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Interrelation between energy and momentum

  1. Feb 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Two mine cars of equal mass are connected by a rope which is initially slack. Car A is imparted a velocity of 4m/s with car B initially at rest. When the slack is taken up the rope suffers a tension impact that imparts a velocity to car B and reduces the velocity of car A.

    a. If 40% of the kinetic energy of car A is lost during the impact calculate the velocity imparted to car B.

    b. Following the initial impact car B overtakes car A and the two are coupled together. Calculate their final common velocity.


    2. Relevant equations

    The definition of kinetic energy:
    KE = [itex]\frac{1}{2}[/itex]*m*[itex]V^{2}[/itex]

    Momentum impulse equation:
    [itex]m_{1}[/itex]*[itex]v_{1}[/itex]+[itex]\int\sum F[/itex] = [itex]m_{2}[/itex]*[itex]v_{2}[/itex]

    3. The attempt at a solution

    part B is easy, you just assume both carts are a system so there is no outside impulse.
    [itex]m_{B}[/itex]*0+[itex]m_{A}[/itex]*4=[itex]m_{A}[/itex]*v+[itex]m_{B}[/itex]*v
    [itex]m_{A}[/itex] = [itex]m_{B}[/itex]
    Since they are both the same mass they each end up going 2 m/s.

    Part A is confusing me. The problem states that 40 % of the kinetic energy of cart A is lost during the impact. That means that 60 % of the kinetic energy is remaining. I should be able to use the definition of kinetic energy to solve for the velocity of cart A after the impact by saying:
    0.6*[itex]KE_{A1}[/itex]=[itex]KE_{A2}[/itex]

    After that I'm not really sure what to do. I presume the change in kinetic energy can be seen as an impulse, where
    M*[itex]V_{2}[/itex]-M*[itex]V_{1}[/itex] = [itex]\int\sum F[/itex]
    With M being the mass of one cart, V1 being the initial speed of cart A (4 m/s), and V2 being the speed of cart A after the impact. Then that impulse would be applied to cart 2 with an initial speed of 0 m/s.

    The problem is that when I take this method I solve for a post-impact speed of cart A being 3.1 m/s and cart B being 0.9 m/s. Obviously this can't be right since the stationary cart is supposed to be overtaking the cart that was originally moving.

    I hope that made sense. I'm not really used to this LATEX formatting. If I can elaborate on anything please let me know.
     
  2. jcsd
  3. Feb 19, 2012 #2
    What is the direction of car A after rope goes taut?
     
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