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Two particles of masses 2m and m are moving along the x−axis with constant velocities u

_{1}

**i**and u

_{2}

**i**respectively, where u

_{1}> u

_{2}. They collide ‘head on’, and the coefficient of restitution of the collision is e. Find the velocities of each of the particles after the collision, and show that the loss in the kinetic energy of the system of two particles is:

$\frac{(1-e^2)m(u_1-u_2)^2}{3}$

so far I have:

the velocities of the particles after impact are

**v**=v

_{1}_{1}

**i**and

**v**=v

_{2}_{2}

**i**

so 2mu

_{1}

**i**-mu

_{2}

**i**=2m

**v**+m

_{1}**v**

_{2}ie 2mu

_{1}-mu

_{2}=2mv

_{1}+mv

_{2}(which is my first equation)

then from the law of restitution:

e(u

_{1}-u

_{2})=v

_{1}-v

_{2}(second equation)

solving the 2 equations I get

v

_{1}=1/3((e+2)u

_{1}-(e+1)u

_{2})

v

_{2}=1/3((2-2e)u

_{1}+(2e-1)u

_{2})

the to get the loss of kinetic energy:

(1/2*2m*(v

_{1})

^{2}+1/2*m*(v

_{2})

^{2})-(m*(u

_{1})

^{2}+1/2*m*(u

_{2})

^{2})

but obviously (v

_{1})

^{2}and (v

_{2})

^{2}gets very messy

Thanks x