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 (-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

Answer 1

Answer:

Explanation:

a. Given that:

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

Let $\bar v_3$ be velocity of 3rd piece=?

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

$m\bar v_1 +m\bar v_2 +3m \bar v_3=0$   #Dividing both sides by $m, m\neq 0$

$\bar v_1+ \bar v_2 +3\bar v_3=0\\3\bar v_3=-(\bar v_1 +\bar v_2)\\\bar v_3=-\frac{1}{3}(\bar v_1 +\bar v_2)$

#Plug the$\bar v_1 ,\bar v_2$  values:

$\bar v_3=-\frac{1}{3}(-23\dot i -23\dot j)=\frac{1}{3}(23\dot i +23\dot j)$

#So the magnitude of the third piece is:

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

Magnitude of the 3rd piece is 10.84 m/s

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

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

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