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Homework Help: Conservation of momentum & energy

  1. Dec 18, 2009 #1
    1. The problem statement, all variables and given/known data
    A block of mass m1 = 2 kg slides along a frictionless table with speed 10 m/s. Directly in front of it, and moving in the same direction with a speed of 3 m/s is a block of mass m2 = 5kg. A massless spring with spring constant k = 1120 N/m is attached to the second block. The spring is compressed a maximum amount x. What is the value of x?


    2. Relevant equations
    m1v1 + m2v2 = m1vf1 + m2vf2
    KEi = KEf + 1/2kx^2


    3. The attempt at a solution
    I knew that there would be a maximum compression of the spring when the first block has velocity = 0 m/s. I used cons. of momentum :
    (2 kg)(10 m/s) + (5 kg)(3 m/s) = (5 kg)(Vf) to find Vf = 7 m/s
    then used cons. of energy:
    (0.5)(5 kg)(7 m/s)^2 = 1/2(1120)x^2
    and got x = 0.4677 m

    but the correct answer is 0.250 m
    can anyone help me find the correct answer
     
  2. jcsd
  3. Dec 18, 2009 #2

    rl.bhat

    User Avatar
    Homework Helper

    After collision both the masses move together until the compression is maximum. At that instant the common velocity is given by
    m1v1 + m2v2 = (m1+m2)vf.
    Applying the conservation of energy, you can write
    m1v1^2 + m2v2^2 = (m1+m2)vf^2 +1/2*k*x^2.
    Now proceed to find x.
     
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