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Velocities after totally elastic collision

  1. Jun 15, 2012 #1
    I found this problem on a website. Is it right? (BTW, this is not homework, but I am just curious if I am thinking right about it.)
    A marble with a mass of 2 grams moves to the left with a velocity of 2 m/s when it collides with a 3 gram marble moving in the opposite direction with a velocity of 2 m/s; if the first marble has a velocity of 1.5 m/s to the right after the collision, determine the velocity of the second marble after the collision.

    I thought it could be solved using the conservation of momentum:

    momentum before = momentum after
    p.moving right + p.moving left = p.moving left + p.moving right

    3(2) + 2(-2) = 3(v1') + 2(+1.5)

    solving for v1' gives -0.3333... I think and ignoring sig.figs.

    But then I checked using the conservation of energy. It didn't check.
    So I guess one of the velocities is wrong. Let's say the velocity of the first marble after the collision is unknown. Then what would v1' and v2' be?
    Someone else on that website said v2' would be 2.333...., but then that doesn't seem right. Why would the larger mass have more energy aft.than before?
    I don't have my books and I forgot the formulas. I tried deriving, but I am not sure how to deal with the signs after equating the momentum factors in the difference of two squares. In fact I don't think I have seen this derivation for probably ten years - I couldn't even find it on the web.
    Finally, could it be that the collision is not totally elastic? How would that work? I guess then the velocity of the 3 gram after could be 1.5, but not sure. Idk, brain fried. I stayed awake trying to figure it out last night. I guess what I really want to know is what do you get for velocities when it is a totally elastic collision?
     
  2. jcsd
  3. Jun 15, 2012 #2

    TSny

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    Google "1d elastic collision equation"
     
  4. Jun 15, 2012 #3

    HallsofIvy

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    The difficulty is that you have set up an impossible situation. If given that the first marble has mass 2 g and velocity -2 m/s, the second marble has mass 4 g and velocity 2 m/s, you cannot simply assign an arbitrary velocity to one marble after the collision.

    Before the collision, the total momentum is 2(-2)+ 3(2)= 2 g m/s and the total kinetic energy is (1/2)(2)(4)+ (1/2)(3)(4)= 10 kg m^2/s^2. Assigning velocities v1 and v2 after the collision we have the two equations 2(v1)+ 3(v2)= 2 and v1^2+ (3/2)v2^2= 10 to solve for both v1 and v2.
     
  5. Jun 15, 2012 #4

    TSny

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    The original statement of the problem did not specify that the collision is elastic. You correctly used conservation of momentum to find the final velocity of the second marble. As you noticed, kinetic energy is not conserved. So, the collision is indeed inelastic. That's ok and, in fact, that's usually the case for everyday objects.

    If you start with the same setup but impose the condition that the collision is elastic, then you are not free to specify the final velocity of either marble. The final velocities are determined by conservation of momentum and conservation of kinetic energy. You can use the equations for a 1d elastic collision to find the final velocities of both marbles (or just set up the conservation equations yourself and work through the algebra).

    If you are bothered by why kinetic energy in not conserved in an inelastic collision, it's because some of the kinetic energy of the marbles is converted into other forms of energy such as heat and sound. If you keep track of all the changes in all of the different forms of energy, it would turn out that the total energy is conserved.
     
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