(b) [image:6_image79.emf]
$v_{2} = \frac{2m_{1}v_{1}}{m_{1} + m_{2}} = \frac{2 \times m \times 9}{m + 2m} = 6\ m/s$
$i.e.$ After elastic collision $B$ strikes to $C$ with velocity of $6\ m/s$. Now collision between $B$ and $C$ is perfectly inelastic
[image:6_image80.emf]
By the law of conservation of momentum
$2m \times 6 + 0 = 3m \times v_{sys}$
$\Rightarrow v_{sys} = 4\ m/s$