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Momentum & energy

  1. Nov 1, 2008 #1
    1. The problem statement, all variables and given/known data

    Two people of different masses are riding in a sleigh on frictionless ice.
    One has mass m1 and the other has mass m2.
    When they jump out of the sleigh each has speed u respect to the sleigh.

    (I). What is the final velocity of the sleigh if they jump simultaneoulsy in the +x direction?
    (II). What is the final velocity of the sleigh if one with mass m1 jumps first and
    the other person with mass m2 jumps a few seconds later?


    2. Relevant equations

    conservation of momentum and energy (I think)


    3. The attempt at a solution

    I tried using momentum and energy but somehow I ended up having u = 0,
    which seems to be wrong. Please help!
     
  2. jcsd
  3. Nov 1, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi kky1638! Welcome to PF! :smile:

    Show us what you tried, and where you're stuck, and then we'll know how to help. :smile:
     
  4. Nov 1, 2008 #3
    I tried using momentum conservation and energy conservation.

    momentum: (m1 + m2 + M)(Vi) = (m1 + m2)(Vf + u) + (M)(Vf)

    (u is the velocity of the people respect to the sleigh
    Vi = initial velocity of sleigh and people combined, Vf = final velocity of the sleigh)

    energy: 1/2(m1 + m2 + M)(Vi)^2 = 1/2(m1 + m2)(Vf + u)^2 +1/2(M)(Vf)^2

    To make calculation a little easier, I let A = m1 + m2
    then did some algebra with the two equations and ended up having u = 0,
    which I think is definitely wrong.

    Part II, I think is doing the same thing I did on Part I twice.
    Tell me if I did anything wrong both conceptually and algebraically. Thanks!
     
  5. Nov 2, 2008 #4

    tiny-tim

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    Hi kky1638! :smile:
    Yup … energy is not conserved … imagine you're standing on a stationary wheeled thing, and you jump off forwards … is the energy conserved then? :rolleyes:

    General principle: energy is never conserved in collisions unless the question says so (or says "perfectly elastic", which is the same thing).

    (But momentum is always conserved)​

    Look at the number of equations … you need two equations, and sometimes they're conservation of energy and momentum …

    so when you don't have conservation of energy, you need one other equation … in this case, the equation for u.

    The other side of the coin is that, if you're given an extra equation, like the one for u, then obviously there must be an equation missing, and that is the conservation of energy. :smile:
     
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