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Wedge problem

  1. Apr 6, 2015 #1
    1. The problem statement, all variables and given/known data:There is a wedge of mass kept on smooth surface. M and block of mass m is sliding in the depression of the wedge.And there is wall in left of the wedge.At which instant velocity of the wedge will be maximum?(surface of contact between wedge and block is smooth)

    motion.jpg
    2. Relevant equations
    free.jpg
    But the horizontal component of normal force is balanced by the wall reaction and vertical force is balanced by normal reaction on wedge .So no motion of wedge.So
    conservation of energy as no dissipative force is there.
    blocks potential energy is converted into it's kinetic energy completely.



    3. The attempt at a solution:
    But when it reaches at this point
    forum.jpg
    After this point the

    DIAGRAM.jpg
    When the block reaches at height h and the block height is achieved at this point,Vertical force is still balanced but horizontal force is not balanced by any force as wall is not on right hand side.
    So block accelerate.Now the block kinetic energy at mid point lets say 1/2 mv^2 of the depression is not completely converted to it's potential energy.As some goes in kinetic energy of the wedge.

    My teacher says velocity of the wedge will be maximum at the below instant
    forum.jpg
    I am not getting why?As can be seen from the image no net force is acting on thee wedge at this instant.Because the only force acting is in vertical direction which is balanced by normal force.I think velocity of the wedge should be maximum at
    MAXIMUM.jpg

    As in this the only force acting is in horizontal so maximum force ,maximum acceleration hence maximum velocity.

     
  2. jcsd
  3. Apr 6, 2015 #2
    You meant the maximum velocity of wedge or the block?
     
    Last edited: Apr 6, 2015
  4. Apr 7, 2015 #3
    maximum velocity of wedge
     
  5. Apr 7, 2015 #4
    Some errors were there in my original post.I am correcting them
    1. The problem statement, all variables and given/known data:There is a wedge of mass kept on smooth surface. M and block of mass m is sliding in the depression of the wedge.And there is wall in left of the wedge.At which instant velocity of the wedge will be maximum?(surface of contact between wedge and block is smooth)
    https://physicsforums-bernhardtmediall.netdna-ssl.com/data/attachments/64/64652-5de495f229ba4bd42a727d2570e0f6fc.jpg [Broken]

    2. Relevant equations
    https://physicsforums-bernhardtmediall.netdna-ssl.com/data/attachments/64/64654-a429551536e8d18cafe546da0c38a3dd.jpg [Broken]
    But the horizontal component of normal force is balanced by the wall reaction and vertical force is balanced by normal reaction on wedge .So no motion of wedge.So
    conservation of energy as no dissipative force is there.
    blocks potential energy is converted into it's kinetic energy completely.



    3. The attempt at a solution:

    https://physicsforums-bernhardtmediall.netdna-ssl.com/data/attachments/64/64659-5168d6ca9d612915662a0111d957a41e.jpg [Broken]
    When the block reaches at height h
    as in the below image(I forgot to specify height "h"in my picture.) the block maximum height in right hand side is "h" ,Vertical force is still balanced but horizontal force is not balanced by any force as wall is not on right hand side.
    So block accelerate.Now the block kinetic energy at mid point lets say 1/2 mv^2 of the depression is not completely converted to it's potential energy.As some goes in kinetic energy of the wedge.

    My teacher says velocity of the wedge will be maximum at the below instant
    https://physicsforums-bernhardtmediall.netdna-ssl.com/data/attachments/64/64656-7d13d77e1bad91857f1a6f7eca50d4bb.jpg [Broken]
    I am not getting why?As can be seen from the image no net force is acting on thee wedge at this instant.Because the only force acting is in vertical direction which is balanced by normal force.
     
    Last edited by a moderator: May 7, 2017
  6. Apr 7, 2015 #5
    Using conservation of momentum, ΣPi = ΣPf.
    ΣPi = 0
    0= MV + mv
    V= -mv/M
    If V = Vmax, then Vmax = -mvmax/M
    So now the question is when is v=vmax. The maximum velocity of the block is when the block has maximum kinetic energy and minimum potential energy by conservation of energy. So that is the point where the block is at the lowest point in the depression.
    Since there is a wall left side of the wedge, when the wedge has a maximum velocity to the left (block has maximum velocity to the right), there is a normal force by the wall on wedge to decelerate it to 0. So the wedge has V=0. But when the wedge has a maximum velocity to the right, there is no wall on the right side of the wedge thus it will move to the right at maximum velocity which happens when the block has a maximum velocity to the left.

    Or you can see it in this way: V = Vmax, then Vmax = -mvmax/M
    Taking right as positive. Since the wedge has to move to the right, Vmax must be positive so vmax must be negative which means the block has to move to the left with vmax.
     
    Last edited: Apr 7, 2015
  7. Apr 7, 2015 #6

    haruspex

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    A moment before the situation shown, where was the block in relation to the wedge? What direction would have been its force on the wedge? Which way would the wedge have been accelerating?
    Same questions for a moment after the position shown.
     
    Last edited by a moderator: May 7, 2017
  8. Apr 7, 2015 #7
    relation.jpg
    FORCE ON.jpg
    In the right direction.
     
  9. Apr 7, 2015 #8

    haruspex

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    Right. So had it reached max speed then?
    What about the situation a moment after that in the diagram?
     
  10. Apr 7, 2015 #9
    Which diagram?This one?
    rnhardtmediall.netdna-ssl.com%2Fdata%2Fattachments%2F64%2F64656-7d13d77e1bad91857f1a6f7eca50d4bb.jpg
     
  11. Apr 7, 2015 #10

    haruspex

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    Yes.
     
  12. Apr 7, 2015 #11
    The wedge doesn't accelerate any more as force is no longer there so it moves with constant velocity.Right?
     
  13. Apr 7, 2015 #12

    haruspex

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    In your teacher's diagram, the block is at the midpoint of the depression and moving left relative to the wedge. Where will it be relative to the wedge a moment later? What direction will its force on the wedge be?
     
  14. Apr 7, 2015 #13
    posituon.jpg
    One Component of normal force on wedge by block is In the left direction but it will be balanced by normal force by wall and the ground balances the other component of normal force on wedge by block.
     
    Last edited: Apr 7, 2015
  15. Apr 7, 2015 #14

    haruspex

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    In the position in your teacher's diagram, it is no longer in contact with the wall.
     
  16. Apr 8, 2015 #15
    Yes. I got your point .Wedge is no longer in contact with the wall,because the wedge has moved further.
    In the left direction, so it will decelerate the wedge.
     
  17. Apr 8, 2015 #16
    But why not the wedge has maximum velocity at this position
    upload_2015-4-8_13-42-47.png
    This is the moment just before the instant when block comes to mid point of depression with velocity in left direction.i.e
    ediall.netdna-ssl.com%252Fdata%252Fattachments%252F64%252F64656-7d13d77e1bad91857f1a6f7eca50d4bb.jpg
     
  18. Apr 8, 2015 #17
    Actually I am not getting what this theta (angle) is?Is horizontal component Nsin theta or N cos theta?
     
  19. Apr 8, 2015 #18

    haruspex

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    As long as the block is to the right of the midpoint of the depression, it is exerting a force to the right on the wedge. So the wedge is accelerating to the right, and is at less than its maximum speed. Maximum speed occurs when acceleration is zero.
     
  20. Apr 8, 2015 #19
    As long as the block is to the right of the midpoint of the depression, it is exerting a force to the right on the wedge.I wanted to ask is the force which causes acceleration of wedge varies as the block moves ?
     
  21. Apr 8, 2015 #20
    Is it right?
     
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