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Applying Conservation of Momentum along the Y axis

  1. Oct 14, 2015 #1
    1. The problem statement, all variables and given/known data

    A block of mass ##m## slides down a wedge of mass ##M## and inclination ##\theta## whose surfaces are all frictionless. Find the velocity of the block when it just reaches the bottom of the wedge.

    2. Relevant equations


    3. The attempt at a solution

    I was told that to solve this problem, we have to apply the concepts of Relative Velocity (between the block and the wedge), Conservation of Mechanical Energy, and Conservation of Linear Momentum. However, I've got a few doubts regarding the conservation of Linear Momentum.

    I know that we can conserve Linear Momentum in any direction as long as the net external force is ##0## in that direction. Now, the given method states that for the system ##Earth+Block+Wedge## I can conserve Linear Momentum in the ##X## direction ONLY. However, I fail to understand why we cannot conserve Linear Momentum in the ##Y## direction too. As per the FBD's I've made, the forces acting on the Wedge are the Normal force ##N_1## due to the Earth, and the Normal force ##N_2## due to the block. Also, the forces acting on the block are ##N_2## due to the wedge. The reaction force acting on the Earth is ##N_3## due to the wedge.

    Thus, the forces operating on the system ##Earth+Wedge+Block## are Internal Forces. Hence the Net External Force is ##0##.

    Thus, can't we apply Conservation of Linear Momentum in the ##Y## direction too? If so, could somebody kindly show me how to? Many thanks in advance!
     

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    Last edited: Oct 14, 2015
  2. jcsd
  3. Oct 14, 2015 #2

    haruspex

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    You can indeed apply conservation of momentum in the Y direction if your system includes the Earth. But you will not find that useful. What is useful is that there is no X direction force from the Earth, so you can apply horizontal momentum conservation to the wedge+block system.
     
  4. Oct 14, 2015 #3
    Thanks for responding Sir. Sir, I can't understand how to apply it though (ie I can't form the equations properly).

    Sir, the initial Momentum in the horizontal direction is ##0##. Thus the Initial Momentum along the Y axis is $$(M_{earth}\times 0)+(M\times 0)+(m\times 0)$$ However, for the final momentum along the Y axis, I get $$(M_{earth}\times 0)+(M\times 0)+(m\times -v\hat{i})$$ which is clearly not equal to initial Momentum. I made the assumption that the Earth has no velocity either initially or finally since till date I've always been taught (perhaps incorrectly) that we assume the Earth to be at absolute rest.

    In any case, when the block ##just ## ##reaches## the bottom of the earth, I assumed that the block had not yet collided with the earth. Hence, I fail to see what could impart any velocity to the Earth.

    Could you please correct my mistake Sir?


    Lastly, could you please explain how to apply the Conservation of Mechanical Energy assuming the System is ##Wedge+Block +Earth##?

    The equation I get (assuming the Earth does not move) is as follows:

    $$\large{PE_{earth}+mgH+Mg\dfrac{h}{3}=\dfrac{1}{2}MV_W^2+\dfrac{1}{2}mv_{block,earth}^2+PE_{earth}}$$ where U=0 at the Earth's surface.
     
    Last edited: Oct 14, 2015
  5. Oct 14, 2015 #4

    haruspex

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    The Earth will move in the Y direction, but by an extremely small amount. Just as the Earth attracts the block, the block attracts the Earth (with the same force). You can ignore the Earth's movement in regards to energy, but not in regards to momentum.
     
  6. Oct 14, 2015 #5
    But Sir, why do we disregard Earth's velocity with respect to Energy? I couldn't understand Sir.
     
  7. Oct 14, 2015 #6

    haruspex

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    The two atractions are equal (law of action and reaction). So the accelerations are in inverse ratio of the masses. If we take the Earth to be some huge number N multiple of the block's mass then the Earth's acceleration is 1/N times the block's acceleration. As a result, at any point in the descent of the block, the Earth's 'upward' velocity is 1/N times the block's downward velocity. That makes the two momenta equal and opposite, resolving the difficulty you had with conservation of momentum.
    But when we look at energy, that goes as mv2, so the Earth's KE is only 1/N times the block's KE. That makes it a tiny amount of energy, and we can safely ignore it.
     
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