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Is this correct- momentum question

  1. Oct 31, 2007 #1

    klm

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    A 20 g ball of clay traveling east at 2.0m/s collides with a 30 g ball of clay traveling 30 degree south of west at 1.0m/s.
    What are the speed of the resulting 50 g blob of clay?

    so this is what i i did:

    mivi=mfvf
    vx= (.02)(2) / (.05) = .8
    vy= (.03)(-1)/ .05 = -.6

    is that much correct so far? and then i can use vy= v sin theta to find v?
     
  2. jcsd
  3. Oct 31, 2007 #2

    mjsd

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    hard to understand what you have done with your notations.
    but this is clearly a conservation of momentum question. so remember this is a 2D problem, so both total momentun in x and y direction must be the same before and after.
     
  4. Oct 31, 2007 #3

    klm

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    sorry for the messy calculations!

    i am basically finding the Vx = ( mass of ball 1)( velocity of ball 1) / ( total mass )
    so Vx = ( 0.02 kg) ( 2 m/s ) / ( 0.05 kg ) = 0.8

    and Vy = (.03 kg) ( -1 m/s) / (.05 kg) = -.6
    (is the bold part in this equation correct? )
    but i am a little confused if i can use this equation to find V now, Vy= v sin(theta)
     
  5. Oct 31, 2007 #4

    mjsd

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    As far as i understand it. you have two momentum vectors to begin with and you need to find the resultant vector. so due to momentum conservation, you get
    [tex]\vec p_1 + \vec p_2 = \vec p_f[/tex]
    where [tex]\vec p_i = m_i v_i[/tex]
    the speed that you are after is really [tex]|\vec v_f|[/tex]
     
  6. Oct 31, 2007 #5

    klm

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    so i dont need the components then? or did i find the components of Vf..?
     
  7. Oct 31, 2007 #6

    mjsd

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    do you know how to handle vectors? do u know how to add them?

    first resolve everything into component forms and get two equations: one for each x and y
     
  8. Oct 31, 2007 #7

    klm

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    okay for the first ball: Vx= 2 and Vy= 0
    second ball: Vx= -1cos(30)
    Vy= -1sin(30)

    would the be correct?
     
  9. Oct 31, 2007 #8

    HallsofIvy

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    Let me see if I can give more detail about what you have done.
    You have one ball of 20 g (0.02 kg) moving east at 2 m/s. It's scalar momentum is 0.02(2)= 0.04 kgm/s so the vector momentum is [itex]0.04\vec{i}[/itex]. I can see no reason to divide by total mass.

    You have one ball of 30 g (0.03 kg) moving south at 1 m/s. Its scalar momentum is 0.03(1)= 0.03 kgm/s so the vector momentum is [itex]-0.03\vec{j}[/itex].

    The total momentum vector before collision is [itex]0.04\vec{i}- 0.03\vec{j}[/itex].

    The combined balls must have the same momentum vector, so letting [itex]\vec{v}[/itex] be its velocity vector, [itex](0.02+ 0.03)\vec{v}= 0.04\vec{i}- 0.03\vec{j}[/itex].

    NOW you can get the velocity vector by dividing that equation by the total mass 0.05 kg: [itex]\vec{v}= 0.80\vec{i}- 0.06\vec{j}[/itex]

    The speed is the length of that vector.
     
  10. Oct 31, 2007 #9

    klm

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    okay i think i understand what you did...but i thought for some reason you should divide by the total mass, but i think we somehow got to same place either way. but one question, should that 0.06 be 0.6 ( just want to make sure, not trying to be picky :)! so the speed would just be .8^2 + .6^2 and then the square root of that number.
     
  11. Oct 31, 2007 #10

    klm

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    because i seem to be getting 1 and that is incorrect...
     
  12. Oct 31, 2007 #11
    That looks good to me! (Assuming +x to be along east direction and +y along north.)
    Go ahead...
     
  13. Oct 31, 2007 #12

    klm

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    ok so for ball 2 Vx= -.866 and Vy= .5 and then the magnitude of that is .999
    and ball 1 the magnitude is just 2
     
  14. Oct 31, 2007 #13

    klm

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    so if i have to do m1v1+ m2v2 :
    (.02)(2) + (.03)(1)
     
  15. Oct 31, 2007 #14
    no... u have to conserve linear momentum in each direction seperately! (remember, momentum is a vector quantity.. so treat it as a vector... in 1-D, since it can be only positive or negative direction, you used to get result even considering as scalar -- but with apprpriate positive and negative signs.)
     
  16. Oct 31, 2007 #15
    Of course it would be 2 for 1st ball, and 1 for 2nd ball. From this only you got these components!
    Maybe it's a typing mistake.. but vy = -.5. Check it.
     
  17. Oct 31, 2007 #16

    klm

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    oh so is it -1
    so .04i - .03j
     
  18. Oct 31, 2007 #17

    klm

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    haha yeah sorry. it is -.5, i forgot the neg sign sorry!
     
  19. Oct 31, 2007 #18
    Conservation of linear momentum
    In x-direction: m1*v1,x + m2*v2,x = (m1 + m2)*vf,x
    In y-direction: m1*v1,y + m2*v2,y = (m1 + m2)*vf,y
     
  20. Oct 31, 2007 #19

    klm

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    okay sorry this might be a dumb question, but why do we need to break up the conservation into components?
     
  21. Oct 31, 2007 #20

    klm

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    okay so i got .2804 for vfx and -.3 for vfy and then i just do square them both and take the square root and get .411 which should be the speed correct?
     
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