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Momentum and angular collision

  1. May 14, 2009 #1
    Q. Two objects A and B are moving on a frictionless horizontal surface. A has a mass of 6kg and speed of 3m/s , B has a *** of 8kg and a speed of 5m/s. if they collide and stick together the speed in m/s of the composite object after the collision is:

    diagram:

    A----------------------> X
    angle = 30->



    B is traveling at an angle towards the x mark


    so what i have been doing is i have been finding the before and after collision as x and y components. then putting the final xcomponent momemtum and y component momentums into this formula :total momentum after = (x^2 + y^2)^.5

    i would then have my total momentum before and then the total momentum after with v unknown. i got 4.02 for it. the question is multiple choice and these are the choices 3.23 16.2 2.71 4.02 or 3.76
     
  2. jcsd
  3. May 14, 2009 #2

    Doc Al

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    Staff: Mentor

    Sounds good.

    Looks like a match to me.

    It's not clear exactly where that 30 degree angle fits in. For example: If A moves along the +x axis, does B move at an angle of 30 degrees above the +x axis? If so, then you're OK.

    (Next time show every key step of your calculation--it makes it easier to follow.)
     
  4. May 14, 2009 #3
    pax= 6v1 pay = 0
    pbx= 8v2cos30 pby = 8v2sin30

    pcx= 6v1 +8v2cos30 = 12.9v pcy = 8v2sin30= 4v

    (12.9v^2 + 4v^2)^.5 = 13.5v

    so 58 = 13.5v

    v= 4.2

    a is moving along the x axis and b is moving below a but is heading towards a. the angle between them is 30
     
  5. May 14, 2009 #4
    any ideas what im doing wrong?
     
  6. May 14, 2009 #5
    sorry i meant i got 4.2 not 4.02 . so im obviously doing something wrong
     
  7. May 14, 2009 #6

    Doc Al

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    OK.

    Careful: v1 and v2 are given, and they are not equal. Redo this step, putting in values for v1 and v2.
     
  8. May 14, 2009 #7
    yes only before the collision are v1 and v2 are different. however the 2 balls ''stick together'' therefore the velocity is the same for both of them after the collision
     
  9. May 14, 2009 #8

    Doc Al

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    That's true. But you're calculating the total momentum before the collision, so you must use the speeds they had before the collision.
     
  10. May 14, 2009 #9
    which i did but im still getting 4.2 . thanks a million for helping me with this by the way. still not sure what im doing wrong tho?
     
  11. May 14, 2009 #10

    Doc Al

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    Redo the steps, using v1 and v2. Show me what you get.
     
  12. May 14, 2009 #11
    AW DUDE! couldnt we just use m1xv1+ m2xv2 = m1xv1+m2xv2 ? i think the angle might of been in the question to put u off???
     
  13. May 14, 2009 #12

    Doc Al

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    No. That only works for collisions in one dimension.
    You were on the right track. Just redo it.
     
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