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Colliding blocks attached to a spring - How is the spring compressed

  1. Dec 4, 2005 #1
    "A block of mass m1 = 2.4 kg slides along a frictionless table with a speed of 10 m/s. Directly in front of it, and moving in the same direction, is a block of mass m2 = 4.6 kg moving at 2.8 m/s. A massless spring with spring constant k = 1160 N/m is attached to the near side of m2, as shown in Fig. 10-35. When the blocks collide, what is the maximum compression of the spring? (Hint: At the moment of maximum compression of the spring, the two blocks move as one. Find the velocity by noting that the collision is completely inelastic to this point.)"

    I cannot seem to get this answer right.

    here is how I approached the problem:

    using the inelastic momentum formula:

    m1v1 + m2v2 = (m1+m2) * vfinal
    (2.4 * 10) + (4.6 * 2.8) = (2.4 + 4.6)*v

    solving for V... i get about 5.269 m/s (this is the velocity of the system)

    THEN, in order to find the amount of compression, i use kinetic energy and potential energy in a spring:

    .5kx^2 = .5mv^2
    .5 * 1160 * x^2 = .5 * (2.4 + 4.6) * (5.269)^2

    Solving for x... i get 0.409 m.

    Unfortunately, this is wrong. Any help on solving this problem would be much appreciated.
     
  2. jcsd
  3. Dec 4, 2005 #2
    when the spring compression is at a maximum, the blocks are at a turning point in the motion, v = 0
    /s
     
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