
Let the masses of the three parts be m′,m′ and 2m′.
Momentum of first part p1=m′×30=30m′ and momentum of second part p2=m′×40=40m′.
Third part will fly in a direction opposite to the direction of resultant of p1 and p2, so that total momentum of the system after explosion remains zero.
Momentum of the third part is p3=p12+p22=(30m′)2+(40m′)2=50m′
Velocity of third part, v=2m′50m′=25ms−1