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 :-(
*1/2m(w^2)=aRd
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)
But they have given it like that in the question, that A reaches the previous point.
To show that u^2 = 2agd[1+(1/e^2)], we need to reconsider the equations you've derived and make some corrections.
Firstly, let's define the positive x-direction as the direction of A's initial velocity, u. Also, let the mass of particles A and B be m.
After the collision, let the velocities of A and B be v1 and v2, respectively. Since A comes back to its previous point, v1 = -u (opposite direction), and B comes to rest, v2 = 0.
Using the conservation of momentum, we have:
mu = mv1 + mv2
mu = m(-u) + m(0)
mu = -mu
From this equation, we can see that m cancels out, and we are left with:
u = -u
This implies that the magnitude of the initial velocity, u, is equal to the magnitude of the final velocity of A, v1.
Now, let's consider the work done against the frictional force, given by the equation:
Work = force x distance
The force of friction can be calculated using the equation:
Frictional force = coefficient of friction x normal force
The normal force, N, is equal to the weight of particle A, which is mg.
Therefore, the frictional force, F, is:
F = aN
F = amg
The work done against the frictional force is:
Work = F x d
Work = amgd
From the work-energy theorem, we know that the work done against the frictional force is equal to the change in kinetic energy:
Work = ΔKE
Work = (1/2)mv1^2 - (1/2)mu^2
Substituting the values we have:
amgd = (1/2)m(-u)^2 - (1/2)mu^2
Simplifying this equation:
amgd = (1/2)mu^2 + (1/2)mu^2
amgd = mu^2
Finally, rearranging the equation to solve for u^2:
u^2 = 2agd
This equation does not match the desired result u^2 = 2agd[1+(1/e^2)], so there may be some steps in your calculations that need to be revised.