Question

Consider a system of two particles in the xy plane: ${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\mathbf{2}\mathbf{kg}$ is at the locationrole="math" localid="1668080105256" $\overrightarrow{{\mathit{r}}_{\mathbf{1}}}\mathbf{=}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{00}\hat{\mathit{i}}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{00}\hat{\mathit{j}}\mathbf{)}\mathbf{m}$ and has a velocity of $\left(3.00\hat{i}+0.500\hat{j}\right)\mathit{m}\mathbf{/}\mathit{s}$;${\mathit{m}}_{\mathbf{2}}\mathbf{=}\mathbf{3}\mathbf{}\mathit{k}\mathit{g}$ is at$\overrightarrow{{\mathbf{r}}_{\mathbf{2}}}\mathbf{=}\left(-4.00\hat{i}-3.00\hat{j}\right)\mathit{m}$ and has velocity $\left(3.00\hat{i}-2.00\hat{j}\right)\mathit{m}\mathbf{/}\mathit{s}$. (a) Plot these particles on a grid or graph paper. Draw their position vectors and show their velocities. (b) Find the position of the center of mass of the system and mark it on the grid. (c) Determine the velocity of the center of mass and also show it on the diagram. (d) What is the total linear momentum of the system?

Short answer

1. The position and velocity vector are shown below.
   
   

(b) The position of the centre of the mass is $\left(-2\hat{i}-1\hat{j}\right)m$.

(c) The velocity of the centre of mass is $\left(3\hat{i}-1\hat{j}\right)\mathrm{m}/\mathrm{s}$.

(d) The linear momentum of the system is$\left(15\hat{i}-5\hat{j}\right)\mathrm{kg}·\mathrm{m}/\mathrm{s}$.

Step-by-step solution

Write the given data from the question.

Mass of first particle, $m_1=2\mathrm{kg}$

Position of first particle, $\overrightarrow{r_1}=1\hat{j}+2\hat{j}$

Velocity of first particle, $\overrightarrow{v_1}=\left(3\hat{i}+0.5\hat{j}\right)\mathrm{m}/\mathrm{s}$

Mass of second particle, $m_2=3\mathrm{kg}$

Position of second particle, $\overrightarrow{r_2}=-4\hat{i}-3\hat{j}$

Velocity of second particle, $\overrightarrow{v_2}=3\hat{i}-2\hat{j}$

Determine the formulas to calculate the position, velocity centre of mass of system and linear momentum of the system.

The equation to calculate the position of centre of the mass is given as follows.

$\overrightarrow{{\mathbf{r}}_{\mathbf{CM}}}\mathbf{=}\frac{{\mathbf{m}}_{\mathbf{1}}\overrightarrow{{\mathbf{r}}_{\mathbf{1}}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}\overrightarrow{{\mathbf{r}}_{\mathbf{2}}}}{{\mathbf{m}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}}$ .............(i)

The equation to calculate the velocity of the centre of the mass is given as follows.

$\overrightarrow{{\mathbf{v}}_{\mathbf{CM}}}\mathbf{=}\frac{{\mathbf{m}}_{\mathbf{1}}\overrightarrow{{\mathbf{v}}_{\mathbf{1}}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}\overrightarrow{{\mathbf{v}}_{\mathbf{2}}}}{{\mathbf{m}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(\mathrm{ii}\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}$

The equation to calculate the linear momentum is given as follows.

localid="1668083398041" $\overrightarrow{{\mathbf{P}}_{\mathbf{CM}}}\mathbf{=}{\mathbf{m}}_{\mathbf{1}}\overrightarrow{{\mathbf{v}}_{\mathbf{1}}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}\overrightarrow{{\mathbf{v}}_{\mathbf{2}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{iii}\mathbf{\right)}$

Draw the position and velocity graph of the particles.

(a)

On the basis of the given positions and velocity vectors of the first and second particle, it can be represented on the graph paper is shown below.

Calculate the position of the centre of mass of the system.

(b)

Calculate the position of the centre of mass.

Substitute 2 kg for $m_1$, $\left(1\hat{i}+2\hat{j}\right)\mathrm{m}$for $\overrightarrow{r_1}$, for $m_2$and $\left(-4\hat{i}-3\hat{j}\right)\mathrm{m}$for $\overrightarrow{r_2}$into equation (iii).

$$\begin{gathered} \overrightarrow{r_{CM}}=\frac{2\left(1\hat{i}+2\hat{j}\right)+3\left(-4\hat{i}-3\hat{j}\right)}{2+3} \\ \overrightarrow{r_{CM}}=\frac{2\hat{i}+4\hat{j}-12\hat{i}-9\hat{j}}{5} \\ \overrightarrow{r_{CM}}=\frac{-10\hat{i}-5\hat{i}}{5} \\ \overrightarrow{r_{CM}}=\left(-2\hat{i}-1\hat{j}\right)\mathrm{m} \end{gathered}$$

The location of position of centre of mass of the system is shown below.

Hence the position of the centre of the mass is $\left(-2\hat{i}-1\hat{j}\right)\mathrm{m}$.

Calculate the velocity of the centre of mass.

(c)

Calculate the velocity of the centre of mass.

Substitute 2 kg for $m_1$, $\left(3\hat{i}+0.5\hat{j}\right)\mathrm{m}/\mathrm{s}$for $\overrightarrow{v_1}$, 3 kg for $m_2$and $\left(3\hat{i}-2\hat{j}\right)\mathrm{m}/\mathrm{s}$for $\overrightarrow{v_2}$into equation (iii).

$$\begin{gathered} \overrightarrow{v_{CM}}=\frac{2\left(3\hat{i}+0.5\hat{j}\right)+3\left(3\hat{i}-2\hat{j}\right)}{2+3} \\ \overrightarrow{v_{CM}}=\frac{6\hat{i}+1\hat{j}+9\hat{i}-6\hat{j}}{5} \\ \overrightarrow{v_{CM}}=\frac{15\hat{i}-5\hat{i}}{5} \\ \overrightarrow{v_{CM}}=\left(3\hat{i}-1\hat{j}\right)\mathrm{m}/\mathrm{s} \end{gathered}$$

The location of the velocity of centre of mass of the system is shown below.

Hence the velocity of the centre of mass is $\left(3\hat{i}-1\hat{j}\right)\mathrm{m}/\mathrm{s}$.

Calculate the linear momentum of the system.

(d)

Calculate the linear momentum of the system,

Substitute 2 kg for $m_1$, $\left(3\hat{i}+0.5\hat{j}\right)\mathrm{m}/\mathrm{s}$for $\overrightarrow{v_1}$, 3 kg for $m_2$and $\left(3\hat{i}-2\hat{j}\right)\mathrm{m}/\mathrm{s}$for $\overrightarrow{v_2}$into equation (iii).

$$\begin{gathered} \overrightarrow{P_{CM}}=2\left(3\hat{i}+0.5\hat{j}\right)+3\left(3\hat{i}-2\hat{j}\right) \\ \overrightarrow{P_{CM}}=6\hat{i}+1\hat{j}+9\hat{i}-6\hat{j} \\ \overrightarrow{P_{CM}}=\left(15\hat{i}-5\hat{j}\right)\mathrm{kg}·\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence the linear momentum of the system is $\left(15\hat{i}-5\hat{j}\right)\mathrm{kg}·\mathrm{m}/\mathrm{s}$.