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Forces two objects apply on each other during inelastic collision

  1. Oct 14, 2011 #1
    1. The problem statement, all variables and given/known data
    11mavpi.png

    Values from previous problem:
    Pre-collision: v2kg=6m/s -->, v3kg=3m/s <--
    Post-collision: v2kg=2m/s <--, v3kg=5.67m/s -->
    Collision time: 0.2 sec

    2. Relevant equations
    F=mA=m(Δv/Δt)

    3. The attempt at a solution
    My instinct to solve this was to use the pre-collision values to find out the forces each puck had on each other, because it seems like post-collision values for velocity would be irrelevant because the pucks are no longer having any force on each other. So, I did the following...

    Force of 2kg puck on 3kg puck: F=2kg(6m/s / 0.2 sec) = 60N
    Force of 3kg puck on 2kg puck: F=3kg(3m/s / 0.2 sec) = 45N

    Is this the correct way to approach this problem? Thanks for your help!
     
  2. jcsd
  3. Oct 14, 2011 #2
    J = Favg*t. Also, Newton's Third Law.
     
  4. Oct 14, 2011 #3
    Can you elaborate a bit on that? I'm not completely familiar with that formula. We probably just use different variables because I don't recognize the J.
     
  5. Oct 15, 2011 #4
    J = magnitude of change in momentum.
     
  6. Oct 15, 2011 #5
    I think watermelonpig is saying Impulse= Force(average) x Time Interval. I=Ft. I think most physics books call change in momentum, impulse, so maybe that will help. Your also given the time....
     
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