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 (Fig. 5.14) is a possible result after collision ?

Answer 1

Case (ii) 

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

or 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: 

= \(\frac{1 }{ 2 }\) mV² + \(\frac{1 }{ 2 }\) (2m)0 

= \(\frac{1 }{ 2 }\) mV²

Case (i) 

Total kinetic energy of the system after collision: 

= \(\frac{1 }{ 2 }\) m × 0 + \(\frac{1 }{ 2 }\) (2m) (\(\frac{v }{ 2 }\))² 

= \(\frac{1 }{ 4 }\) mV²

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

Case (iii)  

Total kinetic energy of the system after collision: 

= \(\frac{1 }{ 2 }\) (3m) \((\frac{v }{ 3 })\) ² 

\(=\frac{ 1 }{ 2} mV^2\)

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