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Momentum Collission Inclined Plane 2

  1. Dec 4, 2011 #1
    Yet another question i was unable to finish last semester which still bugs me.... D:

    A ball drops from (H) with a velocity = 0. with a mass M. It collides with an inclined plane with a mass M1. the ball goes exactly horizontally to the right, the incliend plane goes to the left. Momentum is conserved in the X direction. Energy is conserved. What are the velocities of the plane and the ball after collision? The angle of the inclined plane is 45 degrees.


    here is a picture of the question
    http://s12.postimage.org/a8lmqwqwd/momentum_conserved_in_the_x.png

    My attempt:
    So i set the gravitational potential point at 0 where the height is.
    ET1 = ET2
    EG1 = EK2
    mgh=1/2mv2^2
    V2= √MGH

    This is the velocity before the collision. it is also in the Y direction according to my axis.

    so now.. i'm stuck.. because how can i compare a Y momentum with an x compenent which is conserved... what is the relation i don't see it at all...

    collision formula: mv2y+0 = mv2y' M1vm1y'

    but now i'm left with two unknowns..

    so i compare change in EK between... from where it falls to before the collision.

    ΔEKball = 1/2 mv2^2 - 0
    ΔEKball = 1/2 M1Vm1^2 - 0


    but i i'm not sure if this comparison is possible... :(

    some advice is required please :(
     
  2. jcsd
  3. Dec 5, 2011 #2

    ehild

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    When the ball collides with the slope they interact with a force normal to the inclined plane. This force of interaction has a horizontal and a vertical component. The vertical component of the force exerted on the ball changes its vertical momentum, so it becomes zero after the impact, the horizontal component accelerates the ball to the final horizontal velocity. The horizontal component of the force exerted by the ball on the slope accelerates the slope to a velocity opposite to that of the ball, but the vertical force component is balanced by the normal force from the table. The normal force is just so great so it balances the force from the ball. The normal force from the table is an external force on the ball-slope system. The momentum is not conserved when there is an external force during the collision.

    As there is a vertical external force, the vertical component of the momentum is not conserved. You can write conservation of momentum only for the x components. The sum of the horizontal components of momentum was zero before the collision, so it has to be zero also after it.
    You have an other equation for the speeds: conservation of energy. You can determine the velocities from these two equations: conservation of energy and conservation of the horizontal component of momentum.


    ehild
     
  4. Dec 5, 2011 #3
    You can determine the velocities from these two equations: conservation of energy and conservation of the horizontal component of momentum.


    ehild[/QUOTE]


    Am i allowed to compare the the energies of the inclined plane, and the ball?
    ie;
    ΔEKball = ΔEKPlane

    if not can i compare Changes in kinetic energy.. ie:

    ball at maximum height dropped where it reaches max velocity before hitting plane
    ΔEKball=1/2mv2^2 - 1/2mv2^2
    ΔEKball = 1/2mv2^2

    Balls energy right before it hits the inclined plnae and right afterward. (consequently the same velocity 2 from the equation above)
    ΔEKBall2= 1/2mv3^2 - 1/2mv2^2


    ΔEKBall = ΔEkball2
    1/2mv^2 = 1/2 mv3^2 - 1/2mv2^2


    And from this point.. if this strategy works i can rearrange vectors into a triangle... and just use pythtag theorem to solved for V3
     
  5. Dec 5, 2011 #4
    or is it change in total energy as opposed to kinetic..
     
  6. Dec 5, 2011 #5

    ehild

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    The total energy of the whole system is conserved in the collision. The ball has only potential energy at the beginning, transforms into KE just before the impact with he slope. The slope is in rest and we can take its potential energy zero.
    Let be the mass of the ball m and the mass of the slope M. The speed of the ball before the impact is vi, that of the slope is Vi =0.
    The speed of the ball after the impact is vf, that of the slope is Vf

    So the initial energy of the system is

    Ei=1/2 mvi^2+1/2MVi^2=mgH--->

    Ei=mgH.

    The collision changes the velocities, therefore the kinetic energy of both the ball and the slope, but does not alter the potential energies. So the total energy of the system is Ef=1/2 mvf^2+1/2MV^2. The final energy is the same as the initial energy:

    mgH=1/2 mvf^2+1/2MV^2.

