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Urgent- collision question

  1. Apr 19, 2009 #1

    C.E

    User Avatar

    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 both at an angle of 45 degrees from the initial path of the particle of mass m
    show 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.

    I tried to upload a diagram I had drawn but it would not let me, sorry.

    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
     
  2. jcsd
  3. Apr 20, 2009 #2
    Hello, C.E. !

    I would do some help, though I donot know whether it arrives in time or not.

    Just believe yourself and conduct the calculating, and you are able to work out the problem yourself.

    The collision happens very soon and the instant force is quite large, so the external influence could be ignored and the system satisfies law of conservation of momentum, which should be treated using componential equations along the x-direction and y-direction.

    Let's assume the initial direction of the velocity of $m$ to be the x-axis and along its positive direction, the velocity of $km$ to be the y-axis and along its positive direction. Hence the momentum conserves both along the x-axis and the y-axis:
    $$
    mu=mv_1(cos45')+kmv_2(cos45') (Eq1)
    $$
    $$
    kmu=mv_1(sin45')-kmv_2(sin45') (Eq2)
    $$
    On the other hand, since the collision is elastic, the mechanical energy conserves:
    $$
    1/2 mu^2 + 1/2 kmu^2 = 1/2 mv_1^2 + 1/2 kmv_2^2 (Eq3)
    $$
    In Eq1~Eq3, there are three variables, k, v_1, v_2. Three variables, three independent equations, then we could easily draw $k$ via algebraic methods.

    Squares of Eq1 and Eq2 and deduction of $m$ lead to:
    $$
    2u^2= v_1^2 + k^2 v_2^2 + 2k v_1 v_2 (Eq4)
    $$
    $$
    2k^2 u^2= v_1^2 + k^2 v_2^2 - 2k v_1 v_2 (Eq5)
    $$

    Plus Eq4 and Eq5:
    $$
    u^2 (1+k^2) = v_1^2 +k^2 v_2^2 (Eq6)
    $$
    From Eq3, we have
    $$
    u^2 (1+k) = v_1^2 +kv_2^2 (Eq7)
    $$
    Eq6 minus Eq7 to cancel v_1, we have
    $$
    u^2 (k^2-k) =v_2^2 (k^2-k) (Eq8)
    $$

    hence
    $$
    v_2=u
    $$
    Insert $ v_2=u $ into Eq6 or Eq7, we have
    $$
    v_1=u
    $$
    with v_1 = v_2 = u, Eq4 becomes
    $$
    2u^2 = ( k^2+1 +2k ) u^2
    $$
    Hence
    $$
    k^2+2k-1=0
    $$
    Abandon the negative root, we have k=\sqrt{2}-1
     
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