Chapter 9, Problem 28P

The vector position of a 3.50-g particle moving in the xy plane varies in time according to ${\stackrel{\to }{r}}_1=(3\hat{i}+3\hat{j})t+2\hat{j}{t}^2$, where t is in seconds and $\stackrel{\to }{r}$ is in centimeters. At the same time, the vector position of a 5.50 g particle varies as ${\stackrel{\to }{r}}_2=3\hat{i}-2\hat{i}{t}^2-6\hat{j}t$. At t = 2.50 s, determine (a) the vector position of the center of mass of the system, (b) the linear momentum of the system, (c) the velocity of the center of mass, (d) the acceleration of the center of mass, and (e) the net force exerted on the two-particle system.

To determine

The vector position of the centre of mass of the system.

Answer to Problem 28P

The vector position of the centre of mass of the system is $\text{-2}\text{.89}\widehat{\text{i}}-1.39\widehat{\text{j}}$.

Explanation of Solution

Given information:

The mass of the particle is $3.5\text{g}$, and the position of the particle is ${\overrightarrow{\text{r}}}_1=\left(3\widehat{\text{i}}+3\widehat{\text{j}}\right)t+2\widehat{\text{j}}{t}^2$. The mass of another particle is $5.5\text{g}$ and the position of the particle is ${\overrightarrow{\text{r}}}_2=3\widehat{\text{i}}-2\widehat{\text{i}}{t}^2-6\widehat{\text{j}}t$.

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 centre of mass of the system.
• $m_1$ is the mass of first particle.
• $m_2$ is the mass of second particle.

Substitute $3.5\text{g}$ for $m_1$, $5.5\text{g}$ for $m_2$, $\left(3\widehat{\text{i}}t+3\widehat{\text{j}}t+2\widehat{\text{j}}{t}^2\right)\text{cm}$ for ${\overrightarrow{\text{r}}}_1$ and $\left(3\widehat{\text{i}}-2\widehat{\text{i}}{t}^2-6\widehat{\text{j}}t\right)\text{cm}$ for ${\overrightarrow{\text{r}}}_2$ in equation (I).

$\begin{array}{c}{r}_{\text{CM}}=\frac{\left(3.5\text{g}\right)\left(3\widehat{\text{i}}t+3\widehat{\text{j}}t+2\widehat{\text{j}}{t}^2\right)\text{cm}+\left(5.5\text{g}\right)\left(3\widehat{\text{i}}-2\widehat{\text{i}}{t}^2-6\widehat{\text{j}}t\right)\text{cm}}{3.5\text{g}+5.5\text{g}}\\ =\frac{\left(10.5t+16.5-11t^2\right)\widehat{\text{i}}+\left(7t^2-22.5t\right)\widehat{\text{j}}}{9\text{g}}\end{array}$ (II)

Substitute $2.5\text{s}$ for $t$ in above equation.

$\begin{array}{c}{r}_{\text{CM}}=\frac{\left(10.5\times 2.5\text{s}+16.5-11{\left(2.5\text{s}\right)}^2\right)\widehat{\text{i}}+\left(7{\left(2.5\text{s}\right)}^2-22.5\times 2.5\text{s}\right)\widehat{\text{j}}}{9\text{g}}\\ \text{=-2}\text{.89}\widehat{\text{i}}-1.39\widehat{\text{j}}\end{array}$

Conclusion:

Therefore, the vector position of the centre of mass of the system is $\text{-2}\text{.89}\widehat{\text{i}}-1.39\widehat{\text{j}}$.

To determine

The linear momentum of the system.

Answer to Problem 28P

The linear momentum of the system is $-44.5\widehat{\text{i}}+12.5\widehat{\text{j}}$.

Explanation of Solution

Given information:

The mass of the particle is $3.5\text{g}$, and the position of the particle is ${\overrightarrow{\text{r}}}_1=\left(3\widehat{\text{i}}+3\widehat{\text{j}}\right)t+2\widehat{\text{j}}{t}^2$. The mass of another particle is $5.5\text{g}$ and the position of the particle is ${\overrightarrow{\text{r}}}_2=3\widehat{\text{i}}-2\widehat{\text{i}}{t}^2-6\widehat{\text{j}}t$.

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

$\overrightarrow{\text{v}}=\frac{d{r}_{\text{CM}}}{dt}$

• $\overrightarrow{\text{v}}$ is the velocity of centre of mass of the system.

Substitute $\frac{\left(10.5t+16.5-11t^2\right)\widehat{\text{i}}+\left(7t^2-22.5t\right)\widehat{\text{j}}}{9\text{g}}$ for $r_{\text{CM}}$ in above equation.

$\begin{array}{c}\overrightarrow{\text{v}}=\frac{d\frac{\left(10.5t+16.5-11t^2\right)\widehat{\text{i}}+\left(7t^2-22.5t\right)\widehat{\text{j}}}{9\text{g}}}{dt}\\ =\frac{\left(10.5-22t\right)}{9}\widehat{\text{i}}+\frac{\left(14t-22.5\right)}{9}\widehat{\text{j}}\end{array}$ (III)

Formula to calculate the momentum is given below.

$p=\left(m_1+m_2\right)\times \overrightarrow{\text{v}}$

Substitute $\frac{\left(10.5-22t\right)}{9}\widehat{\text{i}}+\frac{\left(14t-22.5\right)}{9}\widehat{\text{j}}$ for $\overrightarrow{\text{v}}$, $3.5\text{g}$ for $m_1$, $5.5\text{g}$ for $m_2$ in above equation.

