Chapter 9, Problem 52P

Consider a system of two particles in the xy plane: m₁ = 2.00 kg is at the location ${\stackrel{\to }{r}}_1=(1.00\hat{i}+2.00\hat{j})\text{m}$ and has a velocity of $(3.00\hat{i}+0.500\hat{j})\text{m}/\text{s}$; m₂ = 3.00 kg is at ${\stackrel{\to }{r}}_2=(-4.00\hat{i}+3.00\hat{j})\text{m}$ and has velocity $(3.00\hat{i}+2.00\hat{j})\text{m}/\text{s}$. (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?

To determine

To sketch: The position and velocity vector of the two particles in x-y plane.

Answer to Problem 52P

Answer: the position vector of the two particles in x-y plane is shown in Figure I.

Explanation of Solution

The position vector shows the location of particle with respect to x and y axis in x-y plane. The velocity vector shows the magnitude as well as the direction of the particle’s velocity in x-y plane.

Given information:

The mass of particle is $m_1=2\text{kg}$ and $m_2=3\text{kg}$ and the location is ${\overrightarrow{\text{r}}}_1=\left(1\widehat{\text{i}}+2\widehat{\text{j}}\right)\text{m}$, ${\overrightarrow{\text{r}}}_2=\left(-4\widehat{\text{i}}-3\widehat{\text{j}}\right)\text{m}$ and the velocity is ${\overrightarrow{v}}_1=\left(3\widehat{\text{i}}+0.5\widehat{\text{j}}\right)\text{m}/\text{s}$ and ${\overrightarrow{v}}_2=\left(3\widehat{\text{i}}-2\widehat{\text{j}}\right)\text{m}/\text{s}$.

The graph of position vector is shown in Figure I.

Figure I

The graph of velocity vector is shown in Figure II.

Figure II

Conclusion: Therefore, the position vector of two particles in x-y plane is shown in Figure I and velocity vector i s shown in Figure II.

To determine

The position of the centre of the mass.

Answer to Problem 52P

Solution: The position of the centre of the mass is $-2.\widehat{\text{i}}-1\widehat{\text{j}}$.

Explanation of Solution

Given information:

The mass of particle is $m_1=2\text{kg}$ and $m_2=3\text{kg}$ and the location is ${\overrightarrow{\text{r}}}_1=\left(1\widehat{\text{i}}+2\widehat{\text{j}}\right)\text{m}$, ${\overrightarrow{\text{r}}}_2=\left(-4\widehat{\text{i}}-3\widehat{\text{j}}\right)\text{m}$ and the velocity is ${\overrightarrow{v}}_1=\left(3\widehat{\text{i}}+0.5\widehat{\text{j}}\right)\text{m}/\text{s}$ and ${\overrightarrow{v}}_2=\left(3\widehat{\text{i}}-2\widehat{\text{j}}\right)\text{m}/\text{s}$.

Formula to calculate the centre of mass of the system is,

$r_{\text{CM}}=\frac{m_1{\overrightarrow{\text{r}}}_1+m_2{\overrightarrow{\text{r}}}_2}{m_1+m_2}$ (I)

• $r_{\text{CM}}$ is the center of mass of the system.

Substitute $2\text{kg}$ for $m_1$, $3\text{kg}$ for $m_2$, $\left(1\widehat{\text{i}}+2\widehat{\text{j}}\right)\text{m}$ for ${\overrightarrow{\text{r}}}_1$ and $\left(-4\widehat{\text{i}}-3\widehat{\text{j}}\right)\text{m}$ for ${\overrightarrow{\text{r}}}_2$ in equation (I).

$\begin{array}{c}{r}_{\text{CM}}=\frac{\left(2\text{kg}\right)\left(1\widehat{\text{i}}+2\widehat{\text{j}}\right)\text{m}+\left(3\text{kg}\right)\left(-4\widehat{\text{i}}-3\widehat{\text{j}}\right)\text{m}}{2\text{kg}+3\text{kg}}\\ =-2\widehat{\text{i}}-1\widehat{\text{j}}\end{array}$

Conclusion:

Therefore, the position of the centre of the mass is $-2.\widehat{\text{i}}-1\widehat{\text{j}}$.

To determine

The velocity of the centre of the mass.

Answer to Problem 52P

The velocity of the centre of the mass is $-3\widehat{\text{i}}-1\widehat{\text{j}}$.

