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Tranfer of Energy

  1. Oct 26, 2006 #1
    A 50 kg boy runs at a speed of 10.0 m/s and jumps onto a cart originally at rest. The cart, withe the boy on it, then takes off in the same direction in which the boy was running. If the cart with the boy has a velocity of 2.50 m/s what is the mass of the cart?

    I need the equation to find the Mass of the cart. Please help:cry:
     
  2. jcsd
  3. Oct 26, 2006 #2

    OlderDan

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    What physical principle do you think applies to this situation?
     
  4. Oct 26, 2006 #3

    PhanthomJay

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    What is the definition of momentum?

    Think conservation of momentum. What is the initial momentum of the boy and cart? What is final momentum of boy and cart after he jumps on?
     
  5. Oct 26, 2006 #4
    I have the equation that starts

    m1v1 = m1v1' + m2v2'
     
  6. Oct 26, 2006 #5
    I first work it like m2v2 = m1v1 but not sure it that was right
     
  7. Oct 26, 2006 #6

    PhanthomJay

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    It is actually m1v1 +m2v2 = m1v1' + m2v2'. but here, v2 is 0, so you're equation is sort of OK. But what can you say about v1' and v2'? How does each relate to the given v_f = 2.5?
     
  8. Oct 26, 2006 #7
    v1' is initial velocity which equals 0, and v2' is the the combonation of the boy and the cart velocity which equals 2.50
     
  9. Oct 26, 2006 #8

    PhanthomJay

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    No, its total initial momentum = total final momentum. v1 is the initial velocity of the boy(10). v2 is the initial velocity of the cart(0). v1' is the final velocity of the boy, and v2' is the final velocity of the cart. The boy and the cart move together after he jumps on it. So what's v1' and v2'?
     
  10. Oct 26, 2006 #9
    Oh ok I think I am with you, the both are 2.50 m/s
     
  11. Oct 26, 2006 #10
    so would you work it out like

    m1v1 - m1v1'/ v2' =m2
     
  12. Oct 26, 2006 #11

    PhanthomJay

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    Yes, now you just have to solve for m2. Note that since the boy and cart move together at the same speed and direction after he jumps on it, you could have written the equation as

    m1v1 +m2v2 = (m1 +m2)v_f
    50(10) + 0 = (50 +m2)(2.5)
    and get the same result for m2.

    m2 =???
     
  13. Oct 26, 2006 #12
    so m2 = 3 kg

    hopfully that is right
     
  14. Oct 26, 2006 #13

    PhanthomJay

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    You had the right equation, but slipped up on the parentheses and the algebra.
    m2 = (m1v1 - m1v1')/ v2'
    m2 = ((50)(10) - 50(2.5))/2.5
    m2 = (500 - 125)/2.5
    m2 = 375/2.5
    m2 = ????
     
  15. Oct 26, 2006 #14
    ok m2 = 150 kg right?

    I did come up with 150 kg the second time I work it but was not sure.
    so I worked it a different way and came up with 3 kg, what I did was cancel out m1 on each side, but now I see that I cant do that because I didnot have m1 only but also had m2 so i could not cancel out m1

    I had (v1-v1')/ v2 = m2 I guess that is not right.
     
  16. Oct 27, 2006 #15

    PhanthomJay

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    The 150kg mass for m2 is correct. you could have reworked your equation to read
    m1(v1 -v1')/v2' = m2, from which,
    50(v1 - v1')/v2' = m2
    50(10 - 2.5)/2.5 = m2
    50(7.5)/2.5 = m2
    150 = m2
    m2 = 150kg
    In doing the algebra, I often find it easier to plug in as many numerical values into the original equation first, before trying to solve for the unknown, because it is easier to work with numbers than it is with letters.
    So if you have
    m1v1 + m2v2 = m1v1' + m2v2', and you wish to solve for m2, rather than first isolating m2 by itself and easily getting mixed up witht the letters, put in the values first, thus
    50(10) + m2(0) = 50(2.5) + m2(2.5)
    500 + 0 = 125 + 2.5m2
    375 = 2.5m2
    150 = m2
    if u see what i mean.
     
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