1. Not finding help here? Sign up for a free 30min 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!

Elastic collision 4

  1. Aug 4, 2014 #1
    1. The problem statement, all variables and given/known data
    Two identical masses collide. one is at rest. what are their final velocities.

    2. Relevant equations
    The relative velocities before and after the collision are identical:
    ##v_1-v_2=-(u_1-u_2)##

    3. The attempt at a solution
    I draw the final velocities in the same direction:
    [tex]\left\{ \begin{array}{l} mv_1=mu_1+mu_2 \\ -v_1=-(u_1-u_2) \end{array} \right. [/tex]
    It gives $$u_2=0$$ which isn't true, u1 should be 0.
    And more, i can switch between u1 and u2 in the drawing and it will also be good since they are drawn in the same direction, so, how do i know who stops?
    I get this answer also if i use the conservation of energy formula instead of the relative velocities formula.
    If i draw u1 backwards and change the formula in accordance:
    ##\left\{ \begin{array}{l} mv_1=mu_1-mu_2 \\ -v_1=-(u_1-u_2) \end{array} \right.##
    I again get the same result: $$u_2=0$$
    Apart from the question i posed, do i have to assume a direction and then write the formuls accordingly or i have always to write:
    ##\left\{ \begin{array}{l} mv_1=mu_1+mu_2 \\ -v_1=-(u_1-u_2) \end{array} \right.##
     

    Attached Files:

  2. jcsd
  3. Aug 4, 2014 #2

    Doc Al

    User Avatar

    Staff: Mentor

    The left hand side of the second equation should be ##v_1##, not ##-v_1##.
     
  4. Aug 5, 2014 #3
    --

    Thanks, it solved it.
    I used the next set of equations:
    [tex]\left\{ \begin{array}{l} mv_1=mu_1+mu_2 \\ \frac{1}{2}mv^2_1=\frac{1}{2}mu^2_1+\frac{1}{2}mu^2_2 \end{array} \right. [/tex]
    ##\Rightarrow \left\{ \begin{array}{l} v_1=u_1+u_2 \\ v^2_1=u^2_1+u^2_2 \end{array} \right.##
    I isolated u1:
    ##\Rightarrow v^2_1=(v_1+u_2)^2+u^2_2##
    And got the same answer, good. but when i solved the same set of equations by isolating v1 i got a different answer:
    ##(u_1+u_2)^2=u^2_1+u^2_2 \rightarrow u_1 u_2=0##
    Do you know this behavior of sets of equations, that you get different answers by using different ways?
     
  5. Aug 5, 2014 #4

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    No, you will not get different answers by doing it different ways. The statement ##u_1 u_2 = 0## simply tells you that either ##u_1## or ##u_2## is zero after the collision, both of which are perfectly valid roots (##u_2 = 0## corresponds to the balls not having collided yet).

    The equations you have are symmetric in ##u_1## and ##u_2## so if ##u_1 = 0, u_2 = U## is a solution, then also ##u_1 = U, u_2 = 0## must be a solution.
     
  6. Aug 5, 2014 #5

    Doc Al

    User Avatar

    Staff: Mentor

    Realize that these equations are equivalent to those you used earlier.

    You made an error in your substitution. ##u_1 \ne v_1+u_2##.

    This time you did it correctly.

    Only when you make an error. :wink:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Elastic collision 4
  1. Elastic Collisions? (Replies: 3)

  2. Elastic Collisions (Replies: 5)

  3. Elastic collision (Replies: 6)

Loading...