Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Momentum Conservation Between Two Masses

  1. Jun 9, 2013 #1
    I have been studying momentum and began believing that I understood it, but I began thinking...

    Between two objects of the same mass, the first object will collide with another object (let's say the second object is not moving) and the momentum, in an isolated system, will be transferred, stopping the first object. However, why is this? Why is half of the momentum not transferred to object 1 and another to object 2?
     
  2. jcsd
  3. Jun 9, 2013 #2

    Nugatory

    User Avatar

    Staff: Mentor

    You are right that momentum would be conserved if half the momentum ended up with each object (after the collision, they're both moving at speed v/2). But we also have to conserve energy, and in the elastic collision that you are describing no kinetic energy is lost to heat or crushing and deforming the colliding masses, so the kinetic energy after the collision has to be equal to the kinetic energy before.

    Kinetic energy is given by ##E=\frac{mv^2}{2}##, and there's only one solution that conserves this quantity as well as the momentum ##p=mv## and that's the one in which the first object stops dead and the second ends up with all the momentum.
     
  4. Jun 9, 2013 #3
    In this situation, as has already been pointed out, the KE has to stay the same (elastic collision). But keep in mind that the total momentum of the system is always conserved, in elastic and inelastic collisions.
     
  5. Jun 10, 2013 #4
    In a plastic (or completely inelastic) collision of two ball of equal mass this is how the momentum is distributed. They each have half of the original momentum. And they move together.
    Conservation of momentum alone is not sufficient to determine the redistribution of momentum after collision.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook