#### Question

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?

#### Solution 1

Let m be the mass of each ball bearing. Before collision, total K.E. of the system

=1/2mv^{2} + 0 =1/2 mv^{2}

After collision, K.E. of the system is

Case I, E_{1} = 1/2 (2m) (v/2)^{2} = 1/4 mv^{2}

Case II, E_{2} = 1/2 mv^{2}

Case III, E_{3} = 1/2(3m) (v/3)^{2} = 1/6mv^{2}

Thus, case II is the only possibility since K.E. is conserved in this case.

#### Solution 2

It can be observed that the total momentum before and after collision in each case is constant.

For an elastic collision, the total kinetic energy of a system remains conserved before and after collision.

For mass of each ball bearing m, we can write:

Total kinetic energy of the system before collision:

= (1/2)mV^{2} + (1/2)(2m) × 0^{2}

= (1/2)mV^{2}

__Case____ (i) __

Total kinetic energy of the system after collision:

= (1/2) m × 0 + (1/2) (2m) (V/2)^{2}

= (1/4)mV^{2}

Hence, the kinetic energy of the system is not conserved in case (i).

__Case____ (ii)__

Total kinetic energy of the system after collision:

= (1/2)(2m) × 0 + (1/2)mV^{2}

= (1/2) mV^{2}

Hence, the kinetic energy of the system is conserved in case (ii).

__Case____ (iii)__

Total kinetic energy of the system after collision:

= (1/2)(3m)(V/3)^{2}

= (1/6)mV^{2}

Hence, the kinetic energy of the system is not conserved in case (iii).

Hence, Case II is the only possibility.