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Ball explosion

  1. Oct 28, 2011 #1
    An object with total mass mtotal = 14.4 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.9 kg moves up and to the left at an angle of θ1 = 22° above the –x axis with a speed of v1 = 26.8 m/s. A second piece with mass m2 = 5.1 kg moves down and to the right an angle of θ2 = 27° to the right of the -y axis at a speed of v2 = 20.9 m/s. The third mass m3 = 4.4kg.

    What is the x-component of the velocity of the third piece?

    What is the y-component of the velocity of the third piece?


    What is the magnitude of the velocity of the center of mass of the pieces after the collision?

    Calculate the increase in kinetic energy of the pieces during the explosion.

    Any help would be wonderful and greatly appreciated Thank you.
     
  2. jcsd
  3. Oct 28, 2011 #2
    what have you tried? what's not working for you?
     
  4. Oct 29, 2011 #3
    Im not sure where to start this problem. Which equation to use.
     
  5. Oct 29, 2011 #4
    Momentum is going to be conserved, so the momentum of the ball before it explodes is equal to the sum of the momentums of the 3 pieces
     
  6. Oct 29, 2011 #5
    So m1v1+m2v2+m3v3/m1+m2+m3
     
  7. Oct 30, 2011 #6
    If the object was initially at rest then its momentum was zero, that means that:
    [tex]
    m_1v_1 + m_2v_2 + m_3v_3 = 0
    [/tex]
    It also means that the horizontal components of momentum will add to give zero as will the vertical components.
     
  8. Oct 30, 2011 #7
    I already tried zero and it didnt work. I dont know what else to try.
     
  9. Oct 30, 2011 #8

    Doc Al

    User Avatar

    Staff: Mentor

    Show exactly what you did.

    What did you get for the x and y components of the momentum of m1 and m2?
     
  10. Oct 30, 2011 #9
    are you trying to solve a simultaneous system of equations of 2 equations and 2 unknowns?

    Because I think that's where the solution is...you need to take that equation given by JHamm and break it down into x and y components, along with the unknown x and y components of the 3rd mass...don't worry about the angle of it, just yet, just get the x and y components of the velocity and THEN you can do a vector sum of those to get the result.
     
  11. Oct 30, 2011 #10
    My home work is aon a site called smart physics. I can in put an answer and it will tell me if im right or wrong. I put in zero and it didn't work. I dont know which equation to use or how to find which equation to use.
     
  12. Oct 30, 2011 #11
    actually, it's not even two equation two unknowns...it should be one equation one unknown, twice...very simple...why don't you show what you are trying to do?
     
  13. Oct 31, 2011 #12
    I just tried to input zero with any equation. The reason i did this was because i don't know what to do. So i was just tring to see if any of the answers would be zero. I don't know which equation to use to start these problems
     
  14. Oct 31, 2011 #13
    It is not a matter of guessing an answer, punch it in the site and see if it is right or wrong...you actually need to solve a couple of equations...have you understood this part yet?

    You have been told that the sum of the momentum from all 3 pieces was zero before the explosion and will remain zero after the explosion.

    The equation you need to work with has been already posted and you were even told that all you have to do is break it down into x-components and y-components and that each equation still equals zero, from where you should be able to come up with an answer...

    ...this is a lot of explanation...longer that would have taken me to solve the problem.

    hope this helps...
     
  15. Oct 31, 2011 #14
    gsal; I don't think you did a good job explaining. I too am at a loss on what to do and I have found the velocity of the third particle. What I dont know what to do is how to actually break it down into the x-component and the y-component!!!
     
  16. Oct 31, 2011 #15
    For the y component you multiply the velocity by the sine of the angle it makes with the x axis, for the x component you multiply the velocity by the cosine of that angle.
    In other words the first one is moving at 26.8m/s at an angle of 22 degrees, so...
    [tex]
    V_1_x = -26.8\cos{22^o}
    [/tex]
    The negative is there because it is moving towards the negative x direction.
    [tex]
    V_1_y = 26.8\sin{22^o}
    [/tex]
     
    Last edited: Nov 1, 2011
  17. Oct 31, 2011 #16
    Where did you get -28.6? I tried -28.6cos(22)/26.8= x
    This is what you ment right?
     
  18. Nov 1, 2011 #17
    Sorry, it should read 26.8 not 28.6, I've fixed it now
     
  19. Nov 1, 2011 #18
    V_1_x = -26.8\cos{22^0}
    I tried to do this equation in my calculator -26.8 divided by cos(22^0). Is this what you ment? Im sorry for all the questions im tring to really understand physic.
     
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