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Momentum textbook problem

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  1. Feb 22, 2015 #1
    1. The problem statement, all variables and given/known data
    The 75kg boy leaps off the cart A with a horizontal velocity of v' = 3 m/s to the left measured relative to the cart. Determine the velocity of cart A just after the jump. If he then lands on cart B with the same velocity that he left cart A, determine the velocity of cart B just after he lands on it. Carts A and B have the same mass of 50kg and are originally at rest.

    2. Relevant equations
    Relative motion: Va = Va/b + Vb
    Momentum: m1V1 = m2V2

    3. The attempt at a solution
    the subscript j will represent the boy, while a/b represent the respective carts. the coordinate system: left represents negative, right positive

    Relative motion
    Vj = Vj/a + Va
    (1) Vj = -3 + Va

    Momentum
    0 + 0 = ma*Va - mj*Vj
    mj*Vj = ma*Va
    (2) Vj = (ma)/(mj)Va = (50/75)Va

    Substituting (1) and (2)
    (50/75)Va = -3 + Va
    Va(50/75-1)=-3
    Va = 9 m/s (to the right)
    Vj = -3 + 9m/s = 6m/s (to the left)

    Now I find Vb (assume direction right)
    mj*Vj + mb*Vb = (mj+mb)V
    75*(-6) + 50*0 = (50+75)V

    V = (75*6)/(50+75) = 3.6 (left)

    But this is wrong,the answer is 0.720 m/s (left)
     
  2. jcsd
  3. Feb 22, 2015 #2

    haruspex

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    Need to be consistent about signs. Are you measuring all velocities as positive right, or the cart's as positive right and the boy's as positive left?
    What relative velocity would that be?
     
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