1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook