A 1 kg stationary bomb is exploded in three parts having mass 1:1:3 . . .

Question : A 1 kg stationary bomb is exploded in three parts having mass 1:1:3 respectively. Parts having same mass move in perpendicular direction with velocity 30 m/s, then the velocity of bigger part will be 

(1) 10√2 m/s
(2) 10/√2 m/s
(3) 15√2 m/s
(4) 15/√2 m/s

AIPMT 2001

Solution : 
M=1kg
m1=m
m2=m
m3=3m 

v1=v2=30 m/s

Let the velocity of the bigger part be v3 m/s

Applying the law of conservation of linear momentum. 

Total momentum before the explosion = Total momentum after the explosion
M(0) = m(30i)+m(30j)+3m(v3)
0=m[30i+30j+3v3]
0/m=30i+30j+3v3
0=30i+30j+3v3
-30i-30j=3v3
(-30i-30j)/3=v3
-10i-10j=v3
v3=-10i-10j
v3=-10i-10j
|v3| = √[(-10)²+(-10)²]
|v3| = √[100+100]
|v3| = √[200]
|v3| = 10√2 m/s
Hence, (1) 10√2 m/s, would be the correct option. 

Note : We can also solve this question by resolving the vectors into Horizontal and Vertical and then equation them but the above method is short and quick and that's why recommended.