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Conservation of momentum for accelerating object

  1. Feb 3, 2015 #1
    I always see that a ball moving in a constant speed collide with another ball is common conservation of momentum problems. Obviously, I know that the momentum is mass*velocity

    But, what if the ball accelerates and collides with another ball ?
    I think I can't use mass*velocity for the momentum because the velocity is not constant.

    Then, I think that I might have missed the Δ symbol.
    Because I think that the conservation of momentum is derived like below.
    F = -F
    m Δv/Δt = m Δv/Δt
    Because the object A contacts object B at the same time as the object B contacts object A, so we can cancel Δt
    So, m Δv = m Δv

    So, is the formula m Δv ?
    But, I doubt it, because I think non-accelerating objects which has Δv = 0 since its velocity doesn't change also have momentum
    Or, maybe the formula is mass times the velocity the object has when contacting another object.
    But, I don't know how to derive the formula (p = m*v) since F = ma = m Δv/Δt has a delta symbol and momentum formula doesn't have delta symbol.

    So, What is the momentum formula ?
    Is it mass times the velocity the object has when contacting another object or m v ?
    Or is it m Δv ?
    If the formula is m v , Please tell me how to derive that formula.
     
  2. jcsd
  3. Feb 3, 2015 #2
    The momentum is mv. If the velocity increases (the body is accelerating) the momentum increases too. This is what Newton's second law states (in the momentum format): if there is net force acting on the system, the momentum changes (F=Δp/Δt).
    The momentum is conserved only if the net force is zero. If there is a non-zero net force, the momentum is not conserved, of course.

    So for the collision described, if there is an external force acting on one or both colliding bodies, the momentum is not conserved. By external force I mean a force which is not due to the interaction between the colliding bodies but interaction with other bodies, not participating in the collision.
    However, as the collision time is (usually) very short it may be possible to neglect the change in total momentum that occurred during the collision. It depends on the problem if you can or cannot do this.
     
  4. Feb 3, 2015 #3

    A.T.

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    Delta means difference. The difference of two momentum quantities is still a momentum. Just like the difference of two lengths is still a length.
     
  5. Feb 3, 2015 #4

    Svein

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    If just the two balls are involved, the law of conservation of momentum reads mAvA = mBvB.

    This law is valid whether the impact is elastic or non-elastic. Also observe that Newton's original formula reads [itex]\frac{d(mv)}{dt}= \frac{dm}{dt}\cdot v + m\cdot \frac{dv}{dt} [/itex]. Usually the mass is constant, so [itex]\frac{dm}{dt} = 0 [/itex]. When your you are dealing with rockets, you need the complete formula (the mass decreases as the fuel is used up).
     
  6. Feb 4, 2015 #5
    Suppose ball A is accelerating and going to hit ball B which is static. And, both balls have the same mass, and the collision is elastic
    After the collision, the acceleration of the ball B is the same as the initial acceleration of the ball A, right ?
    And, after the collision, ball A's acceleration is the same as B but, with opposite direction, right ?
    The contact force is mass * the acceleration the ball A has before collision, right ?

    I think there is no external force when the ball is moving. The external force is just to get the ball moving (accelerate).
     
  7. Feb 4, 2015 #6

    Svein

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    None of the relevant laws say anything about acceleration. Use the "conservation of momentum" formula. In addition, for an elastic collision, the kinetic energy is conserved.
     
  8. Feb 4, 2015 #7

    A.T.

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    You seem to confuse acceleration and velocity.
     
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