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Elastic collision 3

  1. May 24, 2014 #1
    1. The problem statement, all variables and given/known data
    2 masses moving on a friction less surface. one of 2 kg moving at a speed of 10 and one of 6 kg moving at the speed of 4. at the back of the heavier one is a spring with a constant k=800.
    What is the maximum shortening of the spring

    2. Relevant equations
    Conservation of momentum: [itex]m_1v_1+m_2v_2=m_1u_1+m_2u_2[/itex]
    Conservation of energy: [itex]\frac{1}{2}mv^2=mgh[/itex]
    Elastic energy in s spring: [itex]E=\frac{1}{2}kx^2[/itex]

    3. The attempt at a solution
    At the maximum shortening the velocities are equal:
    Conservation of energy:
    [tex]\frac{1}{2}\cdot 2 \cdot 100=\frac{1}{2}\cdot 800 \cdot x^2+\frac{1}{2}\cdot 8 \cdot u^2[/tex]
    Conservation of momentum:
    [tex]2 \cdot 10=8 \cdot u[/tex]
    Those give x=0.43 while it should be 0.259
     

    Attached Files:

  2. jcsd
  3. May 24, 2014 #2
    6 kg block is in motion .You have not considered its kinetic energy.

    Same mistake .You have not included momentum of the 6 kg block.
     
  4. May 24, 2014 #3

    vela

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    It would help if you would write the equations down in terms of variables first. You should avoid plugging numbers in until later in the exercise-solving process.
     
  5. May 24, 2014 #4
    I took in consideration the 6 kg block in both cases. the number 8 in the kinetic energy and the momentum equation is (2+6)
     
  6. May 24, 2014 #5
    You have incorrectly presumed that 6 kg block is initially stationary .Look carefully in the picture .The 6 kg block is moving with 4m/s before the 2 kg block comes in contact with the spring.

    Include the momentum of 6 kg block in the initial momentum i.e in the left hand side .Same with energy conservation .
     
  7. May 24, 2014 #6
    Equations

    ##\frac{1}{2}m_1v_1=\frac{1}{2}kx^2+\frac{1}{2}(m_1+m_2)u^2##
    ##\Rightarrow \frac{1}{2}\cdot 2 \cdot 100=\frac{1}{2}\cdot 800 \cdot x^2+\frac{1}{2}\cdot 8 \cdot u^2 ##
    ##m_1v_1=(m_1+m_2)u##
    ##\Rightarrow 2 \cdot 10=8 \cdot u##
     
  8. May 24, 2014 #7

    vela

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    Have you read at all what Tanya has written?
     
  9. May 24, 2014 #8
    Items in red are incorrect.
     
  10. May 24, 2014 #9
    Equations

    ##\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2=\frac{1}{2}kx^2+\frac{1}{2}(m_1+m_2)u^2##
    ##\Rightarrow \frac{1}{2}\cdot 2 \cdot 100+\frac{1}{2}\cdot 6 \cdot 16=\frac{1}{2}\cdot 800 \cdot x^2+\frac{1}{2}\cdot 8 \cdot u^2 ##

    ##m_1v_1+m_2v_2=(m_1+m_2)u##
    ##\Rightarrow 2 \cdot 10 +4 \cdot 6=8 \cdot u \rightarrow u=6##

    They give x=0.1, wrong
     
    Last edited: May 24, 2014
  11. May 24, 2014 #10
    Be careful with your calculations .## u = \frac{44}{8} ≠ 6##
     
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