1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Wedge and block initial momentum

  1. Apr 6, 2015 #1
    1. The problem statement, all variables and given/known data
    :block slides on smooth triangular wedge kept on smooth floor.Find velocity of wedge when block reaches bottom.
    (image is attached)
    2. Relevant equations:Let the velocity of the triangular wedge be V.And v be the velocity of block m.

    applying conservation of linear momentum in the horizontal direction we get

    MV +m(v cos theta -V)=0
    And because of conservation of energy
    1/2 MV^2+1/2 m[(v cos theta - V)^2 +v sin theta]^2
    The answer comes out to be


    3. The attempt at a solution:I understood all these except conservation of linear momentum part.

    MV +m(v cos theta -V)=0
    here
    MV +m(v cos theta -V) is final momentum
    and 0 is taken as initial momentum.
    why their initial momentum and velocity is taken as zero?I have learnt in constrains, force responsible for motion of wedge is Nsin theta which is always there as long as block and wedge are in contact.And similarly motion of block is because of mg sin theta which should be always there if block is on the wedge.Then which initial condition are you referring to when these forces are not there so that they have zero velocity?I know we are taking blocks and wedge as our system now so normal forces become internal and cancels out.This problem must be a part of kinematics,Kinematics is the branch of classical mechanics which describes the motion of points, bodies (objects) and systems of bodies (groups of objects) without consideration of the causes of motion.

    But I am confused why initial momentum is zero?
     

    Attached Files:

  2. jcsd
  3. Apr 6, 2015 #2
    The answer comes out to be sorry I am not able to use latex.
     
    Last edited: Apr 6, 2015
  4. Apr 6, 2015 #3
    upload_2015-4-6_11-51-29.png
    Above is the answer
     
  5. Apr 6, 2015 #4
    WEDGE MOTION.jpg
     
  6. Apr 6, 2015 #5
    Nothing is moving initially, right?
     
  7. Apr 6, 2015 #6
    That's what I want to know.What really is initial condition?When block was not placed on wedge?
     
  8. Apr 6, 2015 #7

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The problem text is not clear. Did you copy it correctly?
    The initial condition should be both velocities (that of the wedge and the block) and the position of the block. As the velocity of the block is asked when it reaches the bottom of the wedge, it must start from some height h. The simplest initial condition is that the whole system is in rest when the block is at height h.
     
  9. Apr 6, 2015 #8
    Are you talking about this ?
     
  10. Apr 6, 2015 #9
    But if the block is at height "h"i.e on the block.Then they will not be at rest.Force responsible for motion of wedge is Nsin theta which is always there as long as block and wedge are in contact.And similarly motion of block is because of mg sin theta which should be always there if block is on the wedge.
     
  11. Apr 6, 2015 #10

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Yes, I talk about your problem. The whole system can be initially in rest if you keep it in rest, and it starts to move when you release it.
    By the way, the solution you cited is valid for the initial condition that the block was at height h initially, and the system was in rest.
     
  12. Apr 6, 2015 #11
    Is this height of wedge?
     
  13. Apr 6, 2015 #12
    What does it mean to keep something in rest and to release it?I often see these terms inh physics problems but I really don't understand how to interpret it?
     
  14. Apr 6, 2015 #13
    Imagine that nothing is moving because you are holding the block in place at the top of the wedge. Then imagine letting go of the block, and allowing it to move. That is what it means.
     
  15. Apr 6, 2015 #14

    ehild

    User Avatar
    Homework Helper
    Gold Member

    h is the initial height of the block. The wedge can be of any height, but you put the block on the wedge at a point that is at height h above the ground.
     
  16. Apr 6, 2015 #15
    Yes, or like ehild says, h can be any arbitrary height on the wedge.
     
  17. Apr 6, 2015 #16

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Imagine you hold a pebble in your hand. It is in rest. It starts to fall when you release it, you let it fall out from your hand.
     
  18. Apr 6, 2015 #17
    Am I correct?
    Am I correct?
     
  19. Apr 6, 2015 #18
    But in the problem nothing such is mentioned that block is holded and then released.
     
  20. Apr 6, 2015 #19
    Exactly. That is why I asked for you to elaborate upon whether or not the system is initially at rest. The solution that you provided is only valid under the assumption that the system is initially at rest.
     
  21. Apr 6, 2015 #20

    ehild

    User Avatar
    Homework Helper
    Gold Member

    What was the original text of the problem?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Wedge and block initial momentum
  1. Block on Wedge (Replies: 3)

  2. Block on a wedge (Replies: 2)

  3. Block on a wedge (Replies: 1)

  4. Block on Wedge (Replies: 25)

  5. WEDGE and block (Replies: 3)

Loading...