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Energy and momentum conservation

  1. Dec 27, 2015 #1
    1. The problem statement, all variables and given/known data
    problems_MIT_boriskor_BKimages_10-mass-spring-two-block-collision.png

    A block of mass m is attached to a spring with a force constant k, as in the above diagram. Initially, the spring is compressed a distancex from the equilibrium and the block is held at rest. Another block, of mass 2m, is placed a distance x/2 from the equilibrium as shown. After the spring is released, the blocks collide inelastically and slide together. How far (Δx) would the blocks slide beyond the collision point? Neglect friction between the blocks and the horizontal surface.
    3. The attempt at a solution
    At first I thought that this question is very easy.But now I don't know how I should break the problem up such that I can write neat equations for momentum and energy conservation.
    should I consider the intermediate position when m reaches its equilibrium position and its energy is purely kinetic ? the question asks for delx but the equations contain only velocity terns and why will it even stop post collision there is no friction?

    edit:
    I understood that it will stop because it would have a leftward velocity and would eventually crash into the spring compressing it by delx
     
  2. jcsd
  3. Dec 27, 2015 #2

    PeroK

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    I wouldn't worry about the equilibrium point. So, what is the key stage in the motion after the block is released?
     
  4. Dec 27, 2015 #3
    the inelastic collision .
    but the kinetic energy is not conserved during such a collision.
    so should I write an equation just before the collision and will m's velocity be the same at x/2 (pre collision) as it was at position x
     
  5. Dec 27, 2015 #4

    PeroK

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    You're forgetting that the spring will pull the block back after the equilibrium point.
     
  6. Dec 27, 2015 #5
    actually I thought about the back and forth motion but it seemed too complicated so I thought that the block detaches itself at x

    so How should I think?
    at where it compresses to :we consider the P.E there ?
    Or should I read the chapter on SHM for this ?
     
  7. Dec 27, 2015 #6

    haruspex

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    Not sure what you are saying there, but it doesn't sound right. You do not need to get to the point where anything has a leftward velocity. You are only interested in how far the blocks get before anything goes to the left.
    The question is a little unclear. Do the blocks stick together or merely collide without bouncing at all? I think it means they stick together, otherwise they would start to separate immediately, even though there is no bounce.
     
  8. Dec 27, 2015 #7

    SammyS

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    Yes.


    No.

    Use conservation of energy to determine the velocity of block of mass m immediately before the collision.
     
  9. Dec 27, 2015 #8
    yes they do.

    I am not understanding the situation here @PeroK said that the block will not detach itself from the spring at x(the equilibrium) so it will go back and forth until it hits 2m
    than it will stick to 2m and go "leftwards" causing delx compression which we are supposed to figure out.
    what is going on? what what point should I think about the energies?
     
  10. Dec 27, 2015 #9

    haruspex

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    It won't go back and forth before hitting 2m. If 2m were not there, how far would it go before going leftwards at all?
    The question asks how far they go "beyond the collision point". That is, how far before reversing direction.
     
  11. Dec 27, 2015 #10

    SammyS

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    The problem statement says that the block of mass m is attached to the spring.
     
  12. Dec 27, 2015 #11

    PeroK

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    It says the block is attached to the spring. If it wasn't attached it would be jettisoned at the equilibrium point. But, it is attached so it stays attached.

    Although you may want to calculate the velocity at the collision you don't have to.

    As Haruspex says, the blocks must stay stuck together after the collision.
     
  13. Dec 27, 2015 #12
    so m will go beyond x , collide with m2 at x/2 , stick with it, and then they both move forward .
    we have to calculate how far before spring pulls them back


    I wrote an equation with equilibrium as 0 position
    and equating the initial potential energy of m (##0.5kx^2##)and the final potential energy of 3m (##0.5k((x/2)+(Δx))^2##)
    but I don't get the correct answer on solving this quadratic equation
     
  14. Dec 27, 2015 #13

    SammyS

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    For the inelastic collision:
    There will be a loss of kinetic energy. Right?
     
  15. Dec 28, 2015 #14
    first I considered the situation b/w the launch and the collision
    here the mechanical energy is conserved.
    It gets converted from purely potential to partly kinetic and partly potential(because it moves past the equilibrium point )
    I used this to find the velocity of m right before the collision

    during the INELASTIC collision the momentum is conserved(because of negligible internal forces) and energy is not
    I found the velocity of (now) 3m right after the collision

    then again 3m moves further out until it gains max potential energy
    we know that the post collision -part K.E and part P.E gets converted wholly into potential energy

    the displacement after the the collision gave me the answer . thanks everyone!!!
     
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