\({P_y} = m \times {v_y} = 1 \times 21 = 21\;kg\;m/s\)

Resultant =\(\sqrt {P_x^2 + P_y^2} = 21\sqrt 2 \)kg m/s

The momentum of heavier fragment should be numerically equal to resultant of \({\vec P_x}\) and \({\vec P_y}\).

\(3 \times v = \sqrt {P_x^2 + P_y^2} = 21\sqrt 2 \) \(v = 7\sqrt 2 \)

\(= 9.89 \,m/s\)