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Homework Help: Is my solution correct? About momentum

  1. Aug 22, 2008 #1
    Block 1 (mass 6 kg) is moving rightward at 8.0m/s and block 2 (mass 4 kg) is moving rightward at 2.0m/s. The surface is frictionless and a spring with a spring constant of 8000N/m is fixed to block 2. Eventually block 1 overtakes block 2. At instant block 1 is moving rightward at 6.4m/s, what are (a) the speed of block 2 and (b) the elastic potential energy of the spring?

    a) [tex]\sum[/tex]Pi = [tex]\sum[/tex]Pf
    ------------>>>>> (6)(8) + (4)(2) = (6)(6.4) + 4vf2

    Hence, vf2 = 4.4m/s

    and

    b) I think that kinetic energy before collision is equal to kinetic energy after collision
    I tried with

    1/2(m)(vi2) - 1/2ks2 = 1/2(m)(vf2)
    ------------>>>>> (6)(82) - ks2 = (6)(6.42) [Because I cancel out 1/2 all equation]
    ------------>>>>> ks2 = 138.24
    Hence, potential energy of spring = (1/2)(ks2) = 69.12J

    Am I all correct? For b) problem, do I have to calculate the Block2 kinetic energy?

    Thank you
     
  2. jcsd
  3. Aug 22, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    This is good, since momentum is conserved.

    No. Total energy remains constant, not kinetic energy.
    Not correct. Yes, you must include the KE of block 2 as well.
     
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