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Collision, need assitance

  1. Dec 7, 2006 #1
    block 1 of mass m1 slides along an x-axis on a frictionless floor with speed of v1i=4.00 m/s. Then it undergoes a one dimentional elastic collision with stationary block 2 of mass m2=.500m1. Next, block 2 undergoes a one dimentional elastic collision with stationary block 3 of mass m3=.500m2. (a) What then is the speed of block 3? Are (b) the speed, the kinetic energy, and (d) the momentum of block 3 greater than, less than, or the same as the initial values for block1?

    The lines, the "x" and numbers are kinda croocked, sorry.

    a)We use the principle of conservation of momentum here.

    Initially momentum is (m1)(4)
    After collision with m2, it will be (v)(m1+0.5m1) = (v)(1.5m1)
    After collision with m3, it will be (u)(1.5m1+0.25m1) = (u)(1.75m1)
    this is bevause m3 = 0.5m2 = 0.5(0.5m1) = 0.25m1.

    They must be equal so
    (m1)(4) = (u)(1.75m1)
    u = 2.3m/s roughly!

    (b) Calculate the KE of m3. It is 1/2(2.3)(0.25m1)2
    The KE of m1 is 1/2(m1)(42)

    See which one is greater!

    (d) They are asking the momentum of m3 and not total momentum. Right?This will be (0.25m1)(2.3) where as that of m1 is (m1)(4) so of course that of m1 is greater
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

  2. jcsd
  3. Dec 8, 2006 #2


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    It appears that you are treating the collisions as inelastic collisions. The problem says they are elastic collisions. Both momentum and kinetic energy are conserved. For such collisions there is a velocity difference relationship you might want to use.
  4. Dec 8, 2006 #3
    Oh, How would you do it if it were elastic?
  5. Dec 8, 2006 #4


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    By definition, in an elastic collision, kinetic energy is conserved. Calculate the total kinetic energy before and after collision.
  6. Dec 8, 2006 #5
    How do I know which numbers are before and after? Can you show me some how?
  7. Dec 8, 2006 #6


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    The momentum conservation equations apply to each collision

    m1*v1i = m1*4.00 m/s <== total momentum

    m1*v1i = m1*v1f + m2*v2b <== first collision (b subscropt for between collisions

    m2*v2b = m2*v2f + m3*v3f <== second collision

    If you combine these equations, you have

    m1*v1i = m1*v1f + m2*v2f + m3*v3f

    The total momentum after both collisions have occured is the same as the initial total momentum. But this is not enough for you to find the velocities. You need to apply conservation of kinetic energy to each collision. Kinetic energy involves the squares of the velocities, so the algebra can get bit messy. But it is not too difficult to do the calculation algebraically one time to find a relationship between the differences between the velocities of the two masses before and after the collision. Once you have this relationship, you can use it in all elastic collsion problems without dealing with the squares in the energy equation. If your textbeook does not have this equation, see if you can derive it yourself. If you get stuck, look here towrd the bottom of the page.

    http://www.astrophysik.uni-kiel.de/~hhaertel/MECHANICS/Mathe/mathe.html [Broken]

    For any 1-dimensional two-mass elastic collision
    M1*V1i + M2*V2i = M1*V1f + M2*V2f

    M1*V1i² + M2*V2i² = M1*V1f² + M2*V2f²

    Solve for V1f - V2f in terms of V1i - V2i
    Last edited by a moderator: May 2, 2017
  8. Dec 8, 2006 #7

    I'm sorry, I'm still not really following. I still have to find the momentum of block B and state is the kinetic, the speeed of block 3 greater than, less than, or same as initial.
  9. Dec 8, 2006 #8


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    For any 1-dimensional two-mass elastic collision:

    M1*V1i + M2*V2i = M1*V1f + M2*V2f <== Rearrange this momentum equation

    M1*V1i - M1*V1f = M2*V2f - M2*V2i <== Factor the Ms

    M1(V1i - V1f) = M2(V2f - V2i)

    M1*V1i² + M2*V2i² = M1*V1f² + M2*V2f² <== Rearrange this energy equation

    M1*V1i² - M1*V1f² = M2*V2f² - M2*V2i² <== Factor the Ms

    M1(V1i² - V1f²) = M2(V2f² - V2i²) <== Factor the differences of squares

    M1(V1i - V1f)(V1i + V1f) = M2(V2f - V2i)(V2f - V2i)

    Divide this last equation by the rearranged factored momentum equation above to get

    (V1i + V1f) = (V2f + V2i)

    This is usually rearranged and written as

    (V1i - V2i) = -(V1f -V2f)

    This equation is valid for any elastic collision of two objects in one dimension. Use it with the conservation of momentum equation for each collision in your problem. At least give this a try an post your results.
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