A shell fired vertically up, when reaches its highest point, explodes into three fragments A,B and C of masses `m_(A) = 4kg, m_(a) = 2_kg` and `m_(c ) = 3kg`. Immediately after the explosion, A is observed moving with velocity `v_(A) = 3m//s` towards north and B with a velocity `v_(b) = 4.5 m//s` towards east as shown in the figure. Find the velocity `v_(c )` of the piece C.

Explanation

Explosion takes negligible duration, therefore, impulse of gravity, which is a finite external force, can be neglelted. The pieces fly off acquiring above-mentioned velocities due to internal forces developed due to expanded gases produced during explosion. The forces aplied by the expanding gases are internal forces, hence, momentum of the system of the three pieces remains conserved during the explosion and total momentum before and after the explosion are equal.
Assuming the east as positive x-direction and the north as positive y-direction, the momentum vectors `vecp_A` and `vecp_B` of pieces A and B become
`vec(p)_(A) = m_(A)v_(A)hat(j) = 12 kg-m//s` and `vec(p)_(B) = m_(B)v_(B)hat(i) = 9 kg-m//s`
Before the explosion, momentum of the shell was zero, therefore from the principle of conservation of momentum, the total momentum of the fragments also remains zero.
`vec(p)_(A) + vec(P)_(B)+vec(P)_(c ) = vec(0) rarr " " vec(p)_(c) = -(9hat(i) + 12hat(j))`
From the above equation, velocity of the piece C is `vec(v)_(c ) = (vec(p)_(C ))/(m_(C )) = -(3hati+4hatj)=5m//s, 53`.south west.