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Two body collision with a spring

  1. Jun 23, 2010 #1
    1. The problem statement, all variables and given/known data
    A block of mass m1 = 1.6 kg is initially moving to the right with a speed of 4 ms-1 on a frictionless horizontal track and collides with a spring attached to a second block of mass m2 = 2.1 kg initially moving to the left at a speed of 2.5 ms-1. The spring constant is 600 NM-1.

    a. Find the velocities of the two blocks after the collision.

    2. Relevant equations
    (1) m1v1i+m2v2i = m1v1f+m2v2f

    (2) v1i - v2i = -(v1f - v2f)

    3. The attempt at a solution
    So this is a problem in the book that is worked out for me, but I cannot seem to figure out the reasoning behind the way they combined and manipulated the equations. They first begin by subbing in the known values into EQ (1) which I agree with. They then state to use EQ (2) due to the collision being ellastic, which I don't totally understand this relationship. They sub known values into EQ (2) giving the following:

    6.5 ms-1 = -v1f = v2f

    The book then states that I must multiply the EQ I gave right above by 1.6 kg which gives me the following which I will label as EQ (3): This is what I don't understand, why would I multiply EQ (2) by 1.6kg?

    10.4kgms-1 = (-1.6kg)v1f + (1.6kg)v2f

    The book goes on to state that I now must add EQ (1) and EQ (3) which apparently allows me to solve for v2f, but I don't see where the thought process determines that I add these two EQ. Thanks in advance for the assistance, I am hoping this is a relatively simple algebraic procedure that I am missing.

    Joe
     
  2. jcsd
  3. Jun 23, 2010 #2

    Doc Al

    User Avatar

    Staff: Mentor

    The point here is that EQ (2) is only valid for elastic collisions. One derives that equation by combining EQ (1) with conservation of energy. (The derivation should be in your textbook.) Working with EQ (2) is much easier than working the conservation of energy equation directly.


    It's just a trick. One equation has a +1.6v1 term and the other has a -1.6v1 term. Add them together and the v1 terms cancel leaving only v2.
     
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