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Conservation of Momentum Questions

  1. Aug 2, 2013 #1
    I was wondering if anyone can share their expertise on the following two problems. I was able to solve both of them and arrive at the correct answer. For the first problem, I have additional questions and for the second problem I am hoping that there is an easier way to solve it than the method I used.

    1) An open cart moves horizontally along a frictionless surface at constant speed. A large load of coal is suddenly dumped into the cart. What happens to the speed of the cart?

    Now I understand that this is a simple conservation of momentum problem. Since the mass of the 'cart' is increased, its velocity must decrease to maintain the same momentum. I am trying to understand this however on a deeper level. When the coal is dumped into the cart, we can no longer call the cart an isolated system as it is being acted upon by an external force. So is the reason that momentum is conserved because there is no 'horizontal' net force applied to the cart or because we are including coal in the system? I was thinking that the momentum of the coal is not zero as it is being dumped into the cart so the total momentum of the system before the coal lands in the cart is a combination of two momenta, the carts and the falling coals so why isn't that factored in. The coal has negative vertical momentum and no horizontal momentum so maybe the negative momentum is transferred to the ground and 'lost' from the system while the horizontal momentum is conserved and leads to the decreased velocity of the cart/coal system. Another question I had was according to Newton's Second Law, F = ma, a change in velocity requires a net force. The carts horizontal velocity changed yet what exactly was the horizontal net force that precipitated this change. If it was the coal that led to a negative horizontal force on the cart then I am thinking that this must mean a horizontal impulse was provided to the cart during that interaction leading to a decrease in momentum of the cart (and only the cart). So the momentum lost by the cart is gained by the coal and hence the total momentum of the system is conserved.

    I hope I'm thinking of this in the right way. I love thinking deeper about simple problems. There is so much depth to physics!

    2) A ping-pong ball originally at rest is hit head-on by a bowling ball traveling with initial speed v. The mass of the ping-pong ball is negligible compared to the mass of the bowling ball. It is an elastic collision. What is the speed of the ping-pong ball in terms of v?

    I approached the problem as follows:
    Mv^2 = Mvb^2 + mvp^2 (elastic collision)
    Mv = Mvb + mvp (conservation of momentum)

    but then I had to do quite a bit of algebra and substitution to solve for vp (the velocity of the ping-pong ball) and in the end I got: vp = 2Mv/(m+M) which simplifies to 2v which is the correct answer.

    I was wondering if anyone knows of a simpler, more intuitive way at arriving at this answer. Why is it that when an extremely heavy object collides elastically with an extremely light object at rest that the light object ends up traveling at twice the velocity of the heavier object?

    Thanks so much for your help!

    -Junaid
     
  2. jcsd
  3. Aug 2, 2013 #2

    mfb

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    Staff: Mentor

    1. The horizontal component of the momentum in the system "coal+cart" is conserved, as all external forces (gravity and normal force) are vertical. Before the coal reaches the cart, this horizontal momentum is just the horizontal momentum of the cart.
    The vertical momentum component is not conserved.

    Work in the frame of the bowling ball, and treat it like a wall.
     
  4. Aug 2, 2013 #3
    OK this is the first time I'm trying this approach. Please let me know if my logic is correct:

    I let the bowling ball be the wall. I had a small ball of mass m and velocity v approach it. In this situation the bowling ball is stationary and has no momentum or kinetic energy. Thus the total momentum of the system is -mv and the total KE is (1/2)mv^2. What I would expect to happen is that the ball bounces backwards with the same speed (opposite direction) so the new total momentum is +mv and KE is conserved. This means that Δp = +2mv.

    When I apply the same Δp to the original problem in which the small ball started from rest, I arrive at a final velocity of 2v. Is this correct the way I did it? When you let the bowling ball move with velocity v, the total momentum of the system is conserved. When you use the bowling ball as the frame of reference, the total momentum is not conserved. Or is it that we assume that -2mv Ns of impulse was delivered to the wall and lost from the system?

    Thanks,
    Junaid
     
    Last edited: Aug 2, 2013
  5. Aug 2, 2013 #4

    mfb

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    Yes. I would use Δv=2v instead of Δp = 2mv, but that is just a matter of taste.

    It is conserved in the frame of the original bowling ball motion as well - the bowling ball gets a momentum of -2mv, together with a negligible energy.
     
  6. Aug 2, 2013 #5
    Thanks a lot. It actually makes sense now. I agree, using Δv would be a nicer way to do it.

    -J
     
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