A ‘m’ kg fish swimming at 2 m/s catches a ‘m/9’ kg fish coming towards it from opposite direction swimming at 3 m/s. The velocity of the larger fish immediately after the grab is:

(A) $1.5 \hat i $ m/s
(B) $2.1 \hat i $ m/s
(C) $-1.5 \hat i $ m/s
(D) $-2.1 \hat i $ m/s

Solution

This can be considered as inelastic collision. Using conservation of linear momentum along x-direction,

$2m-3 \frac {m}{9} = (m+\frac {m}{9} )v$

$\Rightarrow 2 – \frac {1}{3} = \frac {10}{9} v $

$\Rightarrow v=1.5 $m/s

Hence, (A)