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: Spring question

  1. Jan 22, 2007 #1


    User Avatar

    1. The problem statement, all variables and given/known data

    A 6 kg mass and 3 kg mass are at rest on a frictionless table. A massless spring is compressed between them and they are held together with a string. When released (string is cut) the 6 kg mass moves to the left at 4 m/s - what is the velocity of the 3 kg mass as it moves to the right?

    2. Relevant equations

    a little unsure - i beleive i need to use the F=kx (force of a spring) and assume that k is constant throughout since its the same spring for both.

    3. The attempt at a solution

    what i did was assume that the force on each mass was the same since each was compressed the same and had the same spring constant, k.

    what i then did was assume that if that force moved the 6 kg mass 4 m/s to the left, that it would have an equal reaction to the 2nd mass which was only 3 kg. thinking in terms of momentum (is that correct?) i said the 2nd mass would move to the right at 8 m/s:

    (6 kg)(4 m/s) = (3 kg)(x m/s)
    x = 8 m/s

    is this correct? does my logic work?

  2. jcsd
  3. Jan 22, 2007 #2
    Conservation of momentum.
  4. Jan 22, 2007 #3


    User Avatar

    so then my answer is correct.

    m1v1 + m2v2 = m1v1 + m2v2

    the first two are 0 because v1=v2=0

    then you have:

    0 = (6)(4) + (3)(x)

    x = -8 (8 m/s in the other direction)


    so basically this spring is considered a "collision"?
  5. Jan 22, 2007 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I'd say it was more of an explosion than a collision, but in either case, in the absence of external forces, momentum is always conserved, and your answer is correct, good job.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook