Question

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 :-(

Answer 1

*1/2m(w^2)=aRd

Answer 2

Sorry, I do not think that A could make it back to the starting point. (even if no friction you would have to put external forces on)

Answer 3

But they have given it like that in the question, that A reaches the previous point.