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Momentum of Skiers in a Completely Inelastic Collision

  1. Dec 12, 2011 #1
    1. The problem statement, all variables and given/known data
    In a movie stunt, a 65 kg skier starts from a rest position at the top of a hill 30m high. She slides down the hill to the bottom, where she collides with a 45 kg stationary skier. The collision is completely inelastic. Find the final velocity of the skiers.


    2. Relevant equations
    M1V1 +M2V2= (M1+M2)V
    ET'=ET'
    ET'=(M1+m2)gh ( ithink this is relevant?)



    3. The attempt at a solution

    ET'= (65+45)(9.8)(30)
    ET'= 32,340

    32,340= 1/2mv2



    Sq RT[(32,340)(2)/(65)] = velocity of skier going down hill?

    =31.5 m/s (100 km/h + seems kinda fast for a skier)
     
  2. jcsd
  3. Dec 12, 2011 #2
    I think you need to first find the speed of the 65kg skier from their starting height. You can then calculate momentum and go from there.
    If the collision is completely inelastic then ONLY momentum is conserved in the collision.
     
  4. Dec 12, 2011 #3
    How would I calculate the velocity of the skier? This is where I am confused.. Thats why I used the ET' calculation..

    I know you dont want to give me the answer.. but any hints would be immensely appreciated.. I mean I know the mass and height of the skier, no idea how to calculate their velocity though..

    thanks for your time and response!
     
  5. Dec 12, 2011 #4
    the skier gains KE as a result of 'falling' a distance of 30m.
    You need to use PE = KE to calculate the speed of the 65kg skier.
    Then it is a 'straight forward' momentum before = momentum after calculation.
    Have a go and if you get stuck let me know where it goes wrong.
     
    Last edited: Dec 12, 2011
  6. Dec 12, 2011 #5

    THANKS again!

    So PE=KE
    mgh=1/2mv2
    √(2)(9.8)(30)= V
    V= 24.25 m/s


    Ek = 1/2mv2
    =1/2(65)(24.3)2
    =19,190.9 J (initial energy) ( dont think I have to do this)

    PT=PT'

    M1V1+M2V2=(M1+M2)V'
    (65)(24.3)+(45)(0)=(110)V'
    V'=(65*24.3)/(110)
    V'=14.36 m/s

    Ek= 1/2(M1+M2)V'2
    =1/2(110)(14.36)2
    =11 341.53


    how does the final velocity of both look? As thats what Im most concerned about.

    Again thank you so much for your help!
     
  7. Dec 12, 2011 #6
    That's it.... you did not need to calculate KE in the question but well done.
     
  8. Dec 12, 2011 #7
    Thank you so much ! Is there a rating system at PF?
     
  9. Dec 12, 2011 #8
    Don't know !!!.... not bothered....
     
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