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 (-23 m/s) and a second piece, also of mass m, moves with velocity (-23 m/s) . The third piece has mass 3m. Just after the explosion, what are the (a) magnitude and (b) direction (as an angle relative to the x axis) of the velocity of the third piece

[tex]m-[/tex] mass of first & second piece, [tex]3m-[/tex]mass of 3rd piece,[tex]\bar v_1[/tex]-velocity of first piece([tex]-23\dot i \ m/s[/tex]) and [tex]\bar v_2[/tex] as velocity of 2nd piece ([tex]-23\dot j \ m/s[/tex])

Let [tex]\bar v_3[/tex] be velocity of 3rd piece=?

#Vessel is at rest before explosion. Considering conservation of linear momentum:

[tex]m\bar v_1 +m\bar v_2 +3m \bar v_3=0[/tex] #Dividing both sides by [tex]m, m\neq 0[/tex]

[tex]\bar v_1+ \bar v_2 +3\bar v_3=0\\3\bar v_3=-(\bar v_1 +\bar v_2)\\\bar v_3=-\frac{1}{3}(\bar v_1 +\bar v_2)[/tex]

#Plug the[tex]\bar v_1 ,\bar v_2[/tex] values:

[tex]\bar v_3=-\frac{1}{3}(-23\dot i -23\dot j)=\frac{1}{3}(23\dot i +23\dot j)[/tex]

#So the magnitude of the third piece is:

[tex]|\bar v_3|=\sqrt{(23/3)^2+(23/3)^2}\\=10.84m/s[/tex]

Magnitude of the 3rd piece is 10.84 m/s

b. To find direction of the magnitude (as an angle relative to the [tex]x[/tex]-axis), we find [tex]\angle \theta[/tex]. The angle is obtained by getting the tan inverse as:

[tex]\theta=tan^-^1(\frac{23/3}{23/3})\\=45\textdegree[/tex]

-The direction of the magnitude (angle relative to the x-axis) is 45°