Explanation

(a) For collision between blocks $A$ and $B$,
$e = \frac{v_{B} - v_{A}}{u_{A} - u_{B}} = \frac{v_{B} - v_{A}}{10 - 0} = \frac{v_{B} - v_{A}}{10}$
$\therefore v_{B} - v_{A} = 10e = 10 \times 0.5 = 5\ \ \ \ \ \ \ldots.(i)$
from principle of momentum conservation,
$m_{A}u_{A} + m_{B}u_{B} = m_{A}v_{A} + m_{B}v_{B}$
Or $m \times 10 + 0 = mv_{A} + mv_{B}$
$\therefore\ \ v_{A} + v_{B} = 10\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots.(ii)$
Adding Eqs. (i) and (ii), we get
$v_{B} = 7.5\ ms^{- 1}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(iii)$
Similarly for collision between $B$ and $C$
$v_{C} - v_{B} = 7.5e = 7.5 \times 0.5 = 3.75$
$\therefore\ \ v_{C} - v_{B} = 3.75ms^{- 1}$ …(iv)
Adding Eqs. (iii) and (iv) we get
$2v_{C} = 11.25$
$\therefore\ \ v_{C} = \frac{11.25}{2} = 5.6ms^{- 1}$