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Another momentum problem

  1. Jul 17, 2005 #1
    A 0.150 kg frame when suspended from a spring, stretches the spring 0.050 m. A 0.200 kg lump of putty is dropped from rest onto the frame from a height of 30.0 cm. I have to find the maximum distance the frame moves downward from its initial position.

    I have calculated the spring constant to be 1.47N/0.050m = 29.4 N/m

    The first stage is when the putty is falling down or:
    Vp2 = sqrt(0.6*g)

    Then I'm not sure. Can I say that momentum is conserved during the collision of the putty and frame?

    The last stage would be conservation of energy when putty and frame travel downwards. But I need to know the total speed.

    Could someone please give me a hint
     
  2. jcsd
  3. Jul 17, 2005 #2

    Pyrrhus

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    Homework Helper

    Yes use conservation of momentum to find the speed of the system putty+frame.
     
  4. Jul 17, 2005 #3
    But since there is a force of gravitation can I really use conservation of momentum?
     
  5. Jul 17, 2005 #4

    Pyrrhus

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    Yes, it would be an impulse approximation. During the collision the force will be greater than the magnitude of the gravitation force, because the time of collision will be small.
     
  6. Jul 18, 2005 #5
    O.K. What I've done is this.

    For the spring I calculated the spring constant

    K=1.47N/0.050m = 29.4 N/m

    For the falling putty I found the final speed of the putty just before it hits the frame.
    Vp2 = sqrt(g*0.6)

    The collision of the putty and frame is
    Vt = speed of putty and frame

    Vt = mpVp2/mt
    (where mp is the mass of the putty, mt is the mass of the putty and frame).

    Then when the frame and putty are travelling downwards I use conservation of energy

    Ug2 + Ue2 = K1 + Ug1 + Ue1 - K2
    (Ug =gravitational potential energy, Ue = elastic potential energy)

    Here I end up with:

    y(squared)*k*0.5 + y*g*mt = 0.5mp(squared) * g*0.6/mt(squared) + 0.0367

    From this I get y = (-3.43 +- sqrt(11.8+10.7) )/29.4

    I should get 23.2 cm But I always get a wrong answer.

    Does anyone see what I'm doing wrong?
     
  7. Jul 18, 2005 #6

    Doc Al

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    Staff: Mentor

    I can't quite follow the details of your calculations, but when you apply conservation of energy after the collision be sure to take into account that the spring has an initial elastic potential energy since it is stretched. You are trying to find the amount of additional stretch.
     
  8. Jul 18, 2005 #7
    After the collision I did:

    Ug2 + Ue2 = K1 + Ug1 + Ue1 - K2
    (Ug =gravitational potential energy, Ue = elastic potential energy)

    mt*g*y + 0.5*k*y(squared) = 0.5*mt*vt(squared) + 0 + 0.5*k*0.050(squared) - 0

    where k is the spring constant

    I take the origin to be where the spring in unstreched and the stretched position as -0.050m.
     
  9. Jul 18, 2005 #8

    Doc Al

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    Staff: Mentor

    What about Ug1? If the origin is the unstretched postion, then Ug1 is not zero.
     
  10. Jul 19, 2005 #9
    Ug1 was the problem. Thanks Doc Al.
     
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