Chapter 8, Problem 42P

(a)

To determine

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

Answer to Problem 42P

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.

(b)

To determine

The position of the centre of the mass.

Answer to Problem 42P

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 $-2.\widehat{\text{i}}-1\widehat{\text{j}}$.

(c)

To determine

The velocity of the centre of the mass.

Answer to Problem 42P

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 $-3\widehat{\text{i}}-1\widehat{\text{j}}$.

(d)

To determine

The total linear momentum of the system.

Answer to Problem 42P

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 $\left(15\widehat{\text{i}}-5\widehat{\text{j}}\right)\text{kg}\cdot \text{m}/\text{s}$.