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Mechanics Question of two balls

  1. Apr 20, 2006 #1
    Mechanics Question :)

    Hi.

    Two small balls A and B have masses 0.5kg and 0.2 kg respectively. They are moving towards each other in opposite directions on a smooth horizontal table when they collide directly. Immediately before the collission, the speed of A is 3m/s and the speed of B is 2 m/s. The speed of A immediately after the collision is 1.5 m/s. The direction of motion of A is unchanged as a result of the collision.

    By modelling the balls as particles, Find;
    (a) the speed of B immediately after the collision.

    (b) the magnitude of the impulse exerted on B in the collision.


    ----
    For part (a) i am getting 2.3 m/s.. however my teacher's worked solution is giving 1.75 m/s.... now.. i dont know which one is correct...

    for part B... i use Ft=mv-mu and get -0.75 ... i think im wrong....

    Please help me out here.
    Thanks
    Faraz
     
  2. jcsd
  3. Apr 20, 2006 #2

    Doc Al

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    Staff: Mentor

    Show how you solved it and we can take a look.
     
  4. Apr 20, 2006 #3

    J77

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    conservation of momentum

    Your teacher seems right...

    0.5*3-0.2*2=0.5*1.5+0.2*X

    where X is the unknown speed
     
    Last edited: Apr 20, 2006
  5. Apr 20, 2006 #4
    HI there..
    why is it 0.5 x3 MINUS 0.2*2 ?
    because my 2.3 comes from adding these two....
     
  6. Apr 20, 2006 #5

    J77

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    They're travelling in opposite directions initially...
     
  7. Apr 20, 2006 #6
    ah.. thanks a lot J77... clear now.

    any idea about part (b)?
    im quite sure my -0.75 is wrong
     
  8. Apr 20, 2006 #7

    J77

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    Well, impulse = change in momentum...

    momentum before = 0.2 * 2

    momentum after = 0.2 * 1.75

    implies impulse = 0.2 * (2 - 1.75) = 0.2 * 0.25 = 0.05
     
  9. Apr 20, 2006 #8

    Doc Al

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    Staff: Mentor

    Why do you have a minus sign? What direction is the impulse on B?

    Also, regardless of direction, they asked for the magnitude of the impulse.
     
  10. Apr 20, 2006 #9
    I= 0.2 x (2+1.75)= 0.75Ns
    isnt this the answer?

    I just looked at the mark scheme and it confirms that.... :confused:
     
  11. Apr 20, 2006 #10
    Momentum is a vector, so you must consider the direction. Based on your working for (a), B moves in the positive direction after the collision, and in the negative direction before the collision.

    Do you now understand how the correct answer of 0.75 Ns is obtained?

    P/S Impulse on B = Final momentum of B - Initial momentum of B
     
  12. Apr 20, 2006 #11
    Thanks a lot... i get it now...
    :)
     
  13. Apr 20, 2006 #12

    Doc Al

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    That's correct. Your only mistake (in your earlier answer) was including the minus sign.

    PS: J77 made an error in post #7--he had the wrong sign for the initial momentum. (This is the same error that he pointed out to you in your solution to part a.)
     
  14. Apr 21, 2006 #13

    J77

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    :redface: :redface: :redface:

    oops!
     
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