Incomplete question as we have not told to find what.So the complete question is here
A vessel at rest at the origin of an xy coordinate system explodes into three pieces. Just after the explosion, one piece, of mass m, moves with velocity (-60 m/s)i and a second piece, also of mass m, moves with velocity (-60 m/s)j. The third piece has mass 3m.Just after the explosion, what are the (a) magnitude and (b) direction of the velocity of the third piece?
Answer:
[tex]V_{3}=(20i+20j)m/s[/tex]
Explanation:
Given data
The vessel at rest
Piece one,of mass m,moves with velocity=(-60 m/s)i
Piece two,of mass m,moves with velocity=(-60 m/s)j
Piece three,of mass 3m
As the linear momentum is conserved in this system,Because the system is closed and no external force acting on it
So momentum is given as
[tex]p_{initial}=p_{final}[/tex]
As the vessel at rest so the initial momentum is zero
So
[tex]m_{1}V_{1}+m_{2}V_{2}+m_{3}V_{3}=0\\m_{3}V_{3}=-m_{1}V_{1}-m_{2}V_{2}\\V_{3}=\frac{-m_{1}V_{1}-m_{2}V_{2}}{m_{3}} \\V_{3}=\frac{-m_{1}(-60m/s)i-m_{2}(-60m/s)j}{3m}\\V_{3}=(20i+20j)m/s[/tex]
