Explanation

Initially the body is at rest, therefore the total momentum of the system before explosion is zero.

Let the masses of the fragments be $2k$, $2k$ and $3k$.

$$ 2k + 2k + 3k = 14 \Rightarrow 7k = 14 \Rightarrow k = 2 $$

Thus, the fragment masses are $4\,kg$, $4\,kg$ and $6\,kg$.

The two equal fragments of mass $4\,kg$ each move perpendicular to each other with speed $18\,m/s$.

Momentum of each lighter fragment is

$$ p = mv = 4 \times 18 = 72\,kg\,m/s $$

Since the momenta are perpendicular, their resultant momentum is

$$ p_{res} = \sqrt{72^2 + 72^2} = 72\sqrt{2} $$

To conserve momentum, the heavier fragment must have momentum equal in magnitude and opposite in direction.

If $v$ is the speed of the heavier fragment of mass $6\,kg$, then

$$ 6v = 72\sqrt{2} $$

$$ v = 12\sqrt{2}\,m/s $$

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