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Two Skaters Collide (inelastic collision)

  1. Nov 10, 2011 #1
    I'm trying to understand the concept of the following situation:

    A 75.0-kg ice skater moving at 10.5 m/s crashes into a stationary skater of equal mass. After the collision, the two skaters move as a unit at 5.25 m/s. Suppose the average force a skater can experience without breaking a bone is 3,867 N. If the impact time is 0.112 s, does a bone break?

    By using F= (MV)/t = [(75 kg) (5.25 m/s)] / 0.112s I found what I believe may be the correct answer (3515.63 N) so not a single bone breaks, but what I'm not understanding is the concept behind this solution.
    Any help would be much appreciated.
     
  2. jcsd
  3. Nov 10, 2011 #2
    Welcome to Physicsforums ArielleDubois. I'm not to sure where you got the equation of (MV)/t = F

    but F = dp/dt

    What you need is the equation for an inellastic collision which is

    M1v1 + M2v2 = (M1 + M2)v3

    This represents a conservation of momentum. I mean obviously there is a force acting on both persons when they collide

    You could use the change in momentum divided by the contact time to calculate the force i'm guessing since (dp/dt) is an infentesimal change in momentum w.r.t. a infinitesimal change in time

    I'm not totally sure conisdering i've never saw a problem like this without using calculus.

    I'd say to use

    M1v1 + M2v2 = (M1 + M2)v3

    were v2 = 0 since the second person is stationary.

    I'm thinking

    {M1(v3-v1)}/t will get you the force for person in motion and

    {Mv33}/t will get you the force for the stationary person

    I'm not 100% sure maybe someone will be able to confirm this
     
  4. Nov 10, 2011 #3
    To the above, MV/t appears to equal the Impulse, given that Momentum (p) = mass * velocity where M and V are velocity, and impulse is the change in momentum with respect to time.

    The idea here is, as the person above has also said, to remember momentum is conserved for an inelastic collision. In the beginning, we have two separate entities, a skater moving, with a momentum equal to her mass times her velocity, and a stationary skater with no initial momentum.

    As a result, the final momentum must be equal to the moving skater's momentum, since final momentum = initial momentum. Since the collision is perfectly inelastic (the once independent masses join as one), we can express final momentum as vf (m1 + m2).

    Since you know m1, m2 and vf, you can get the final momentum! From there, use the impulse equation above! Your only error is you have forgotten to add the mass of the other skater! The rest is correct, but your answer will change.

    Hope this helps!
     
  5. Nov 13, 2011 #4
    Thank you for both of your replies! Here is the problem, as "Rome_Leader," noted in my calculations I only used one of the masses. Since the correct answer for this question is,"no, not a single bone breaks," then in my calculations I should yield a total force that is less than 3867 N, I only obtain such an answer when using one of the masses and not combining both m1 and m2, that is why I am confused. You are also correct, "Rome_Leader," I did use the formula for impulse but instead of finding the impulse I found the force, I used this formula because the question is interested in the resultant force of the collision.
     
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