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A cart and a water hose

  1. Dec 7, 2012 #1
    1. The problem statement, all variables and given/known data
    A friend asking:

    A cart and a water hose:

    A water hose is spraying water at rate m_dot kg/sec, and at a given velocity V, on a cart with a mass M.
    The water hit the cart, and they are bounced back from it in an elastic collision (same velocity in relation to the cart, but opposite direction).
    What is the velocity of the cart as a function of time, v(t)?

    2. Relevant equations


    3. The attempt at a solution
    "My way of solution was this:

    Lets have a coordinate system that moves with the same velocity with the cart at a certain time. Delta_m would signify the small amount of water. Then we write the momentum at time t, and t+dt:

    P(t)=Delta_m*(V-v(t))
    P(t+dt)=Delta_m*(v(t)-V)+M*dv


    P(t)=P(t+dt), so:
    dv/(V-v(t)) = (2*Delta_m/M)


    Delta_m = m_dot*dt, so:

    dv/(V-v(t))=2m_dot*dt/M --> ln(V-v(t))=-2m_dot*t/M + C, and then we can get v(t) by using exp.

    However, the final answer I've seen is v(t) = (2m_dot*t/M)/(1+2m_dot*t/M),
    Which is obviously not an end result of an exp/ln function.

    What is wrong with my solution?
    How do I arrive at the right solution?"
     

    Attached Files:

  2. jcsd
  3. Dec 7, 2012 #2

    TSny

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    Welcome to PF, sapz!
    Think about this statement. Remember, you're in a reference frame that's moving relative to the hose.
     
  4. Dec 7, 2012 #3
    Still no bells ringing. Could you explain your hint?
     
  5. Dec 7, 2012 #4

    TSny

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    The mass of water that exits the hose in time ##dt## is ##\dot{m}\:dt##, but that's not the mass that hits the cart in time ##dt##.
     
    Last edited: Dec 7, 2012
  6. Dec 7, 2012 #5
    How so?
     
  7. Dec 7, 2012 #6

    TSny

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    If I'm standing on the ground throwing balls at a rate of 1 per second toward the cart, then the cart would not receive balls at a rate of 1 per second because the cart is moving away from me. It's like the Doppler effect. The frequency of reception is less than the frequency of the source.
     
  8. Dec 7, 2012 #7
    Oh, I see. So how can Delta m be calculated?
     
  9. Dec 7, 2012 #8

    TSny

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    For the stream of water, the rate at which mass strikes a wall is proportional to the speed of the stream relative to the wall. What would be the ratio of the rate at which mass strikes the moving cart to the rate at which mass would strike a stationary wall?
     
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