Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Problem with inelastic collision

  1. Mar 22, 2014 #1

    ShayanJ

    User Avatar
    Gold Member

    Two masses [itex] m_1 [/itex] and [itex] m_2 [/itex] are closing each other with speeds [itex] v_1 [/itex] and [itex] v_2 [/itex]. The coefficient of restitution is e. Calculate the amount of kinetic energy loss after caused by the collision.
    I solved it in the center of mass coordinates([itex] v_{cm}=u_{cm}=0 [/itex]). The relative speed before and after the collision are [itex] v_r=-(v_1+v_2) [/itex] and [itex] u_r=u_1+u_2 [/itex] respectively. Using conservation of momentum, we know that [itex] m_1v_1=-m_2v_2 [/itex] and [itex] m_1u_1=-m_2u_2 [/itex]. Solving these equations for [itex] v_1,v_2,u_1,u_2 [/itex], we'll have:
    [itex]
    v_1=-\frac{m_2}{m_2-m_1}v_r\\
    v_2=\frac{m_1}{m_2-m_1}v_r\\
    u_1=\frac{m_2}{m_2-m_1}u_r\\
    u_2=-\frac{m_1}{m_2-m_1}u_r
    [/itex]
    Substituting the above results into [itex] m_1v_1^2+m_2v_2^2=m_1u_1^2+m_2u_2^2+2Q [/itex] and using [itex] u_r=e v_r [/itex], We'll have:
    [itex]Q=\frac{m_1m_2}{2} \frac{m_1+m_2}{(m_1-m_2)^2} (1-e^2) v_r^2[/itex]
    But as you can see, this is saying that for [itex]m_1=m_2[/itex] , Q becomes infinite which has no meaning and so something must be wrong. But I can't find what is that. What is it?
    Thanks
     
    Last edited: Mar 22, 2014
  2. jcsd
  3. Mar 22, 2014 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    You use two different conventions for the speeds/velocities - for the relative speed, you use them as absolute values to add them, but in the conservation of momentum, you use them as vectors (which can be negative).

    It is easier to use them as velocity, then your relative speed is wrong.
     
  4. Mar 22, 2014 #3
    Your relative speeds are wrong. They should be vr = v2 - v1, and ur = u2 - u1 respectively

    EDIT: I see that mfb beat me to the punch
     
  5. Mar 22, 2014 #4

    ShayanJ

    User Avatar
    Gold Member

    Ohh...yeah...thanks man.
    Sometimes I really think I have some problems in the basics!!!

    That's when you're dealing them as vectors. When you're dealing with their components, negative signs may appear which may alter that formula.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Problem with inelastic collision
  1. Inelastic Collisions (Replies: 31)

Loading...