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    When an object of mass M and velocity u collides head-on elastically with a stationary object of mass m, show that mass m, moves off with velocity V2 given by:

    V2 = (2M/M+m) u


    I done:

    Mu + m = MV1 + mV2

    and making V2 subject of formula...

    V2 = (Mu + m - MV1)/m

    but from here got stuck.


    any help will be appreciated.
    +rep offered to most thorough answer.

    Thanks!
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    Elastic collision means energy is conserved
    so \frac{1}{2}Mu^2 = \frac{1}{2}Mv_1^2 +\frac{1}{2}mv_2^2

    momentum consideration gives:
     Mu = Mv_1 + mv_2

    Rearrange the momentum equation to get v1 in terms of u and v2. then use the energy equation to eliminate v1 from the system.
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    It's been a while since I did collisions but.... since its elastic then e=1.

    Mu=MV1+mV2

    V1=(Mu-mV2)/M

    e = Seperation /apprach

    So 1 = (V2 - ((Mu-mV2)/M) /u

    u = (MV2-Mu+mV2)M

    Mu = MV2-Mu+mV2

    MV2+mV2 = 2Mu

    V2 = (2Mu/M+m)


    And you just factorise and take the u out.
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    In your first equation you did Mu+m....the seconf object is stationary so its momentum is 0, its just Mu not Mu +m...
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    So before the collision, energy and momentum of the system is:

     E = \frac{1}{2} M u^2
     P = Mu

    After the collision, the mass M now has velocity v_1 and the mass it hit, m, has velocity v_2:

     E = \frac{1}{2} M v_1^2 + \frac{1}{2} m v_2^2
     P = Mv_1 + mv_2

    Equating this two equations together before and after the collision since the collision is elastic:

     Mu = Mv_1 + mv_2 \to v_1 = \frac{Mu-mv_2}{M}
     Mu^2 = Mv_1^2 + mv_2^2 \to Mu^2 = M\left[\frac{Mu-mv_2}{M}\right]^2 + mv_2^2 ---- (1)

    expanding out the bracket:
    M\left[\frac{Mu-mv_2}{M}\right]^2 = Mu^2 - 2muv_2 + \frac{m^2 v_2^2}{M}
    So eq (1) becomes:

     Mu^2 = Mu^2 - 2muv_2 +\frac{m^2 v_2^2}{M} +mv_2^2 \to v_2 = \frac{2Mu}{M+m}
 
 
 
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