Two identical ball bearings in contact with each
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:

Answer: Case (ii)

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:

$=\frac{1}{2} m V^{2}+\frac{1}{2}(2 m) 0$

$=\frac{1}{2} m V^{2}$

Case (i).

Total kinetic energy of the system after collision:

$=\frac{1}{2} m \times 0+\frac{1}{2}(2 m)\left(\frac{V}{2}\right)^{2}$

$=\frac{1}{4} m V^{2}$

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

Case (ii)

Total kinetic energy of the system after collision:

$=\frac{1}{2}(3 m)\left(\frac{V}{3}\right)^{2}$

$=\frac{1}{6} m V^{2}$

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