Chapter 10, Problem 67PQ

(a)

To determine

The position of center of mass at $t=3.00\text{s}$.

Answer to Problem 67PQ

The position of center of mass at $t=3.00\text{s}$ is $\underset{\_}{\overrightarrow{r_{CM}}=\left(1.43t^2-3t+2.14\right)\widehat{i}+\left(-1.43t^2+0.429t\right)\widehat{j}}$.

Explanation of Solution

Write the expression for velocity of center of mass.

  $\begin{array}{c}{\overrightarrow{r}}_{{}_{CM}}=\frac{{\displaystyle \sum _{j=1}^n{m}_j\overrightarrow{r_j}}}{M}\\ =\frac{m_1\overrightarrow{r_1}+m_2\overrightarrow{r_2}}{m_1+m_2}\end{array}$                                                                                               (I)

Here, $m_1$ is the mass of first particle, $m_2$ is the mass of second particle, $r_1$ is the position of the first particle, and $r_2$ is the position of second particle.

Conclusion:

Substitute, $1.00\text{kg}$ for $m_1$, $2.50\text{kg}$ for $m_2$, $\left(\left(2t+5t^2\right)\widehat{i}+4t\widehat{j}\right)\text{m}$ for $r_1$, and $\left(\left(3-5t\right)\widehat{i}+\left(-t-2t^2\right)\widehat{j}\right)\text{m}$ for $r_2$ in equation (I).

  $\begin{array}{c}{\overrightarrow{r}}_{{}_{CM}}=\frac{\left(1.00\text{kg}\right)\left(\left(2t+5t^2\right)\widehat{i}+4t\widehat{j}\right)\text{m}+\left(2.50\text{kg}\right)\left(\left(3-5t\right)\widehat{i}+\left(-t-2t^2\right)\widehat{j}\right)\text{m}}{1.00\text{kg}+2.50\text{kg}}\\ =\left(\left(1.43t^2-3t+2.14\right)\widehat{i}+(-1.43t^2+0.429t)\widehat{j}\right)\text{m}\end{array}$     (II)

Substitute, $3.00\text{s}$ for $t$ in equation (II).

  $\begin{array}{c}{\overrightarrow{r}}_{{}_{CM}}=\left(1.43{\left(3.00\text{s}\right)}^2-3\left(3.00\text{s}\right)+2.14\right)\widehat{i}+\left(-1.43{\left(3.00\text{s}\right)}^2+0.429\left(3.00\text{s}\right)\right)\widehat{j}\\ =\left(6.00\widehat{i}-11.6\widehat{j}\right)\text{m}\end{array}$

Therefore, the position of center of mass at $t=3.00\text{s}$ is $\underset{\_}{\overrightarrow{r_{CM}}=\left(6.00\widehat{i}-11.6\widehat{j}\right)\text{m}}$.

(b)

To determine

The velocity of the centre of mass.

Answer to Problem 67PQ

The velocity of the centre of mass is $\underset{\_}{{\overrightarrow{v}}_{{}_{CM}}=\left(5.58\widehat{i}-8.15\widehat{j}\right)\text{m/s}}$.

Explanation of Solution

Velocity is the time derivative of position.

  ${\overrightarrow{v}}_{{}_{CM}}=\frac{d{\overrightarrow{r}}_{{}_{CM}}}{dt}$                                                                                                        (III)

Here, ${\overrightarrow{v}}_{{}_{CM}}$ is the velocity of center of mass.

Substitute equation (II) in equation (III).

  $\begin{array}{c}{\overrightarrow{v}}_{{}_{CM}}=\frac{d\left(\left(1.43t^2-3t+2.14\right)\widehat{i}+(-1.43t^2+0.429t)\widehat{j}\right)}{dt}\\ =\left(2.86t-3\right)\widehat{i}+\left(-2.86t+0.429\right)\widehat{j}\end{array}$                                      (IV)

Conclusion:

Substitute, $3.00\text{s}$ for $t$ in equation (IV).

  $\begin{array}{c}{\overrightarrow{v}}_{{}_{CM}}=\left(2.86\times 3.00\text{s}-3\right)\widehat{i}+\left(-2.86\times 3.00\text{s}+0.429\right)\widehat{j}\\ =\left(5.58\widehat{i}-8.15\right)\widehat{j}\end{array}$

Therefore, velocity of the centre of mass is $\underset{\_}{{\overrightarrow{v}}_{{}_{CM}}=\left(5.58\widehat{i}-8.15\widehat{j}\right)\text{m/s}}$.

(c)

To determine

The total linear momentum of the system.

Answer to Problem 67PQ

The total linear momentum of the system is $\underset{\_}{\left(19.5\widehat{i}-28.5\widehat{j}\right)\text{kg}\cdot \text{m/s}}$.

Explanation of Solution

Write the expression for momentum.

  $\overrightarrow{p}=\left(m_1+m_2\right){\overrightarrow{v}}_{{}_{CM}}$                                                                                               (V)

Conclusion:

Substitute, $1.00\text{kg}$ for $m_1$, $2.50\text{kg}$ for $m_2$, and $\left(5.58\widehat{i}-8.15\widehat{j}\right)\text{m/s}$ for ${\overrightarrow{v}}_{{}_{CM}}$ in equation (V).

  $\begin{array}{c}\overrightarrow{p}=\left(1.00\text{kg}+2.50\text{kg}\right)\left(5.58\widehat{i}-8.15\widehat{j}\right)\text{m/s}\\ \text{=}\left(19.5\widehat{i}-28.5\widehat{j}\right)\text{kg}\cdot \text{m/s}\end{array}$

Therefore, the total linear momentum of the system is $\underset{\_}{\left(19.5\widehat{i}-28.5\widehat{j}\right)\text{kg}\cdot \text{m/s}}$.