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Totally Inelastic Collisions question

  1. Oct 23, 2014 #1
    1. The problem statement, all variables and given/known data
    A 2000-kg truck is sitting at rest (in neutral) when it is rear-ended by a 1000-kg car going 28m/s .After the collision, the two vehicles stick together.
    #1What is the final speed of the car-truck combination?
    #2What is the kinetic energy of the two-vehicle system before the collision?
    #3What is the kinetic energy of the system after the collision?
    #4Based on the results of the previous parts, what can you conclude about which type of collision this is?
    #5Calculate the coefficient of restitution for this collision.

    2. Relevant equations

    3. The attempt at a solution
    Okay, so for the record I got all these answers correct (masteringphysics) if you've had to use this horrendous program :p

    #1So for the final speed of the car-truck combination:
    1000(28)+2000(0)=m12(V12xf
    =9.33m/s
    #2 The kinetic energy of the two vehicle system before the collision:
    Kcar=0.5(mv2)
    .5(1000)(28)^2
    =392000J
    #3 KE after the collision:
    Ktruckandcar=.5(1000+2000)(9.33)2
    #4 what type of collision is it?
    I know the answer is totally inelastic but I cannot figure out why? By my books definition the final velocity of the truck and car must equal 0? Am I wrong here or can someone explain this for me?
    #5 Since the answer to the previous question was totally inelastic I can assume e=0, but when I calculate it out this is not the case. I need an explanation on this to help my understanding of this! Thanks in advance!
     
  2. jcsd
  3. Oct 24, 2014 #2

    NascentOxygen

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    Staff: Mentor

    Hi CoS. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

    from Wikipedia:-
    In a perfectly inelastic collision, i.e., a zero coefficient of restitution, the colliding particles stick together. In such a collision, kinetic energy is lost by bonding the two bodies together. This bonding energy usually results in a maximum kinetic energy loss of the system.
     
    Last edited by a moderator: May 7, 2017
  4. Oct 24, 2014 #3
    Thank you for your reply, but I'm wondering when calculating the coefficient of restitution which is e=(V12xf)/(v12xi) isn't zero or am I missing something? Isn't the velocity of the two cars 9.33m/s? Shouldn't it actually be zero for e to =0?
     
  5. Oct 24, 2014 #4

    NascentOxygen

    User Avatar

    Staff: Mentor

    coefficient of restitution involves relative velocities. When the bodies stick together their relative velocity is zero. After the collision they are often moving but locked together, so while some K.E. may be lost, it is not all lost.
     
    Last edited: Oct 24, 2014
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