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$\mathbf{(}\mathbf{-}\mathbf{30}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{)}\hat{\mathbf{i}}$ and a second piece, also of mass m, moves with velocity$\mathbf{(}\mathbf{-}\mathbf{30}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{)}\hat{\mathbf{j}}$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?

Short answer

1. The magnitude of the velocity of the third piece is, $v_3\approx 14m/s$
2. The direction of the velocity of third piece is counterclockwise from +x direction.

Step-by-step solution

Step 1: Given Data

The mass of onepieceis,$m_1=m$.

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

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

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

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

Determining the concept

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,$m\overrightarrow{v_0}=m_1\overrightarrow{v_1}+m_2\overrightarrow{v_2}+m_3\overrightarrow{v_3}$

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

(a) Determining the magnitude of the velocity of the third piece

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

According to conservation of linear momentum,

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

Hence, its magnitude$v_3$is given by,

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

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

(b) Determining the direction of the velocity of third piece

Now,

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

Hence,the direction of the velocity of third pieceis$\theta =45°$, 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.