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Conservation of Momentum of a body of mass

  1. Oct 22, 2011 #1
    If a body of mass m and velocity u collides with another body of mass m which is stationary,
    and then the two bodies move off together, then conservation of momentum suggests that the since the combined mass is 2m, the new velocity will be 1/2 u.

    Total momentum before collision = mu + m0 = mu
    Total momentum after collision = m(1/2u) + m(1/2u) = mu

    However energy before collision = 1/2mu2
    and energy after collision = 1/2m(1/2u)2 + 1/2m(1/2u)2 = 1/4mu2

    So it seems the energy is halved or am I making an error in the calculations ?
     
  2. jcsd
  3. Oct 22, 2011 #2

    A_B

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    If, as you say, the object stick together after collision, then indeed kinetic energy is lost. The energy has gone to internal energy of the colliding masses. A collision where the objects move away together is NOT elastic. In an elastic collision (one that does conserve kinetic energy) with two equal masses, one of which is stationary, the incoming object will hit the stationary one and come to rest while the stationary object moves off at a speed equal to that of the incoming object.

    some info and nice graphics: http://en.wikipedia.org/wiki/Elastic_collision
     
  4. Oct 22, 2011 #3
    Does conservation of momentum only take place with elastic collisions ? If the collision is inelastic and the masses stick together, the combined mass is 2m so to conserve momentum, the velocity would be half and energy after collision = 1/4mu2 or exactly half the energy before the collision.
     
  5. Oct 22, 2011 #4

    AlephZero

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    From your OP you seem to understand the difference between energy and momentum, so I think you make a typo there.

    ENERGY is only conserved with perfectly elastic collisions. In fact that's the definition of what a perfectly elastic collision is. There are no PERFECTLY elastic collisions in real life.

    MOMENTUM is always conserved, in any sort of collision, because the forces acting on the two objects are always equal and opposite to each other.
     
  6. Oct 22, 2011 #5

    Andrew Mason

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    Not trying to be picky, but ENERGY is always conserved too. It is just not conserved as KINETIC energy.

    AM
     
  7. Oct 22, 2011 #6
    Going back to the original example, the body with velocity u and mass m strikes the stationary body of mass m. The collision is inelastic and the bodies stick together.
    Am I correct that in saying that the total momentum before the collision is mu and after the collision (2m)(1/2u) = mu and energy before collision is 1/2mu2 and after collision 1/2(2m)(1/2u)2 or 1/4mu2 ? Does exactly half of the energy remain as kinetic energy ?
     
  8. Oct 22, 2011 #7

    Andrew Mason

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    On those facts, yes. In the centre of mass frame, both masses are approaching the centre of mass at speed u/2. The energy before the collision is 2 x .5m(u/2)^2 = mu^2/4 and after the collision it is 0.

    AM
     
  9. Oct 23, 2011 #8
    Total momentum after collision = m(1/2u) + m(1/2u) = mu = MV

    where M = 2m and V = u/2 are mass and velocity of the composite body.

    energy after collision = 1/4mu2 = 1/2 2m (u/2)2 = 1/2 MV2

    Everything is right.

    What happens is that your collision is not elastic and does not conserve kinetic energy (total energy is still conserved). It can be showed that for a system of particles with the same mass, an elastic collision interchange the velocities. That is your stationary particle would acquire velocity u and the moving particle would stop. This is not the case which mean that your collision is not elastic
     
    Last edited: Oct 23, 2011
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