91Ó°ÊÓ

The mass of onepieceis,$m_1=m$.

The velocityof onepiece is,${\stackrel{→}{v}}_1=-30\hat{i}$.

The mass of second piece is,$m_2=m$.

The velocityofsecond piece is,${\stackrel{→}{v}}_2=-30\hat{j}$.

The mass of the third piece is, $m_3=3m$.

By applying conservation of linear momentum, find the magnitude and direction of the velocity of the third piece. According to conservation of linear momentum, momentum that characterizes motion never changes in an isolated collection of objects.

Formulae are as follow:

The conservation of linear momentum,$m\stackrel{→}{v_0}=m_1\stackrel{→}{v_1}+m_2\stackrel{→}{v_2}+m_3\stackrel{→}{v_3}$

The angle from +x axis is,$\mathrm{θ}={\tan }^{-1}\left(\frac{y}{x}\right)$

Let, the mass of onepiece be$m_1=m$, its velocity berole="math" localid="1661486686938" $\stackrel{→}{v_1}=-30\hat{i}$,the mass of second piece be$m_2=m$, its velocity be$\stackrel{→}{v_2}=-30\hat{j}$, the mass of the third piece be.

According to conservation of linear momentum,

$$\begin{gathered} m\stackrel{→}{v_0}=m_1\stackrel{→}{v_1}+m_2\stackrel{→}{v_2}+m_3\stackrel{→}{v_3} \\ 0=m\left(-30\hat{i}\right)+m\left(-30\hat{j}\right)+3m\stackrel{→}{v_3} \\ \stackrel{→}{v_3}=\frac{m\left(30\hat{i}\right)+m\left(30\hat{j}\right)}{3m} \\ \stackrel{→}{v_3}=\left(10\hat{i}\right)+\left(10\hat{j}\right)m/s \end{gathered}$$

Hence, its magnitude$v_3$is given by,

$$\begin{gathered} v_3=\sqrt{{10}^2+{10}^2} \\ {v}_3=10\sqrt{2} \\ {v}_3≈14m/s \end{gathered}$$

Hence, the magnitude of the velocity of the third piece is 14 m/s.

Now,

$$\begin{gathered} \mathrm{θ}={\tan }^{-1}\left(\frac{10}{10}\right) \\ \mathrm{θ}={\tan }^{-1}\left(1\right) \end{gathered}$$

Hence,the direction of the velocity of third pieceis$\mathrm{θ}=45°$, counter clockwise from +x direction.

Therefore, by usingthelinear conservation of momentum, the magnitude of velocity and the direction of the third piece can be found.