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Block collision attached to spring

  1. Aug 10, 2015 #1
    1. The problem statement, all variables and given/known data


    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 distance x 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.

    2. Relevant equations
    U_spring = 1/2*k*x^2

    K = 1/2*m*v^2

    F = m*a

    v_f^2 = v_0^2+2*a*x

    3. The attempt at a solution
    I used conservation of energy to find the velocity where the two blocks collide at x/2. I'm not sure about this though, because I don't think that conservation of momentum applies and I used 3m for the mass (assuming that the blocks had collided). Then I used N2L to find the acceleration based on the force (-kx) and mass (3m). I plugged that into the kinematics equation for final velocity and solved for delta x. I'm really not having any luck.

    E_i = 1/2*k*x^2 = E_f = 1/2*m*v^2 + 1/2*k*x_f^2

    at x/2 where m collides with 2m:

    1/2*k*x^2 = 1/2*3m*v^2 + 1/8*k*x^2

    k*x^2 = 3m*v^2 + 1/4*k*x^2

    3/4*k*x^2 = 3*m*v^2


    v = sqrt(k*x^2/(4*m))

    by newtons second law a = -k*x/(3*m)

    and using kinematics (v_f^2 = v_0^2+2*a*x)

    0 = k*x^2/(4*m) - 2*(k*x/(3*m))*delta(x)

    x/4 = 2/3*delta(x)

    delta(x) = 3/8*x

    Am I way off base with this approach?
  2. jcsd
  3. Aug 10, 2015 #2

    Doc Al

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    Staff: Mentor

    You can use conservation of energy to find the velocity of mass m just before the collision. But energy is not conserved during the collision itself. (But what is conserved?)
  4. Aug 10, 2015 #3
  5. Aug 10, 2015 #4
    According to what I did understand of your method to solve the problem.You have conserved energy in an inelastic collision which I don't think is valid.
    Conservation of linear momentum however will hold true.I suggest you use that to find the final velocities of the blocks.

    Edit:I'm sorry,I didn't see the earlier replies to the post.
  6. Aug 10, 2015 #5

    Doc Al

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    Staff: Mentor

  7. Aug 10, 2015 #6
    so according to conservation of momentum mass m has velocity sqrt(3*k*x^2/(4*m)) before the collision and according to conservation of momentum the blocks have velocity sqrt(3*k*x^2/(4*m))/3 after the collision. Can I find the acceleration and distance traveled with the same approach I tried before?
  8. Aug 10, 2015 #7
    so if I find the velocity of the masses after the collision can I use N2L and kinematics to find distance traveled?
  9. Aug 10, 2015 #8
    I'm assuming not because I got another wrong answer
  10. Aug 10, 2015 #9

    Doc Al

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    Staff: Mentor

    I believe you mean conservation of energy here.


    No. That method was flawed since you assumed constant acceleration.

    During the collision, energy is not conserved. But before and after the collision it is.

    No, but see above for a hint.
  11. Aug 10, 2015 #10
    ahhh so I can use conservation of energy again
  12. Aug 10, 2015 #11
    so is it 1/2*m*v^2 + 1/2*k*(x/2)^2 = 1/2*k*delta(x)^2

    = x^2/12+x^2/4 = delta(x)^2

    delta(x) = .57735?
  13. Aug 10, 2015 #12
    Nope. So what am I doing wrong now?
  14. Aug 10, 2015 #13

    Doc Al

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    Staff: Mentor

    Make sure you have the right mass and speed for the KE term. And express the final PE term in terms of Δx properly. Δx is the distance beyond the collision point, not the final distance from equilibrium.
  15. Aug 10, 2015 #14
    Got it, thanks.
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