$\begin{array}{c}p=\left(3.5\text{g}+5.5\text{g}\right)\times \frac{\left(10.5-22t\right)}{9}\widehat{\text{i}}+\frac{\left(14t-22.5\right)}{9}\widehat{\text{j}}\\ =\left(10.5-22t\right)\widehat{\text{i}}+\left(14t-22.5\right)\widehat{\text{j}}\end{array}$

Substitute $2.5\text{s}$ for $t$ in above equation.

$\begin{array}{c}p=\left(10.5-22\times 2.5\text{s}\right)\widehat{\text{i}}+\left(14\times 2.5\text{s}-22.5\right)\widehat{\text{j}}\\ =-44.5\widehat{\text{i}}+12.5\widehat{\text{j}}\end{array}$

Conclusion:

Therefore, the linear momentum of the system is $-44.5\widehat{\text{i}}+12.5\widehat{\text{j}}$.

To determine

The velocity of the centre of mass.

Answer to Problem 28P

The velocity of the centre of the mass is $-4.94\widehat{\text{i}}+1.39\widehat{\text{j}}$.

Explanation of Solution

Given information:

The mass of the particle is $3.5\text{g}$, and the position of the particle is ${\overrightarrow{\text{r}}}_1=\left(3\widehat{\text{i}}+3\widehat{\text{j}}\right)t+2\widehat{\text{j}}{t}^2$. The mass of another particle is $5.5\text{g}$ and the position of the particle is ${\overrightarrow{\text{r}}}_2=3\widehat{\text{i}}-2\widehat{\text{i}}{t}^2-6\widehat{\text{j}}t$.

The equation of velocity of centre of mass is,

$\overrightarrow{\text{v}}=\frac{\left(10.5-22t\right)}{9}\widehat{\text{i}}+\frac{\left(14t-22.5\right)}{9}\widehat{\text{j}}$

Substitute $2.5\text{s}$ for $t$ in above equation.

$\begin{array}{c}\overrightarrow{\text{v}}=\frac{\left(10.5-22\times 2.5\text{s}\right)}{9}\widehat{\text{i}}+\frac{\left(14\times 2.5\text{s}-22.5\right)}{9}\widehat{\text{j}}\\ =-4.94\widehat{\text{i}}+1.39\widehat{\text{j}}\end{array}$

Conclusion:

Therefore, the velocity of the centre of the mass is $-4.94\widehat{\text{i}}+1.39\widehat{\text{j}}$.

To determine

The acceleration of the centre of mass.

Answer to Problem 28P

The acceleration of the centre of the mass is $-2.44\widehat{\text{i}}+1.56\widehat{\text{j}}$.

Explanation of Solution

Given information:

The mass of the particle is $3.5\text{g}$, and the position of the particle is ${\overrightarrow{\text{r}}}_1=\left(3\widehat{\text{i}}+3\widehat{\text{j}}\right)t+2\widehat{\text{j}}{t}^2$. The mass of another particle is $5.5\text{g}$ and the position of the particle is ${\overrightarrow{\text{r}}}_2=3\widehat{\text{i}}-2\widehat{\text{i}}{t}^2-6\widehat{\text{j}}t$.

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

$\overrightarrow{\text{a}}=\frac{d\overrightarrow{\text{v}}}{dt}$

• $\overrightarrow{\text{a}}$ is the acceleration of centre of mass of the system.

Substitute $\frac{\left(10.5-22t\right)}{9}\widehat{\text{i}}+\frac{\left(14t-22.5\right)}{9}\widehat{\text{j}}$ for $\overrightarrow{\text{v}}$ in above equation.

$\begin{array}{c}\overrightarrow{\text{a}}=\frac{d\frac{\left(10.5-22t\right)}{9}\widehat{\text{i}}+\frac{\left(14t-22.5\right)}{9}\widehat{\text{j}}}{dt}\\ =-\frac{22}{9}\widehat{\text{i}}+\frac{14}{9}\widehat{\text{j}}\\ =-2.44\widehat{\text{i}}+1.56\widehat{\text{j}}\end{array}$

Conclusion:

Therefore, the acceleration of the center of the mass is $-2.44\widehat{\text{i}}+1.56\widehat{\text{j}}$.

To determine

The net force exerted on the two particle system.

Answer to Problem 28P

The net force exerted on the two particle system is $-22\widehat{\text{i}}+14\widehat{\text{j}}$.

Explanation of Solution

Given information:

The mass of the particle is $3.5\text{g}$, and the position of the particle is ${\overrightarrow{\text{r}}}_1=\left(3\widehat{\text{i}}+3\widehat{\text{j}}\right)t+2\widehat{\text{j}}{t}^2$. The mass of another particle is $5.5\text{g}$ and the position of the particle is ${\overrightarrow{\text{r}}}_2=3\widehat{\text{i}}-2\widehat{\text{i}}{t}^2-6\widehat{\text{j}}t$.

Formula to calculate the force on the system is,

$\overrightarrow{F}=\left(m_1+m_2\right)\overrightarrow{\text{a}}$

• $\overrightarrow{F}$ is the net force exerted on the two particle system.

Substitute $3.5\text{g}$ for $m_1$, $5.5\text{g}$ for $m_2$ and $-2.44\widehat{\text{i}}+1.56\widehat{\text{j}}$ for $\overrightarrow{\text{a}}$ in above equation.

$\begin{array}{c}\overrightarrow{F}=\left(3.5\text{g}+5.5\text{g}\right)\left(-2.44\widehat{\text{i}}+1.56\widehat{\text{j}}\right)\\ =-22\widehat{\text{i}}+14\widehat{\text{j}}\end{array}$

Conclusion:

Therefore, the net force exerted on the two particle system is $-22\widehat{\text{i}}+14\widehat{\text{j}}$.