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Spring energy

  1. Apr 21, 2015 #1
    1. The problem statement, all variables and given/known data
    A block of mass m = 0.1 kg is connected to a spring of unknown spring constant k. It is compressed to a distance x from its equilibrium position and released from rest.
    After approaching half the distance (x/2) from equilibrium position, it hits another block and comes to rest momentarily, while the other block moves with a velocity 3 m/s. The total initial energy of the spring is:
    1.5 J
    0.6 J
    0.3 J
    0.8 J

    2. Relevant equations
    Spring potential energy is 1/2 kx2
    In this question we have to do basically energy balance

    3. The attempt at a solution
    Isn't this question wrong?
    We should be given other block mass and told that it is an elastic collision?
     
  2. jcsd
  3. Apr 21, 2015 #2
    Why don't you post your work so we can see what you have so far. It is solvable.
     
    Last edited: Apr 21, 2015
  4. Apr 21, 2015 #3
    Initial energy = 1/2 kx2
    At x/2 distance,
    1/2 kx2 = 1/2 kx2/4 + 1/2 mv2
    Now we know the mass of the block which is colliding but don't know velocity at that instant of colliding.
    If we have been told that other block has same mass and it is an elastic collision then v could be 3m/s ?
     
  5. Apr 21, 2015 #4
    If the first block stops moving after colliding with the second block, and the second block continues forward with a velocity, then I think it is safe to assume an elastic collision. Although it would have been nice if they had specified that it was an elastic collision in the problem statement.

    Your work looks good so far. What do you know about elastic collisions?
     
  6. Apr 21, 2015 #5
    In elastic collisions momentum and energy are balanced initially and finally.
    If two blocks should have same velocity one transferring other, then masses should be same.
    But they have not given that.
     
  7. Apr 21, 2015 #6
    Bingo!

    Set up the conservation of momentum equation and the conservation of energy equation. You have two unknowns, and two equations relating those two unknowns.
     
  8. Apr 21, 2015 #7
    Momentum conservation applying
    0.1 v = 3m
    Applying energy balance
    1/2 * 0.1v2 = 9/2 m
    Yeah v = 3m/s
    But as I say, how we are assuming elastic collision?
     
  9. Apr 21, 2015 #8
    Well, they really should have specified. They obviously want you to treat the collision as perfectly elastic, because they don't mention any energy being lost due to non-conservative forces (such as friction). But really, I guess we don't know. It is the fault of whoever made this question.
     
  10. Apr 21, 2015 #9
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