use energy conservation, I don\'t need to add the bullet\'s kinetic energy? why

Question

use energy conservation, I don't need to add the bullet's kinetic energy? why only the bob is energy conserved.

Problems 285 f 30. As shown in nown in Figure P9.30, a bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod (not a string) of length £ and negli- gible mass. What is themini- mum value of v such that the pendulum bob will barely swing through a complete vertical circle? V/2 Figure P9.30

Explanation / Answer

The bob will complete the circle if its velocity after the collision just takes it to the topmost point of the circle.

Let v' be the velocity of the bob after the collision with the bullet.

The angular momentum of the (bob + bullet) system is conserved about the center of the circle since no external torque acts on the system.

mvL = m(v/2)L + Mv'L

=> mv'L = MvL/2

=> v' = Mv/2m

The kinetic energy of the bob should just take it to the topmost point of the circle. So, the kinetic energy of the bob converts to its potential energy as it moves up to the topmost point.

M(Mv/2m)2/2 = Mg(2L)

=> (Mv/2m)2 = 4gL

=> Mv/2m = 2(gL)1/2

=> v = 4m(gL)1/2/M

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