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Is momentum conserved?

  1. Dec 2, 2009 #1
    I kind of confused about this physics question.

    Assume for example a 1 kg cart is moving at a constant speed of 5 m/s on a frictionless service. Then, a 1 kg weight is dropped onto the cart. Would its initial momentum be conserved? What would be it's final velocity?

    At first, the cart has a momentum of 5 kg m /s. If momentum were conserved, its final velocity would be 2.5 m/s. However, I do not really understand this. According to Newton's First Law of Motion, an object will remain in equilibrium unless acted on by an external force. The cart is apparently instantaneously decelerated to 2.5 m/s. How is this possible? What is the external force applied on the cart?

    If the cart remained the same velocity (5 m/s before and after the weight was added), then conservation of momentum wouldn't hold.

    Any help and insight would be greatly appreciated. It might just be that I don't really understand the laws of motion and momentum. Thanks in advance.
  2. jcsd
  3. Dec 2, 2009 #2
    You would view this as an inelastic collision. Yes, momentum is conserved. The collision will occur over some amount of time giving you what's called an impulse. force = momentum/time. So, the force between the box and cart acts over some delta time. Equal and opposite means force on box and cart is the same. Since both forces act over the same amount of time, box and cart change momentum by the same amount. force*(delta time) = delta momentum. All we see though is the delta momentum. The specific force and time are unknown in general, and are specific to the properties of the two objects and how they interact (ie large force/small time, small force large time). However, it is important for things like car crashes. If the impact time is small, that means the force is large, which means the people inside the car (or the cart) will probably die. This gives you ways to engineer objects so collisions are "safe".
  4. Dec 2, 2009 #3
    Oh, okay, thanks Longstreet, now I understand. If we assume that the weight is set (not thrown) on the cart, we could say that there is an inelastic collision occurring with the initial velocity of the weight being negligibly small (giving it a negligibly small initial momentum).

    Pi = Pf
    Pci + Pwi = P(c+w)f
    (1 kg)(5 m/s) + 0 = (x m/s)(2 kg)
    x = 2.5 m/s

    It also makes sense force-wise, the cart is putting force on the weight, speeding the weight up. Meanwhile, the weight is putting an equal and opposite force on the cart, slowing the cart down.

    However, if we hadn't set the weight down on the cart, but rather first accelerated it to the same speed as the cart and then set it down, it wouldn't change the final velocity.

    Again, thanks for clearing that up.
    Last edited: Dec 2, 2009
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