero.
     
  5. Apr 30, 2015 #4
    Sorry ignore that, I realised I wasn't thinking correctly.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Momentum problem -- The collision of two particles with different masses
Loading...