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Linear Momentum and Collision

  1. Nov 7, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-11-7_20-18-2.png
    Block 1 collides with a stationary block 2, block 1 bounces back going in the opposite direction. I assume by this type of picture it means that block 2 also is going to the right because it's not saying it's at rest any further.

    A) Rank the horizontal forces in both of the free body diagrams during the collision and explain?
    B) During the collision, is the net force exerted on block 1 to the left, right, or zero?
    C) Is the momentum of block 1 conserved during the collision, what about block 2? Is the change in momentum of block 1 to the left, to the right, or zero?

    2. Relevant equations
    Fnet=Δp/Δt
    3. The attempt at a solution
    A) I believe that at the point of collision, there is a normal force and a weight force. There would also have to be a force pushing back from the other block. So for say block 1, there would have to be a force pointing in the left direction. The same for block 2, just in the right direction. So... I want to say these horizontal forces are equal since the free body diagrams would appear the same.

    B)The net force exerted on block A would be the horizontal force, and it would be to the left because I believe that's the only force horizontally, and the vertical components would not be moving so they wouldn't be part of it.

    C) The momentum of block 1 will be conserved since we start with m1v1 and end with m1v1f. I, however, think the momentum for block 2 would not be conserved because we start with 0 and end up with m2v2f.

    The change in momentum of block 1 would be.. to the left because..
    m1(-vf)-m1v1, would make for a negative result, which would mean we would be going in the left direction, i think..
     
  2. jcsd
  3. Nov 7, 2015 #2

    TSny

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    Hello.

    There is a more specific reason why the magnitudes of the horizontal forces on 1 and 2 are the same. Hint: review all three of Newton's laws of motion.

    OK. But I'm not sure exactly what you mean when you say, "and the vertical components would not be moving so they wouldn't be part of it." Can you clarify that?

    Note that first you say that the momentum of block 1 is conserved. That means that the momentum of block 1 does not change. Then you say that there will be a change in the momentum of block 1 to the left.
     
  4. Nov 7, 2015 #3
    A) The forces would be equal but opposite in direction.
    B) I just meant that the sum of the force in the vertical direction=0 because of the lack of movement in the y direction, and so would not contribute to our net force. Our net force would ∑F= -Fcollision for block 1.
    C) And ah yes, I'm thoroughly confused by this part, as you can see..I believe the momentum is conserved because I think.. momentum is always conserved in collisions? So in which case there would be a zero change in momentum for block 1? The part that's confusing me is that block 2 doesn't have an initial momentum, so block 2 can't have conserved momentum, and the change in momentum must be m2v2f.. So we'd have
    m1v1+0=m1v1f+m2v2f
    So.. if the momentum is conserved as a whole..the initial momentum of m1v1 would have to be greater than m1v1f because after the collision we have m1v1f and m2v2f.. I'm probably making this way too complicated and confusing
     
  5. Nov 7, 2015 #4

    TSny

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    But, why?

    OK, there is no acceleration of block 1 in the y direction, so the forces in the y direction add to zero according to Newton's 2nd law.

    There is no law of physics that says that the momentum of each object separately must be conserved in a collision.

    Since ##\vec{p}_1 = m\vec{v}_1## for block 1, the only way its momentum can be conserved is for its final velocity vector to equal its initial velocity vector.

    Yes.

    Yes, the final momentum of block 1 will not be equal to the initial momentum of block 1.
     
  6. Nov 7, 2015 #5
    Total momentum is conserved, yes. So the momentum lost by one of the objects has to equal the momentum gained by the other. Remember that when you have a change in direction like you do for Block 1, you have to take that into account. So, like, if the momentum went from +2 to -2, that would be a loss of 4.
     
  7. Nov 7, 2015 #6
    Ok that makes a lot more sense, so block 1 would definitely have a greater change in momentum out of the two since it changes direction and will have a loss of momentum, correct? So, if block 1 does not have conserved momentum, is the change in momentum in the left direction.. because.. the change in momentum would be the change in velocity, since that's the only factor of the block 1's momentum that changes and our final velocity is a negative, so we'd have m1(-vf-vi), so we would have a negative change in momentum.. which would indicate the left direction..?:nb)
     
  8. Nov 7, 2015 #7
    Yup. Note that we write the change in momentum as ##m_1(v_f-v_i)## where ##v_f## is a negative number.
     
  9. Nov 8, 2015 #8

    TSny

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    If you are taking momentum toward the right as positive and momentum toward the left as negative, then block 1's momentum will change from positive to negative. Block 2's momentum will change from 0 to some positive amount.

    As an example, let's suppose block 1's momentum changed from +10 Ns to -5 Ns. Thus, the change in block 1's momentum would be Δp1 = (-5) - (+10) = -15 Ns. Would the change in block 2's momentum be less than, greater than, or equal to +15 Ns?
     
  10. Nov 8, 2015 #9
    Well put that way.. I think block 2's final momentum would be 15 also because..
    10 Ns= -5Ns+ 15 Ns
    to make both of the side equal, which would be what needs to happen. So, block 2's change in momentum would be equal to block 1's, i think

    If it's ok, I'd like to ask another question, hopefully that I don't make a mess of, and make sure my thought process about this whole collision thing is correct, am I correct to believe that the net force on this system of blocks while the collision is happening would be zero because the net force of block 1 would be ΣF=-Fcollision and block 2 would be ΣF=Fcollision, and when you add them together to find the net force on the system, it would come up to zero. And thinking about these two as a "system", since we know momentum is conserved, the change in momentum of this system would be zero..
     
  11. Nov 8, 2015 #10
    Equal in magnitude, opposite in sign (direction).

    When two equal-but-opposite forces act on different objects, like they do on these blocks, we don't describe it as a net force of zero. We reserve that language for cases when forces on a single object add up to zero, resulting in a net external force of zero.

    In a case like this, where the two blocks exert forces on each other, we say we have a system consisting of two objects. And the forces they exert on each other are internal forces. They are an equal-but-opposite pair in accordance with Newton's Third Law.

    If we had an equal-but-opposite pair of forces acting on a single object, we would say they add up to zero. We would do the same if they were acting on a system, but they would be external forces.

    If you get out of your car and push on the bumper to move it, that's an external force. If you stay in the car and push on the dashboard that's an internal force, it will of course never cause the car to move.

    When the net external force on a system is zero, the system's momentum is conserved.
     
  12. Nov 8, 2015 #11

    TSny

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    Yes, that's right. Block 1 changes its momentum to the left while block 2 changes its momentum by the same amount, but to the right. That is, the magnitude of the change in momentum of each block is the same, but the changes have opposite direction. So, the total change in momentum for the system is zero. In other words, the total momentum of the system does not change (it's conserved).

    Yes, that's correct. The net force on the system is zero and therefore the total momentum is conserved.
     
    Last edited: Nov 8, 2015
  13. Nov 8, 2015 #12
    Ok, great! Thanks so much you guys! This really helped.
     
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