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Elastic collision

  1. May 15, 2014 #1
    1. The problem statement, all variables and given/known data
    A body of mass 3 kg slides on a friction less surface with velocity 4 m/s and collides elastically with a resting mass of 2 kg. calculate the final velocities.

    2. Relevant equations
    Conservation of momentum: [itex]m_1v_1+m_2v_2=m_1u_1+m_2u_2[/itex]
    Conservation of energy: [itex]\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2=\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2[/itex]

    3. The attempt at a solution
    I solved those two equations and got a correct answer. but my question is as follows:
    If we solve, as a general case, the two above equations we get a third equation:
    [tex]v_1-v_2=-(u_1-u_2)[/tex]
    Which states that the relative velocities before and after the collision remain the same.
    When i solved this third equation together with the first equation, that of the conservation of momentum i got, of course, a wrong answer.
    Can you explain? both ways seem correct.
     
  2. jcsd
  3. May 15, 2014 #2
    It might help to show your work. I don't see why the latter approach should give you a wrong answer.
     
  4. May 15, 2014 #3
    My work

    When i use the first method, the long way by solving fully the 2 equations, right from the start there is a difference: from the Conservation of momentum:
    [tex]3\cdot 4=3u_1+2u_2 \Rightarrow u_1=4-\frac{2}{3}u_2[/tex]
    The second equation in both methods is the conservation of energy which remains the same.
    In the second method, from the equation:
    [tex]v_1-v_2=-(u_1-u_2) \Rightarrow 3=-(u_1-u_2) \Rightarrow u_1=u_2-3[/tex]
    The two u1 differ.
     
  5. May 15, 2014 #4
    Based on sign convention, if positive is towards right, then it should be
    [tex]u_{1} - u_{2} = 4[/tex]
    and that gives same answer as the previous approach since [itex]u_{2}[/itex] is zero.
    You might have mixed up some numbers in your attempt (eg. the 3 you have here instead of 4).
     
  6. May 15, 2014 #5
    Thanks, i wouldn't have found it, but i think you have a mistake, if i take it as you wrote:
    [tex]u_1=u_2+4[/tex]
    the answer would be wrong. and why should i change signs? i shouldn't take into consideration the positive direction since the equation:
    [tex]v_1-v_2=-(u_1-u_2)[/tex]
    Is general, it doesn't take in consideration directions, no?
    So, if i use:
    [tex]u_1=u_2-4[/tex]
    then it comes out right
     
  7. May 15, 2014 #6
    I think we might have been using different letters for initial and final. I assumed that u referred to the initial velocity. If so, then u1 = 4 is the correct answer since u2 = 0 and the direction of the initial block is taken to be positive.
     
  8. May 15, 2014 #7
    Yup, seems that you mean final velocities by u. Then the two equations that you mentioned earlier:
    [tex]u_1=4-\frac{2}{3}u_2[/tex]
    [tex]u_1=u_2-4[/tex]
    are correct.

    There is no disagreement between them; they form a pair of simultaneous equations that allow you to determine both [itex]u_{1}[/itex] and [itex]u_{2}[/itex]
     
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