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](https://img.qammunity.org/2021/formulas/physics/college/t06yataw5vlfnnctq8t9z9kzwzglfe4w8g.png)
Step-by-step 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)](https://img.qammunity.org/2021/formulas/physics/college/i0c4v9cfzuusg1udgl3v9lfscxc9u8ac7c.png)
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)=(-m_(1)V_(1)-m_(2)V_(2))/(m_(3)) \\V_(3)=(-m_(1)(-60m/s)i-m_(2)(-60m/s)j)/(3m)\\V_(3)=(20i+20j)m/s](https://img.qammunity.org/2021/formulas/physics/college/xp3t8aczztvrds7otcvv3i8rg72aa3r7s3.png)