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2 blocks and a spring

  1. Aug 9, 2015 #1
    1. The problem statement, all variables and given/known data
    A system consists of two blocks, of masses m and 2m, attached to the ends of a massless spring with a force constant k. The system is placed on a horizontal frictionless surface. Initially, the spring is relaxed. The blocks are then pulled apart an “extra” distance x and simultaneously released from the state of rest.

    problems_F12_MRI_images_MRI14.png

    Find the speed v1 of the block of mass m at the instant the spring is relaxed again. Answer in terms of m, k, and x.

    2. Relevant equations
    U_spring = 1/2*k*x^2
    K = 1/2*m*v^2
    m_1*v_1+m_2*v_2 = m_1*v_1' + m_2*v_2'

    3. The attempt at a solution
    Momentum is conserved and initially 0 so

    m*v_1 = 2m*v_2

    and

    v_1 = 2v_2

    Potential energy in the compressed spring is equal to 1/2*k*x^2 and when the spring reaches equilibrium the potential is 0 and the kinetic energy is equal to 1/2*m*v_1^2+1/2*(2m)*(2v_1)^2 so

    1/2*k*x^2 = 1/2*m*v_1^2+4*m*v_1^2

    = k*x^2 = m*v_1^2+8*m*v_1^2

    = k*x^2 = 9*m*v_1^2

    and finally

    v_1 = sqrt(k*x^2/(9*m))

    Where am I going wrong?
     
  2. jcsd
  3. Aug 9, 2015 #2

    olivermsun

    User Avatar
    Science Advisor

    If ##v_1 = 2v_2##, then why is ##\frac{1}{2}kx^2 = \frac{1}{2} mv_1^2+4mv_1^2##?
     
  4. Aug 9, 2015 #3
    You substituted 2v1 for v2.
     
  5. Aug 9, 2015 #4
    Yes!!! Thank you
     
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