Three balls A,B and C $(m_{A} = m_{C} = 4 m_{B})$ are placed on a smooth horizontal surface. Ball B collides with ball C with an initial velocity v as shown in Fig. Find the total number of collisions between the balls (all collisions are elastic)

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answer is 2.

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Detailed Solution

For the first collision, $e = 1, v = v_{1} + v_{2}$

$\Rightarrow v_{2} = v - v_{1} ...... (i)$
By momentum conservation
$\begin{array}{l} m_{B} v = - m_{B} v_{1} + m_{C} v_{2} \\ m_{B} v = - m_{B} v_{1} + 4 m_{B} v_{2} \end{array} v_{2} = \frac{v_{1} + v}{4} ........ (i i)$

From Eqs. (i) and (ii), $v_{1} = \frac{3}{5} v a n d v_{2} = \frac{2}{5} v$
For the second collision, e = 1

$v_{1} = {v^{'}}_{1} + v_{3} \Rightarrow v_{3} = v_{1} - {v^{'}}_{1} ..... (i i i)$

By momentum conservation, $- m_{B} v_{1} = m_{B} {v^{'}}_{1} - m_{A} v_{3}$
Or $- m_{B} v_{1} = m_{B} {v^{'}}_{1} - 4 m_{B} v_{3} (∵ m_{A} = 4 m_{B})$
$v_{3} = \frac{{v^{'}}_{1} + v_{1}}{4} ......... (i v)$
From Eqs. (iii) and (iv), ${v^{'}}_{1} = \frac{3}{5} v_{1} = \frac{3}{5} (\frac{3}{5} v) = \frac{9}{25} v$
Clearly $\frac{9}{25} v < \frac{2}{5} v$
Therefore, $' B'$ cannot collide with $' C'$ for the second time
Hence, the total number of collisions is 2