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What are the velocities of two objects after they collide?

  1. Jul 13, 2014 #1
    Let's say two equally massive objects move toward each other at equal velocities (let's use 30 kg and 5 m/s for the sake of having some numbers).

    Will the objects bounce off of each other or will they both stop in place? If they bounce off each other, at what velocity?

    The momentum conservation principle suggests that the overall momentum (mass*velocity) of the two objects should remain the same before and after collision (i.e., momentum gained by one object should be equal to the momentum lost by the other).

    This still leaves room for ambiguity in resulting velocities. For instance, here's a solution where one object gains 150 units of momentum and the other loses 150 units, causing both objects to stop.

    Code (Text):
            [B]MOMENTEM[/B]     [B]MOMENTUM[/B]
            [B]BEFORE COLLISION[/B] [B]AFTER COLLISION[/B]
    [B]OBJECT A[/B] (30kg)*(+5m/s)=150  (30kg)(0m/s)=0
    [B]OBJECT B[/B] (30kg)*(-5m/s)=-150 (30kg)(0m/s)=0
    [B]OVERALL[/B]      0           0
    Here's another valid solution where one object gains 225 units of momentum and the other loses 225 units, causing the objects to bounce off each other and head back from where they came from.

    Code (Text):
            [B]MOMENTEM[/B]     [B]MOMENTUM[/B]
            [B]BEFORE COLLISION[/B] [B]AFTER COLLISION[/B]
    [B]OBJECT A[/B] (30kg)*(+5m/s)=150  (30kg)(-2.5m/s)=-75
    [B]OBJECT B[/B] (30kg)*(-5m/s)=-150 (30kg)(2.5m/s)=75
    [B]OVERALL[/B]      0           0
    But there's only one right solution, right? What is it? What other contraint besides momentum conservation am I missing?

    How would one mathematically calculate the actual resulting velocities?

    Thanks a ton!
     
  2. jcsd
  3. Jul 13, 2014 #2

    DrGreg

    User Avatar
    Science Advisor
    Gold Member

    Energy.

    Is kinetic energy conserved, or is some energy lost as heat, sound, etc.? Without that information, there is no unique answer.

    See
     
  4. Jul 13, 2014 #3
    Aha! Thanks! So it depends on whether the collision is elastic or inelastic, which tells you whether kinetic energy is conserved or not. Assuming a perfectly elastic collision, our new velocities for the collision described in my original post would be would be

    [itex]
    v_{1}=\frac{u_1(m_1-m_2)+2m_2u_2}{m_1+m_2}=v_{1}=\frac{+5(30-30)+2*30*(-5)}{30+30}=-5\\

    v_{2}=\frac{u_2(m_2-m_1)+2m_1u_1}{m_1+m_2}=\frac{-5(30-30)+2*30*(+5)}{30+30}=+5
    [/itex]

    which makes sense intuitively: the objects bounce back at the same speed at which they came.

    For a perfectly inelastic collision, the coefficient of restitution is 0, causing both objects to just stop when they hit each other.

    For a collision that's neither perfectly inelastic or elastic, the objects bounce back at some speed less than 5 m/s
     
  5. Jul 13, 2014 #4
    Okay, so I got a FOLLOWUP QUESTION

    What force do the objects exert on each other in the different kinds of collisions?
     
  6. Jul 13, 2014 #5
    Here's my attempt at answering my own questions:

    [itex]Impulse = \int F dt = change In Momentum = mv_1-mu_1[/itex]

    For an elastic collision, the change in momentum for one of the objects is is [itex]30*5-30*(-5)=300[/itex]

    For an inelastic collision, the change in momentum for one of the objects is is [itex]30*5-30*(0)=150[/itex]

    In either case, the force depends on how long the collision is. Well, how long is it? I have no idea. Also, force throughout the contact might not necessarily be constant either right?
     
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