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Math Help - Really urgent mechanics question

  1. #1
    C.E
    C.E is offline
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    Really urgent mechanics question

    Hi I am really stuck on the following and have an exam on it in 2 days. Can somebody please help me.

    A particle of mass m collides elastically at an angle of 90 degrees with a lighter particle of mass km. Initially both particles had a speed u. After the collision the particles move apart as shown in the attachment show (by the way the angles are both 45 degrees if it is unclear) that k=sqrt(2)-1.

    I tried using conservation of linear momentum in horizontal directions giving me 2 equations.

    u=(v1^2 ^kv2^2)/sqrt(2)
    ku=(v1 -kv2^2)/ sqrt(2)
    where v1 is the final velocity of the particle of mass m and v2 is the final velocity of the particle of mass km.

    And conservation of kinetic energy tells us that (1+k)u^2=v1^2+v2^2 but I can't put it all together. Please help me
    Attached Thumbnails Attached Thumbnails Really urgent mechanics question-collision.jpg  
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  2. #2
    MHF Contributor chisigma's Avatar
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    Indicating with \overrightarrow{v_{1}} = v_{1x}\cdot \overrightarrow{i} + v_{1y}\cdot \overrightarrow{j} and \overrightarrow{v_{2}}=v_{2x}\cdot \overrightarrow{i} + v_{2y}\cdot \overrightarrow{j} the velocities of the particles after impact, the conservation of momentum and kinetics energy leads to the equations...

    |\overrightarrow{v_{1}}|^{2} + k \cdot |\overrightarrow{v_{2}}|^{2} = (1+k) \cdot u^{2}

    \overrightarrow{v_{1}} + k \cdot \overrightarrow{v_{2}} = u \cdot (\overrightarrow{i} + k\cdot \overrightarrow{j}) (1)

    ... which is equivalent to the scalar equations system...

    v_{1x}^{2} + v_{1y}^{2} + k\cdot v_{2x}^{2} + k\cdot v_{2y}^{2} = (1+k)\cdot u^{2}

    v_{1x} + k\cdot v_{2x}= u

    v_{1y} + k\cdot v_{2y}= k\cdot u (2)

    We have now three equation and four unknown variables, so that we need another equation. This equation is supplied by the supplementar condition that indicates that \overrightarrow{v_{1}} and \overrightarrow{v_{2}} are orthogonal, so that is...

    v_{1x}\cdot v_{2y} = - v_{2x}\cdot v_{1y} (3)

    ... and the problem can be solved...

    My question [for experts...] is : the condition (3), necessary to have a well defined problem, isn't in some way arbitrary?...

    Kind regards

    \chi \sigma
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