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Simultanious kinetic energy transfer between 3 objects

  1. Nov 8, 2011 #1
    So, what I'm wondering is if it is possible for a kinetic energy transfer between 3 objects.

    Much like a pair of billiard balls, I would like to know if one can have two mobile objects collide with a still object at precisely the same time. Having all of the kinetic energy in both moving objects transfer into the one still object.

    I can't really see why this isn't possible. However I imagine the scenarios where all the energy can be transferred and in the same direction would be few if any.

    I was trying to figure out what the formula for this process would be. At first I thought it was simple adding of vectors. However I am concerned how exactly one would deal with 3 dimensional differences in direction.

    For example if two billard balls strike another at precisely the same time. In a manner where the two moving balls followed a parallel path. Then the 3 ball would, I think still move straight forward ( as in along the same parallel) . Thou not at 2 times the speed of the original balls. That statement also implies the two moving balls have the same starting distance and equal speed at all times.

    Now that example was only a 2 dimensional problem. If we have have a three dimensional collision for example:

    For balls rolling down tubes at equal speeds and from equal distances, simultaniusly strike a ball in another tube. ( they would collide at the open ends of there respective tubes.) The four balls are arranged so there force is evenly distributed on a parallel path. ( think a square with an appropriately sized circle, being struck at the four corners if the square.) Once again I imagine the ball would move strait forward, but how much energy would it have? Also I guess a tube wouldn't work for the 5th ball as that would force the direction of the energy.

    I'm not a student nor do I have much if any schooling in physics. So i aprecciate any time taken to consider my questions.
     
  2. jcsd
  3. Nov 8, 2011 #2
    I hope I understand your question correctly.

    If I do, then your 3D problem is actually always a 2D problem, due to conservation of momentum. Let me try to explain: imagine the three balls moving in 3D in outer space (neglect gravity). Take your left index finger as the velocity of ball 1, your right index finger as the velocity of ball 2. Choose any orientation in space that you wish. Place your index finger tops together; that is where they will collide, by definition the place where ball 3 is waiting. You see that your two index fingers define a plane (cal it the xy plane). After the collision, ball 1 and 2 stop (that's what you're looking for, right?) and ball 3 takes all the momentum and energy. Because momentum is conserved, and because the total initial momentum was in the xy plane, the final momentum (i.e. the momentum of ball 3) must also be in the xy plane. As such everything happens in a fixed plane.

    To see if this is actually possible, you need to solve some simple equations (or show that they cannot be solved, I haven't done the calculations myself): it's a 2D problem. Take the initial velocity of ball 1 along the x-axis, and let the collision be at the origin. The initial velocity of ball 2 is lying on a random line through the origin. Call theta the angle between velocity of ball 1 and of ball 2. The initial speeds of the two balls are also variables. The other two remaining variables are the final speed of ball 3 (initially zero) and an angle phi determining the direction of the final velocity of ball 3.
    You can express the speed of ball 3 directly in terms of the initial speeds of ball 1 and 2 (conservation of kinetic energy!). The remaining variables are speed 1, speed 2, theta and phi. Write down the vector equation for conservation of momentum (this gives two equations) and see whether there exists a choice of speed 1, speed 2, theta and phi that fulfill the momentum conservation.
     
  4. Nov 10, 2011 #3
    Okay so what you wrote was a little off what I was asking about, however it was very use full. Over the last day and a bit I learned quite a bit.

    The only question I have is when two perfectly circular objects colide on a 2d level how much force will be in the same direction as the original moving object.

    So one object say α is sitting still and another object β is moving moving in a strait line with a force of 10 whatever's assume no friction losses. If β hits α but not directly then the force transferred to α would be the Cos of the angle θ times the force of 10. Where θ is the angle between β's trajectory and a straight line that is from the center of α to the center of β at the point of impact.

    If that part is true, which is what I am not certain of. Then the force that is in the original direction would be the Cos of θ times the force value found in the previous equation. See in my would be experiment this happens twice at the same time. Where two balls moving in the direction y hit α at the same time. And the force applied in the x direction would be 0 because the two collisions cancel each other out in the x direction.

    I did all the math for this and if I am correct then the force of up to three balls can be transferred into on ball at a 75% eficciency.

    This is all based on my meager understandings and what I could throw together based on what seemed logical to me at the time. So I may be horribly wrong and if so please show me how I can do this correctly.
     
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