Question

A bomb initially at rest explodes by itself into three equal mass fragments. the velocities of two fragments are (3i^+2j^)m/s and (−i^−4j^)m/s. the velocity of the third fragment is (in m/s ):

Answer 1

mv1+mv2+mv3=0
V1+v2+v3 =0
3i+2j-i-4j= -v3
V3=-2i+2j

Answer 2

Answer:

The velocity of the third fragment is (-2i+2j) m/s.

Explanation:

Given that,

Velocity of first fragment = (3i+2j) m/s

Velocity of second fragment = (-i-4j) m/s

A bomb initially at rest explodes by itself into three equal mass fragments.

Initial velocity = 0

We need to calculate the velocity of third fragment

Using conservation of momentum

$mu=m_{1}v_{1}+m_{2}v_{2}+m_{3}v_{3}=$

Put the value of all velocities

$m(v_{1}+v_{2}+v_{3})=0$

$m(3i+2j-i-4j+v_{3})=0$

$v_{3}=(-2i+2j)\ m/s$

Hence, The velocity of the third fragment is (-2i+2j) m/s.