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Conservation of Momentum- Inelastic Collisions

  1. Oct 30, 2013 #1
    1. This was a hard test question that I took partial credit on. I want to fully understand what I did wrong so that I’m fluid with the concept. I’m also new to this forum. I love constructive criticism too! lol
    Zombie Apocalypse has arrived and the war has begun. Your task as a physics student is to gather intel on the Kinetic energy lost due to the collision of darts on their vulnerable heads. Your crossbow can be modeled as a spring with a 1500N/m constant that can be drawn back to a 32cm maximum. You shoot your 35.0 gram crossbow dart at the zombie's 5.1kg head, striking it horizontally level as fired from your position. The dart buries itself in the zombie's head, and the head slides back across the level table. The CIA needs the fractional amount of Kinetic energy lost in the collision compared to the initial kinetic energy, that is, [delta KE]/KEi




    2. Relevant equations:
    Crossbow Dart KE= 1/2kx^2
    Conservation of Momentum
    Before Collision- Pi= m1v1 + m2v2= kg/m/s
    KEi= 1/2m1v1^2 + 1/2m2v2^2= Joules
    After Collision- P’= m1v1’ + m2v2’= kg/m/s
    KEf= 1/2m1v1^2’ + 1/2m2v2^2’= Joules
    …m1v1=(m1+m2)v2





    3. I tried a few different things, more or less throwing mud on the wall to see what sticks at minimum. But here goes…
    1. Crossbow Dart KE= ½(1500N/m)(0.32m)^2= 76.8J
    2. Solve for V1, ½(0.035kg)(v)^2=(1500N/m)(0.32)^2
    ….Result 66.25 m/s
    3. Solve for V2, (0.035kg)(66.25m/s)= (0.035kg + 5.1kg)v2
    ….Result 0.451557 m/s …0.45 m/s
    4. This is where I think I’m lost…delta KE/KE? Up to this point I hope this is right?


    Thank you all for the guidance
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 30, 2013 #2
    What is the final kinetic energy of the system? How much is lost?
     
  4. Oct 30, 2013 #3
    The final KEf should be 1/2(m1+m2)v2^2, which we would subtract from the initial KEi= 1/2m1v1^2, then divide by the KEi?
     
  5. Oct 30, 2013 #4
    Correct.
     
  6. Oct 30, 2013 #5
    Awesome...so I'm looking at approx 74.728J/76.809J= 0.9729 which is my loss in Kinetic Energy. That wasn't too bad I suppose for a University Physics course student :-)

    Thanks "voko"...I do appreciate the guidance.
     
  7. Oct 30, 2013 #6
    You are welcome. And welcome to Physics Forums, too!
     
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