A big olive ($\mathbf{m}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{50}\mathbf{kg}$) lies at the origin of an XYcoordinates system, and a big Brazil nut ( $\mathbf{M}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{kg}$) lies at the point $\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{,}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{)}\mathbf{m}$ . At $\mathbf{t}\mathbf{=}\mathbf{0}$ , a force ${\mathbf{F}}_{\mathbf{0}}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{0}\widehat{\mathbf{j}\mathbf{)}}$ begins to act on the olive, and a force localid="1657267122657" ${\mathbf{F}}_n\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{0}\widehat{\mathbf{j}\mathbf{)}}$begins to act on the nut. In unit-vector notation, what is the displacement of the center of mass of the olive–nut system at$\mathbf{t}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{s}$ with respect to its position at$\mathbf{t}\mathbf{=}\mathbf{0}$?

The mass of the big olive is $m=0.50kg$

The big Brazil nut is $M=1.5kg$

The coordinates of the Brazil nut is$(x,y)=(1.0m,2.0m)$

The force acts on the olive is${\overrightarrow{F}}_0=\left[(2.0)\hat{i}+(3.0)\hat{j}\right]N$

The force acts on the nut is role="math" localid="1657267783011" ${\overrightarrow{F}}_n=\left[(-3.0)\hat{i}-(2.0)\hat{j}\right]N$

The time for the system is$t=4.0s$

We can use the concept of center of mass of the system and Newton’s second law. The second kinematic equation of motion can be used to find the displacement of the center of mass.

Formula:

$$\begin{gathered} F_{net}=ma \\ S=v_{0}t+\frac{1}{2}a{t}^2 \end{gathered}$$

The total force on the nut-olive system is,

$$\begin{gathered} {\overrightarrow{F}}_{net}={\overrightarrow{F}}_0+{\overrightarrow{F}}_n \\ {\overrightarrow{F}}_{net}=\left[(2.0m)\hat{i}+3.0m)\hat{j}\right]N+\left[(-3.0)\hat{i}-(2.0)\hat{j}\right]N \\ =(-1.0)\hat{i}+(1.0)\hat{j}N \end{gathered}$$

According to the Newton’s second law,

$$\begin{gathered} {\overrightarrow{F}}_{net}=m\overrightarrow{a} \\ =(M+m){\overrightarrow{a}}_{com} \\ \left[(-1.0)\hat{i}+(1.0)\hat{j}\right]N=(1.5kg+0.50kg){\overrightarrow{a}}_{com} \\ \left[(-1.0)\hat{i}+(1.0)\hat{j}\right]N=(-2.0kg)\overrightarrow{a_{com}} \\ {\overrightarrow{a}}_{com}=\left[\left(\frac{-1.0}{2.0}\right)\hat{i}+\left(\frac{1.0}{2.0}\right)\hat{j}\right]m/s^2 \end{gathered}$$

The initial velocity of the system is zero, hence according to the second kinematical equation,

$\begin{array}{l}S=v_{0}t+\frac{1}{2}a{t}^2\\ =\frac{1}{2}a{t}^2\\ {\overrightarrow{S}}_{com}=\frac{1}{2}{\overrightarrow{a}}_{com}{t}^2\\ =\frac{1}{2}\times \left[\left(\frac{-1.0}{2.0}\right)\hat{i}+\left(\frac{1.0}{2.0}\right)\hat{j}\right]\times {(4.0s)}^2\\ =(-4.0m)\hat{i}+(4.0m)\hat{j}\end{array}$

Therefore, the displacement of the center of mass of the olive-nut system at , with respect to its position at $t=0$ is $\begin{array}{l}{\overrightarrow{S}}_{com}=(-4.0m)\hat{i}+(4.0m)\hat{j}.\end{array}$