Asked by Aanya
                Two particles A and B has the same mass and are being kept on a rough horizontal table "d" distance apart.B is kept still and A is given a velocity of u towards B.After the collision,if A comes to its previous point show that,
u^2 = 2agd[1+(1/e^2)]
a- coefficient of friction
I took the velocities of A and B after collision as,w towards A's previous point and v away from B's previous point.
Then I got eu=v+w
when considering the system as it have no unbalanced forces,applying conservation of momentum, I got,
mu=mv-mw
u=v-w
and w =1/2[u(e-1)]
then applying law of conservation of energy,as the lowering of kinetic energy is used to do work against frictional forces,
1/2m(w^2)=aR(R=normal force)
R=mg
1/2m*(1/4)*[u^2(e-1)^2)]=amg
Then I am getting,
u^2=8agd/(e-1)^2 ,which isn't the result we have to show :-(
            
        u^2 = 2agd[1+(1/e^2)]
a- coefficient of friction
I took the velocities of A and B after collision as,w towards A's previous point and v away from B's previous point.
Then I got eu=v+w
when considering the system as it have no unbalanced forces,applying conservation of momentum, I got,
mu=mv-mw
u=v-w
and w =1/2[u(e-1)]
then applying law of conservation of energy,as the lowering of kinetic energy is used to do work against frictional forces,
1/2m(w^2)=aR(R=normal force)
R=mg
1/2m*(1/4)*[u^2(e-1)^2)]=amg
Then I am getting,
u^2=8agd/(e-1)^2 ,which isn't the result we have to show :-(
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