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Distance traveled during a head on collision

  1. Sep 29, 2015 #1
    1. The problem statement, all variables and given/known data
    During the head on collision of a 1500kg car moving 20 m/s and a 2400 kg truck moving 12.5 m/s does the car move farther than, a shorter distance than, the same distance as the truck, or is this indeterminate with the given information.

    2. Relevant equations
    m1(v1)+m2(v2)=(m1+v1)vf=0
    average acceleration of car = 20/t
    average acceleration of truck = 12.5/t
    3. The attempt at a solution
    the final velocity of both cars is 0 m/s. So if the duration of the collision is .5 seconds, then the average acceleration of the car is 40 m/s^2 and the truck is 24 m/s^2. the area under the graph -40x+20 (the velocity of the car) is 5 m. the area under the graph -24x+12.5 (velocity of truck) is 3.25 m. So this would mean the car travels a greater distance. My gut says that the car moves the same distance as the truck but that is not an adequate explanation for my professor. How do I calculate this??
     
  2. jcsd
  3. Sep 29, 2015 #2

    gneill

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    Staff: Mentor

    Why do you conclude that the final velocities of both vehicles is zero? Are there enough details provided to determine the post-collision situation?
     
  4. Sep 29, 2015 #3
    the vehicle's final velocities are zero because 1500*(20)+2400*(-12.5) = (1500+2400)*0

    This is ALL of the information in the problem and my professor grades extremely hard on explanations, so I'm trying to make sure that I can say the correct thing, here.
     
  5. Sep 29, 2015 #4
    Look when uh find the momentum of car and truck they both are same,regardless of car having less mass than truck the car has enough velocity to equalise the momentum with truck so if two different bodies with different mass but same momentum collide both bodies travel same distance or uh could say come at rest cancel each other's momentum. And for final velocity take it as zero
     
  6. Sep 29, 2015 #5

    phinds

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    That is an astounding piece of math. You might want to rethink the way you have expressed it. :smile:
     
  7. Sep 29, 2015 #6

    gneill

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    How do you know that the collision is perfectly inelastic? It wasn't mentioned in the problem statement. Nor were any details given about the structure of the vehicles regarding rigidity (safety crumple zones, etc.).
     
  8. Sep 29, 2015 #7
    I know that the question says nothing about the elasticity of the collision but head on car collisions are generally inelastic so I took that as an assumption. This is the first homework assignment regarding forces and momentum and we really haven't gone over any of the equations except f=ma. Basically everything I've done has come from personal research. My professor makes us figure things out on our own and then grades homework like its an upper level physics course. So I am aware of all of the conditions I just am trying to think like an experienced physicist, even though I'm not, so that I can try to get some decent homework grades. If it is indeterminate I have to be able to THOROUGHLY explain why.
     
  9. Sep 29, 2015 #8
    So could not each vehicle rebound with exactly the same velocity they collide with but in the opposite direction?

    Why are you assuming the two objects become one object?
     
  10. Sep 29, 2015 #9
    As in:
    m1(v1i)+m2(v2i)=m1(v1f)+m2(v2f)
     
  11. Sep 29, 2015 #10
    so this would mean that the car would travel farther than the truck? or is it indeterminate because we dont know from the question whether the collision is elastic or inelastic?
     
  12. Sep 29, 2015 #11
    Very good. If it is fully elastic, what is the answer?
     
  13. Sep 29, 2015 #12

    gneill

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    You'll want to lay out all your assumptions and justify them, then draw conclusions from there. If you're going with an inelastic collision as a model, state why and how well you think it will approximate real life. Definitely mention the initial total momentum and what you can conclude from that if the collision is inelastic.

    How much of a vehicle has to be moving to say that the whole vehicle is moving? Most cars these days have lots of glass and plastic bits that shatter and scatter in a collision :smile: They are also built with crumple zones to absorb and dissipate the energy of a collision. Trucks tend to be less forgiving in their structure.
     
  14. Sep 29, 2015 #13
    i would say that the car would travel farther than the truck if it was fully elastic because it has a greater velocity
     
  15. Sep 29, 2015 #14

    gneill

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    What will stop either vehicle? And does motion several minutes after the collision still count?
     
  16. Sep 29, 2015 #15
    Good.

    Do you not have an equation for the conservation of the energy?

    Perhaps you should write out and apply this equation?
     
  17. Sep 29, 2015 #16
    Good point. It is distance not velocity that they asked for.
     
  18. Sep 29, 2015 #17
    We haven't discussed frictional forces very much or how to calculate them so I know friction will stop both vehicles but I don't know how far either will go, I just know that the car will travel farther, having less mass and higher velocity.
     
  19. Sep 29, 2015 #18
    If the collision is perfectly inelastic, the car will travel the same distance as the truck, because two bodies with equal momentum that undergo inelastic collision will accelerate to zero velocity at the point that the collision occurred. If the collision is fully elastic, then the kinetic energy before the collision will equal the kinetic energy after the collision; KE = .5*m*v^2 . This means that the car will have the same velocity before and after the collision pointed in opposite directions, thus traveling a greater distance than the truck during the collision.


    this is my explanation.
     
  20. Sep 29, 2015 #19
    Hmm, keep working on it.

    Did you write out your Kinetic Energy equation yet?

    How many variables do you have?
    How many knowns?
    How many unknowns?
    How many equations do you have?
     
  21. Sep 29, 2015 #20
    We you given any information to calculate frictional forces with?

    Which vehicle has Michelin All-weather tires?

    Is one of the roads iced up?

    One of the vehicles may have a higher velocity, but without any information on the forces slowing the vehicles, how can you calculate any distance?
     
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