Question

A bomb at rest explodes into 3 fragments of equla mass.two frgaments fly off at right angles with velocities 9 and 12ms.find sped of 3rd fragment

Answer 1

Let m be the mass of each fragment.
therfore, linear momentum of first fragment( le it be p1)= m x 9.
linear momentum of second fragment( let it be p2)= m x 12.

As first fragment and second fragment are perpendicualr to each other. Therefore resultant momeutm of the two fragm,ents 
p= under root p1 square + p2 square =15m

accordind to principle of conservation of momentum.
p+p3=0
p3=-p
​m x v3=-15

Answer 2

Total momentum will be conserved
So v= underroot 9*9+12*12=underroot 81+144=underroot 225=5*3=15