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Rocks thrown off of a skateboard

  1. Jan 28, 2014 #1
    1. The problem statement, all variables and given/known data

    A person sits on a skateboard and wants to propel themselves. The persons mass is 150 lbm whihile the skateboard mass can be ignored. The person has a bucket with 1 kg rocks totaling 10 kg. The rocks are ejected at a rate of 1 kg/s and a velocity of 108 mph with respect to the skateboarder. The rocks leave the hand at an angle of 15 degrees above the horizontal plane. Assume drag and friction are negligible.

    a) Calculate the final velocity of the skateboard
    b) From the skateboards perspective, each rock has the same relative velocity as it is thrown. However, since the skateboard is speeding up, it appears to an inertial observer that the velocity of each rock diminishes. Consequently, the kinetic energy of each rock decreases. However, the skateboard gains a constant amount of kinetic energy as each rock is thrown. Explain this paradox and use equations as necessary.

    2. Relevant equations

    Conservation of linear momentum

    m1v1i + m2v2i = m1v1f + m2v2f

    3. The attempt at a solution

    Object 1 applies to the person and the skateboard. Object 2 applies to the rocks. m1=68.04 kg (150 lbm), m2=10 kg. I understand that v1i=0 and v2i=108 mph, but I don't understand how to interpret the final speed of the rocks (v2f) in order to find the final speed of the skateboard (v1f) and I also don't know how the 15 degree angle comes into the mix. Any help or advice is greatly appreciated!
     
  2. jcsd
  3. Jan 28, 2014 #2

    HallsofIvy

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    To begin with this is wrong. It is the horizontal speed that is relevant here, 108cos(15).

    initially, if the person and skate board are stationary, both v1i and v2i are 0. So your equation is m1v1f+ m2v2f= 0. Set v2f equal to 108 cos(15) and solve for v1f.
     
  4. Feb 4, 2017 #3
    that would neglect that the velocities sought are likely with respect to a stationary observer. the rock leaves the hand at a horizontal velocity of 108 cos (15) _relative to the skateboarder_ who's velocity is not yet known. so the final rock velocity is 108cos(15)-v2f, where v2f is the velocity of the skateboarder with respect to a stationary observer.
     
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