    As I wrote in my previous post, the horizontal component of momentum is conserved, but the vertical component does not. Initially the velocity of the stone is vertical, the horizontal component of its momentum is zero. So the x component of the total momentum is zero before impact and the same after impact.

    mvf+MVf=0

    You have two equation for the two unknowns, vf and Vf.

    ehild
     
  7. Dec 6, 2011 #6
    Thank you very much fir your time it is much appreciated
     
  8. Dec 7, 2011 #7

    ehild

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    You are welcome. Could you solve the problem at the end?

    ehild
     
  9. Dec 7, 2011 #8
    It's a multistep problem the ball actually collides with another plane and moves another distance up of h2. So I'm attempting to solve the end of the problem I'll post what I have tonight after school I'm just I class
     
  10. Dec 7, 2011 #9
    So... we have

    mo = mass of ball - vi = 0

    So.. we have ET1=ET2 to solve for the velocities ball and we get

    V2=√2GH1

    Now as you assisted with me:


    Ei= 0 + 1/2MoV2^2
    Ef = 1/2mov3^2 + 1/2MVM1^2

    Ei=Ef

    MogH1=1/2mov3^2 + 1/2MxVM1^2

    i simplify and rearrange for..

    Vm1 = √[Mo(2gh1-V3^2)/M]

    Px=P'x

    0 = MoV3x + MVm1x

    Vm1 = -mov3/M

    VM1 = Vm1

    √[Mo(2gh1-V3^2)/M] = -mov3/M -- square both sides.. then simplify

    V3 = √[2MgH1/(M+mo)

    Now that i have the velocity of the ball after the collision, the ball accellerates with a constant V until it hits the second plane which is the same mass M, and has the same angle of 45 degrees. Momentum is conserved in the X direction once again.

    Px=P'x
    mov3 = MVm2
    Vm2 = Mov3/M

    Ei2= 1/2mov3^2
    Ef2=1/2moV4^2 + 1/2 MxVm2^2

    Ei2=Ef2

    Vm2 = √[mov3^2-mov4^2/M]

    Vm2=Vm2 -Solve for V4

    √[mov3^2-mov4^2/M] = Mov3/M

    V4 = √[Mv3^2-mov3^2/M]


    Now the ball bounces up a height of H2, its final velocity is zero, because that is its maximum height..

    I used kinematics to solve for this

    d = Vf^2 - Vi^2/2g

    H2 = -Vi^2/2g

    i simplified for...

    H2 = H1 [ (M-mo)/(M+Mo) ]

    This answer makes sense because It is the height compared to the height, and all of the units within the parenthesis cancel out, so there for the only unit remaining is H1 which is a Metre.


    I am fairly certain i simplified correctly but there were a few tricky negatives in there.



    Thanks a lot for your time man. I'm glad that i can add this problem to my collection, and also all the tools i gained with it... I've had this problem for like 8 weeks haha.
     
    Last edited: Dec 7, 2011
  11. Dec 7, 2011 #10
    If you do have more time, i would like to ask you to give me a hint for another problem i'm still working on... You can never have too many problems if you know what i mean :D
     
  12. Dec 8, 2011 #11

    ehild

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    Hi Plutonium,

    It is quite hard to follow your notations...
    Why not use only m for the ball instead of mo? And you use the notations both M and Mo for the slope. M would be enough.
    And you can indexing variables by hitting the x2 key. Does not look vm1 better than vm1? Or just v1....


    Apart from the missing bracket, it is correct.

    It does not accelerate. Moves or travels..

    You meant Vm2 = mov3/M, did you not?

    Yes, it makes sense and even more, it is correct! congratulation. But take more care on typing in. Do not forget parentheses, try to use indices, choose simpler notations, and what abut using tex? It is very easy here in the Forums.
    And you are welcome with other problems.

    ehild
     
    Last edited: Dec 8, 2011
  13. Dec 8, 2011 #12
    Yea my bad I need to work on my typed physics skills also I tend to overuse subscripts and superscripts. I don't have paint right now but I'll see if I can word my next question properly. If u don't understand I'll paint it when I get home.
     
  14. Dec 8, 2011 #13
    You have a circle with a small ball on top at rest. You have the diameter of the big circle. The small ball rolls down the circle at a specific point the small ball detaches from the big circle. What is the distance between the top of the circle and the point the ball detaches? Given info; mass of ball and diameter
     
  15. Dec 8, 2011 #14

    ehild

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    Hi Plutonium,

    This is a very typical problem, I think I understand it. But it is better to start a new thread with this entirely different problem, and you should show your attempt for the solution. The cue of solution is that the normal force on the surface of the big sphere can act only outward.

    ehild
     
  16. Dec 8, 2011 #15
    Ah i put the problem up in another thread.
    https://www.physicsforums.com/showthread.php?t=558306
     
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