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Momentum and rebounding

  1. Oct 21, 2013 #1
    1. The problem statement, all variables and given/known data
    At the 2010 Winter Olympics, one of the most exciting events was the bobsleigh, in which teams of either two or four people on sleds hurtle down an icy half-pipe course at speeds that would be illegal on most highways. The track was 1450m long, with a total vertical drop of 152.

    In an ideal run, the sled would never touch the sides of the course (in reality, glancing collisions are common). Assume a four-man sled, of total mass (sled+crew) 590kg is moving at 135km/hr down the course. It hits the side of the course at a shallow angle of 3.0°, and bounces off at the same angle, with its speed unchanged. The gouge marks on the side of the course (indicating the distance over the collision took place) are 35cm long. During the collision:

    a. What is the change of momentum of the sled? (indicate both magnitude and direction of Δp)
    b. What is the average force exerted on the sled (magnitude and direction)?
    c. What is the acceleration of the sled, measured in units of g?

    2. Relevant equations
    Δp=mΔu=I


    3. The attempt at a solution
    I've finished the exercise but i'm not convinced of my answer. For a, i have that there are two momentum vectors in which the sled rebounds. By finding the resultant momentum (using vector subtraction), i figured out that the total momentum is 44000Ns and that its direction was to the right).

    For b, i calculated the time at which the sleigh rebounded (quotient between skid marks and the velocity), and finally used I = FΔt to calculate the average force to be 4700kN (isn't this an extremely high value though?).

    For c, i found the acceleration of the sleigh using the equation F = ma, calculating the final acceleration to be 880g (however, what about the acceleration due to gravity? Are we taking that into account?

    Thank you very much.
     
  2. jcsd
  3. Oct 21, 2013 #2
    a) you want the change of momentum. The mometum after the collision is nearly in the same direction as the momentum after the collision, and their difference is much smaller than the momentum before or after the collision.

    b) The force is much too large, because your answer to a) is much too large.

    c) Your answer is still much too large for the same reason as a) and b). When the sled moves at 135 km/h wich is likely its terminal speed, the force of gravity will be canceled by friction.
     
  4. Oct 22, 2013 #3
    How are the momentum vectors drawn? I've drawn one momentum vector pointing at 3.0 angle to the surface, and another one that comes out at a 3.0 angle with respect to the horizontal. By vector subtraction, we would have that the new change in momentum is actually just an horizontal component. Is this method right? (i'm not convinced by it). Or whhich are the momentum vectors that i should draw?
     
  5. Oct 22, 2013 #4
    What do you call horizontal here? The change in momentum is perpendicular to the wall.
     
  6. Dec 10, 2013 #5
    I got 42.9g. Can we think of the initial angle as 87 degrees and the final as 93, such that the sleigh rebounds from the wall perpendicularly? In that case, the direction of the impulse is 180
     
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