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Two masses and one spring

  1. Jun 14, 2013 #1
    I think it belong to Introductory Physics homehork, it's my first course college's physics course, I think it's pretty basic.

    We have particle 1 with a mass of 1.6 KG (left one), and particle 2 with an attached spring to it's left with a mass of 2.1 KG (right one).
    [Broken]
    They're both in a frictionless floor.

    The particle 1 has a speed of 4 m/s, and the particle 2 has a speed of -2.5 m/s (they are in the same line).
    The spring has a constant of 600 N/m

    If someone knows a good way to make sketchs involving springs in the computer I'll try to draw it, if the problem isn't clear I'll scan the sketch.

    (a) After the spring starts to shrink, at certain time the particle 1 has a speed 3 m/s. What speed has the particle 2 at the same time?
    (b) How much has the spring shrink at that time?.

    I know how to solve problems involving an immovable spring, but in this problem I always end up making an assumption I don't know how to justify in order to solve it.

    I know:
    [itex]\Delta U_e+\Delta U_c+\Delta U_g=0[/itex]
    [itex]U_e=k x^2[/itex]
    [itex]U_c=\sum_{i=1}^{n} \frac{1}{2}v_i ^2m_i[/itex]

    [itex]U_g=0[/itex] this isn't used in this problem.

    I can calculate how much will the spring shrink at the most because I can think the mass 2 as point of reference and mass 1 with a speed of 6.5 m/s, but I don't know how to solve this without making, for example, the assumption that the spring absorvs energy at the same rate from both sides.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jun 14, 2013 #2
    Are the particles connected to the spring? Not entirely clear from the description.
     
  4. Jun 14, 2013 #3

    gneill

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    Staff: Mentor

    I think you can assume they are connected; after all, the surface is frictionless so no energy is lost before contact is made. The fun begins when the spring starts to compress.
     
  5. Jun 14, 2013 #4
    I think it's correct but I edited the first post to clarity it, so people who read the thread for first time can understand what I meant in the first post.
    I hope I made it clear because after this post I'll be unable to edit the first one.
    PD: I think I should used rectangles instead circles, and I forget to state that the spring is massless.
     
    Last edited: Jun 14, 2013
  6. Jun 14, 2013 #5

    gneill

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    You should consider employing conservation principles. What's conserved no matter what the (inertial) frame of reference?
     
  7. Jun 14, 2013 #6
    It is always useful to think about the conservation laws. Which ones would apply here?
     
  8. Jun 14, 2013 #7
    Momentum? I didn't use it because I thought it wasn't valid when the spring absorvs a part of the energy, if I can still use it then I think I have all the equations I need.
    [itex]m_{1}v_{1i}+m_{2}v_{2i}=m_{1}v_{1f}+m_{2}v_{2f}[/itex]
    [itex]1.6 KG \cdot 4 \frac{m}{s} + 2.1 KG \cdot (-2.5) \frac{m}{s}=1.6 KG \cdot 3 + 2.1 KG \cdot v_{2f}[/itex]
    Then [itex]v_{2f}=-\frac{73}{42} \frac{m}{s}\cong -1.74 \frac{m}{s}[/itex]
    Now we calculate the cinetic energy and
    we can know how much has the spring compressed, with I think it's ~1.5 cm.
     
    Last edited: Jun 14, 2013
  9. Jun 14, 2013 #8

    gneill

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    Yup. Momentum is always conserved. Careful: you dropped the sign on your final result. And be sure to keep the units!
     
  10. Jun 14, 2013 #9
    I edited the message.
    Now we calculate the cinetic energy and
    we can know how much has the spring compressed, with I think it's ~1.5 cm.
     
  11. Jun 14, 2013 #10

    gneill

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    Can you show your work? That value looks too small to me.
     
  12. Jun 14, 2013 #11
    I also think it's pretty small, so here we go:
    [itex]\Delta U_e+\Delta U_c+\Delta U_g=0[/itex]
    [itex]U_e=k x^2[/itex]
    [itex]U_c=\sum_{i=1}^{n} \frac{1}{2}v_i ^2m_i[/itex]
    [itex]U_g=0[/itex]
    Then
    [itex]\Delta U_e+\Delta U_c=0[/itex]
    [itex]U_c=\frac{1}{2}((4 \frac{m}{s})^2 - (3 \frac{m}{s})^2) \cdot 1.6 KG+\frac{1}{2}((2.5 \frac{m}{s})^2 - (1.74 \frac{m}{s})^2) \cdot 2.1 KG=8.99 KG \frac{m^2}{s^2}[/itex]
    I think I made a mistake calculating it the first time, and...
    [itex]U_e=k x^2[/itex] isn't the formula I use, it's [itex]U_e=\frac{1}{2}k x^2[/itex] (as k is a constant I guess it's a matter of convention so no one noticed the mistake in the first post).
    [itex]8.99 KG \frac{m^2}{s^2}=\frac{1}{2}600(N/m) x^2[/itex]
    Then I got 17.3 cm this time, which doesn't seems so small.
     
  13. Jun 14, 2013 #12

    gneill

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    Staff: Mentor

    Yup. Much better :smile:
     
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