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(B) It can be observed that the total momentum before and after the collision in each case is constant.For an elastic collision,the total kinetic ener

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Only $(ii)$

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(B) It can be observed that the total momentum before and after the collision in each case is constant.For an elastic collision,the total kinetic energy of a system remains conserved before and after the collision.For mass of each ball bearing $m,$ we can write:Total kinetic energy of the system before collision:$= \frac{1}{2} m V^{2} + \frac{1}{2}(2m)(0)^{2} = \frac{1}{2} m V^{2}$Case $(i)$:Total kinetic energy of the system after collision$= \frac{1}{2} m (0)^{2} + \frac{1}{2}(2m) \left( \frac{V}{2} ight)^{2} = \frac{1}{4} m V^{2}$Since $\frac{1}{4} m V^{2} 
eq \frac{1}{2} m V^{2}$,the kinetic energy is not conserved.Case $(ii)$:Total kinetic energy of the system after collision$= \frac{1}{2}(2m)(0)^{2} + \frac{1}{2} m V^{2} = \frac{1}{2} m V^{2}$Since $\frac{1}{2} m V^{2} = \frac{1}{2} m V^{2}$,the kinetic energy is conserved.Case $(iii)$:Total kinetic energy of the system after collision$= \frac{1}{2}(3m) \left( \frac{V}{3} ight)^{2} = \frac{1}{6} m V^{2}$Since $\frac{1}{6} m V^{2} 
eq \frac{1}{2} m V^{2}$,the kinetic energy is not conserved.Thus,only case $(ii)$ represents a possible result for an elastic collision.

Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed $V$. If the collision is elastic,which of the following figure is a possible result after collision?

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