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Collisions mechanics question

  1. Apr 16, 2012 #1
    A smooth spherical particle with mass 2.5 kg collides with a second smooth spherical
    particle of mass 6kg. Before the collision, the first particle has velocity 9ms^1
    in the positive x-direction while the second particle has velocity 6ms^1 in the positive y-direction. At the instant of collision, the particles are so that the normals to their surfaces at the point of contact are ±n, where n
    is a unit vector inclined at -pi/4 from the x-axis as shown. You may assume that the coefficient of restitution e = 2/3 and that no other forces are involved. Determine the velocities v1,1 and v2,1 of the first and second particles respectively, immediately after the collision.
     
  2. jcsd
  3. Apr 16, 2012 #2

    tiny-tim

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    welcome to pf!

    hi ly667! welcome to pf! :wink:

    show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
     
  4. Apr 16, 2012 #3
    I dont really have much of an idea to be honest, nor do my classmates. We have been looking at it all night :(
    Basically what I have so far is.. I have constructed my diagram.
    Particle 1: m1=2.5, v1=9
    Particle 2: m2=6, v2=6

    Particles interact along line, to n=v1-v2
    we also need m, a vector perpendicular to n
    m=v1+v2 (so m.n=0)

    Physical assumptions: momentum conservation
    coeff of rest 2/3
    smooth collisions
     
  5. Apr 16, 2012 #4

    tiny-tim

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    this is an inelastic collision

    so you need conservation of momentum in two perpendicular directions (either x and y, or n and z×n)

    and one more equation, which will be the coefficient of restitution equation

    (the coefficient of restitution is the ratio of the relative speed after to the relative speed before the collision, see the pf library)

    show us what you get :smile:
     
  6. Apr 16, 2012 #5
    Okay, so far I have

    Mathematically condition i.) gives
    m1,v1,0+m2v2,0=m1,v1,1+m2v2,1

    ie. 2.5(v1,1)+6(v2,1)=22.5+36 = 58.5

    while ii.) gives(v2,1-v1,1)= e(v2,0-v1,0).n

    and iii.) gives (v2,1-v1,1).m= (v2,0-v1,0).m

    =0 since n=v2,0-v1,0

    Its easiest to work in components

    let v2,1=9u2+6w2
    and v1,1=9u1+6w1

    momentum cons becomes

    2.5(9u1)+2.5(6w1)+6(9u2)+6(6w2)=58.5

    I am stuck and dont know what to do from here.. Is what I have done correct?
     
  7. Apr 16, 2012 #6

    tiny-tim

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    ah, no wonder you're stuck, your professor obviously hasn't explained to you the vector nature of conservation of momentum :redface:

    your 58.5 is the sum of the x and y components of momentum …

    you can't do that!!

    momentum is a vector, and obeys the laws of vector addition

    you can't add components in different directions

    write out two equations, one for conservation of momentum in the x direction, and one for conservation of momentum in the y direction :smile:
     
  8. Apr 16, 2012 #7
    So should I use my velocity value as vectors, ie instead of v1=9, let it equal 9i?
     
  9. Apr 16, 2012 #8

    tiny-tim

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  10. Apr 16, 2012 #9
    Okay :) thank you!! Will do that now! Could you please help me with my collisions question? I have gotten a good bit of it done but am stuck at a point!
     
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