Three particles are placed in the xy plane. A 40-g particle is located at (3, 4) m, and a 50-g particle is positioned at (−2, −6) m. Where must a 20-g particle be placed so that the center of mass of this three-particle system is located at the origin?

Question

Quizplus · Accepted Answer

The center of mass of this three-particle system is located at the origin if the weighted average of the position vectors of the three particles is the zero vector:

$\frac{m_1\mathbf{r_1}+m_2\mathbf{r_2}+m_3\mathbf{r_3}}{m_1+m_2+m_3}=\mathbf{0}$

Substituting known values:

$\frac{40\begin{pmatrix}3 \ 4\end{pmatrix}+50\begin{pmatrix}-2 \ -6\end{pmatrix}+20\begin{pmatrix}x \ y\end{pmatrix}}{40+50+20}=\begin{pmatrix}0 \ 0\end{pmatrix}$

Simplifying:

$\begin{pmatrix}3x-0.5 \ 4y-1.8\end{pmatrix}=\begin{pmatrix}0 \ 0\end{pmatrix}$

Solving for $x$ and $y$:

$x=-0.5/3=-1/6$ and $y=1.8/4=9/20$

So the 20-g particle must be located at $(x,y)=\boxed{	ext{(D) } (-1,7)}$ meters.

Three Particles Are Placed in the Xy Plane