Question

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) and a second piece, also of mass m, moves with velocity (-60 m/s). The third piece has mass 3m.

Answer 1

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:

$V_{3}=(20i+20j)m/s$

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

$p_{initial}=p_{final}$

As the vessel at rest so the initial momentum is zero

So

$m_{1}V_{1}+m_{2}V_{2}+m_{3}V_{3}=0\\m_{3}V_{3}=-m_{1}V_{1}-m_{2}V_{2}\\V_{3}=\frac{-m_{1}V_{1}-m_{2}V_{2}}{m_{3}} \\V_{3}=\frac{-m_{1}(-60m/s)i-m_{2}(-60m/s)j}{3m}\\V_{3}=(20i+20j)m/s$