Q.241. Elastic collision between a particle and a rod

Question 241.
A rod of mass M and length L rests on a frictionless horizontal surface. A particle also of mass M moving with a velocity of V strikes the rod at a distance of L/4 from the center and comes to rest. Then by the conservation of linear momentum, the velocity of the center of mass of rod becomes V. And by the conservation of angular momentum ( M * V * L/4 = I * W) we can determine W where W is the angular velocity gained by rod and I is the moment of inertia about the center of mass. Thus the total K.E of rod is 1/2 * M * V^2 + 1/2 * I * W^2. Suppose the particle struck exactly at the center of the rod and comes to rest, then too by conservation of linear momentum, the velocity of the center of mass of the rod is V but it has no angular velocity. Then the total K.E of rod = 1/2 * M * V^2. We can see that the K.E in the first case is more. But that should not be since energy is conserved in this case. Can anybody explain where am I mistaken in my assessment ?

Answer 241.
Just because some information is given in the question, it need not be taken as correct. In the first case, it is not possible for the ball to come to a stop which has resulted in non-acceptable conclusion. The particle can continue to move in the same direction with reduced velocity imparting less than velocity V to the rod in such a way that both the momentum and kinetic energy are conserved given elastic collision. I will try to work out the velocity of the particle as well as the rod after the collision.
 The second case is acceptable.

I have tried to analyse the first case as under.

Let
u = velocity of the center of the rod after the collision,
v = reduced velocity of the particle in the same direction after the collision, and
w = angular velocity of the rod.

By the law of conservation of linear momentum,
MV = Mu + Mv
=> V = u + v ... (1)

By the law of conservation of angular momentum,
MVL/4 = MvL/4 + (1/12)ML^2 w
=> 3V = 3v + Lw ... (2)

By the law of conservation of energy,
(1/2)MV^2 = (1/2)Mv^2 + (1/2)Mu^2 + (1/2) * (1/12)ML^2 w^2
=> V^2 = v^2 + u^2 + (1/12)L^2 w^2 ... (3)

Plugging V from (1) in (2),
3(u + v) = Lw + 3v
=> Lw = 3u ... (4)

Plugging v = V - u from (1) and w = 3u/L from (4) in (3),
V^2 = (V - u)^2 + u^2 + (1/12)L^2 (3u/L)2
=> 0 = -2Vu + 2u^2 + (3/4)u^2
=> u = (8/11)V

Plugging this value of u in (1),
v = V - u = V - (8/11)V = (3/11)V

Thus, in case 1, the particle will continue to move in the same direction with velocity = (3/11)V imparting velocity (8/11)V to the center of mass of the rod and also imparting an angular velocity w = 3u/L = (24V)/(11L) to the rod.

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Verification:
(1) Linear momentum balance:
Initial linear momentum = MV
Final linear momentum = M(3V/11) + M(8V/11) = MV

(2) Angular momentum balance:
Initial angular momentum = MVL/4
Final angular momentum
= MuL/4 + (1/12)ML^2 w^2
= (ML/4)*(3V/11) + (1/12)ML^2 (24V/11L)
= MVL (3/44 + 8/44) = MVL/4

(3) Kinetic energy balance:
Initial kinetic energy = (1/2)MV^2
Final kinetic energy
= (1/2)Mu^2 + (1/2)Mv^2 + (1/2) * (1/12)ML^2 w^2
= (1/2)M [ 8V/11)^2 + (3V/11)^2 + (1/12)L^2 * (24V/11L)^2]
= (1/2)MV^2 [64/121 + 9/121 + 48/121]
= (1/2)MV^2.

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