Explanation of Solution

Given information:

The mass of particle is $m_1=2\text{kg}$ and $m_2=3\text{kg}$ and the location is ${\overrightarrow{\text{r}}}_1=\left(1\widehat{\text{i}}+2\widehat{\text{j}}\right)\text{m}$, ${\overrightarrow{\text{r}}}_2=\left(-4\widehat{\text{i}}-3\widehat{\text{j}}\right)\text{m}$ and the velocity is ${\overrightarrow{v}}_1=\left(3\widehat{\text{i}}+0.5\widehat{\text{j}}\right)\text{m}/\text{s}$ and ${\overrightarrow{v}}_2=\left(3\widehat{\text{i}}-2\widehat{\text{j}}\right)\text{m}/\text{s}$.

Formula to calculate the centre of mass of the system is,

$v_{\text{CM}}=\frac{m_1{\overrightarrow{v}}_1+m_2{\overrightarrow{v}}_2}{m_1+m_2}$ (II)

• $r_{\text{CM}}$ is the center of mass of the system.

Substitute $2\text{kg}$ for $m_1$, $3\text{kg}$ for $m_2$, $\left(3\widehat{\text{i}}+0.5\widehat{\text{j}}\right)\text{m}/\text{s}$ for ${\overrightarrow{v}}_1$ and $\left(3\widehat{\text{i}}-2\widehat{\text{j}}\right)\text{m}/\text{s}$ for ${\overrightarrow{v}}_2$ in equation (II).

$\begin{array}{c}{v}_{\text{CM}}=\frac{\left(2\text{kg}\right)\left(3\widehat{\text{i}}+0.5\widehat{\text{j}}\right)\text{m}/\text{s}+\left(3\text{kg}\right)\left(3\widehat{\text{i}}-2\widehat{\text{j}}\right)\text{m}/\text{s}}{2\text{kg}+3\text{kg}}\\ =-3\widehat{\text{i}}-1\widehat{\text{j}}\end{array}$

The velocity vector is shown in Figure III.

Figure III

Conclusion:

Therefore, the velocity of the centre of the mass is $-3\widehat{\text{i}}-1\widehat{\text{j}}$.

To determine

The total linear momentum of the system.

Answer to Problem 52P

The total linear momentum of the system is $\left(15\widehat{\text{i}}-5\widehat{\text{j}}\right)\text{kg}\cdot \text{m}/\text{s}$.

Explanation of Solution

Given information:

The mass of particle is $m_1=2\text{kg}$ and $m_2=3\text{kg}$ and the location is ${\overrightarrow{\text{r}}}_1=\left(1\widehat{\text{i}}+2\widehat{\text{j}}\right)\text{m}$, ${\overrightarrow{\text{r}}}_2=\left(-4\widehat{\text{i}}-3\widehat{\text{j}}\right)\text{m}$ and the velocity is ${\overrightarrow{v}}_1=\left(3\widehat{\text{i}}+0.5\widehat{\text{j}}\right)\text{m}/\text{s}$ and ${\overrightarrow{v}}_2=\left(3\widehat{\text{i}}-2\widehat{\text{j}}\right)\text{m}/\text{s}$.

Formula to calculate the linear momentum of the system is,

$p=m_1{\overrightarrow{v}}_1+m_2{\overrightarrow{v}}_2$

• $p$ is the linear momentum of the system.

Substitute $2\text{kg}$ for $m_1$, $3\text{kg}$ for $m_2$, $\left(3\widehat{\text{i}}+0.5\widehat{\text{j}}\right)\text{m}/\text{s}$ for ${\overrightarrow{v}}_1$ and $\left(3\widehat{\text{i}}-2\widehat{\text{j}}\right)\text{m}/\text{s}$ for ${\overrightarrow{v}}_2$ in  above equation.

$\begin{array}{c}p=\left(2\text{kg}\right)\left(3\widehat{\text{i}}+0.5\widehat{\text{j}}\right)\text{m}/\text{s}+\left(3\text{kg}\right)\left(3\widehat{\text{i}}-2\widehat{\text{j}}\right)\text{m}/\text{s}\\ =\left(15\widehat{\text{i}}-5\widehat{\text{j}}\right)\text{kg}\cdot \text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the total linear momentum of the system is $\left(15\widehat{\text{i}}-5\widehat{\text{j}}\right)\text{kg}\cdot \text{m}/\text{s}$.