Question

Centre of mass of three particles of masses $1kg,2kg$ and $3kg$lies at the point $(1,2,3)$ and the centre of mass of another system of particles of total mass $3kg$ lies at the point $(-1,3,-2)$. Where should we put a particle of mass $5kg$ so that the centre of mass of the entire system lies at the centre of mass of the first system?
$\begin{align}
  & A)(0,0,0) \\
 & B)(1,3,2) \\
 & C)(-1,2,3) \\
\end{align}$
$D)$None of these

Answer 1

Hint: Centre of mass of a mass distribution refers to the point where all the masses of the distribution concentrate. Clearly, the total mass of the first system of particles ($1kg,2kg$ and $3kg$) is concentrated at $(1,2,3)$. Similarly, the total mass of the second system ($3kg$) is concentrated at $(-1,3,-2)$. A random point is chosen to be the centre of mass of the third system, whose total mass is $5kg$. All these masses in all the three systems are combined together into a single system, whose centre of mass is given to be the same as that of the first system of particles.
Formula used:
$\begin{align}
  & X=\dfrac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}+{{m}_{3}}{{x}_{3}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}} \\
 & Y=\dfrac{{{m}_{1}}{{y}_{1}}+{{m}_{2}}{{y}_{2}}+{{m}_{3}}{{y}_{3}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}} \\
 & Z=\dfrac{{{m}_{1}}{{z}_{1}}+{{m}_{2}}{{z}_{2}}+{{m}_{3}}{{z}_{3}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}} \\
\end{align}$

Complete answer:
Centre of mass of a mass distribution refers to the point at which the total mass of the distribution is concentrated. We are given that centre of mass of three particles of masses $1kg,2kg$ and $3kg$ lies at the point $(1,2,3)$ and the centre of mass of another system of particles of total mass $3kg$ lies at the point $(-1,3,-2)$. We are required to determine the point at which a particle of mass $5kg$ needs to be placed so that the centre of mass of the entire system lies at the centre of mass of the first system.
Using the concept of the centre of mass of a mass distribution, we can divide the problem into three systems of particles.
The first system of particles involves masses $1kg,2kg$ and $3kg$, whose total mass is given by
$1kg+2kg+3kg=6kg$
It is given that the centre of mass of this system of particles lies on the point $(1,2,3)$. Let us call the centre of mass of the first system $({{x}_{1}},{{y}_{1}},{{z}_{1}})$ and the total mass of the first system ${{m}_{1}}$. Clearly, total mass of the system $({{m}_{1}}=6kg)$ concentrates on $({{x}_{1}},{{y}_{1}},{{z}_{1}})=(1,2,3)$.
Let this be statement 1.
Similarly, the second system of particles involves a total mass of $3kg$, whose centre of mass lies on the point $(-1,3,-2)$. Let us call the centre of mass of the second system $({{x}_{2}},{{y}_{2}},{{z}_{2}})$ and the total mass of the second system ${{m}_{2}}$. Clearly, the total mass of the second system $({{m}_{2}}=3kg)$concentrates on $({{x}_{2}},{{y}_{2}},{{z}_{2}})=(-1,3,-2)$
Let this be statement 2.
Also, let us consider the mass $5kg$ to be the total mass of a third system whose centre of mass lies on a point, say $(x,y,z)$. Let us call the centre of mass of the third system $({{x}_{3}},{{y}_{3}},{{z}_{3}})$ and the total mass of the third system ${{m}_{3}}$. Clearly, total mass of the third system $({{m}_{3}}=5kg)$concentrates on $({{x}_{3}},{{y}_{3}},{{z}_{3}})=(x,y,z)$.
Let this be statement 3.
Now, we are told that centre of mass of all the masses given in statement 1, statement 2, and statement 3 is equal to the centre of mass of the first system i.e. if the centre of mass of the entire system of particles is denoted by $(X,Y,Z)$, then, $(X,Y,Z)$ is equal to $(1,2,3)$. Therefore, we can say that total mass of the entire system $(={{m}_{1}}+{{m}_{2}}+{{m}_{3}}=6kg+3kg+5kg=14kg)$ concentrates on $(X,Y,Z)=(1,2,3)$.
Let this be statement 4.
Clearly, centre of mass of the entire system is given by
$\begin{align}
  & X=\dfrac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}+{{m}_{3}}{{x}_{3}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}} \\
 & Y=\dfrac{{{m}_{1}}{{y}_{1}}+{{m}_{2}}{{y}_{2}}+{{m}_{3}}{{y}_{3}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}} \\
 & Z=\dfrac{{{m}_{1}}{{z}_{1}}+{{m}_{2}}{{z}_{2}}+{{m}_{3}}{{z}_{3}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}} \\
\end{align}$
Substituting the values of $X,Y,Z,{{x}_{1}},{{x}_{2}},{{x}_{3}},{{y}_{1}},{{y}_{2}},{{y}_{3}},{{z}_{1}},z{}_{2},{{z}_{3}},{{m}_{1}},{{m}_{2}}$ and ${{m}_{3}}$ in the above expression, we have
$\begin{align}
  & X=\dfrac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}+{{m}_{3}}{{x}_{3}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}\Rightarrow 1=\dfrac{6(1)+3(-1)+5x}{6+3+5}\Rightarrow 1=\dfrac{3+5x}{14}\Rightarrow x=\dfrac{11}{5} \\
 & Y=\dfrac{{{m}_{1}}{{y}_{1}}+{{m}_{2}}{{y}_{2}}+{{m}_{3}}{{y}_{3}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}\Rightarrow 2=\dfrac{6(2)+3(3)+5y}{6+3+5}\Rightarrow 2=\dfrac{21+5y}{14}\Rightarrow y=\dfrac{7}{5} \\
 & Z=\dfrac{{{m}_{1}}{{z}_{1}}+{{m}_{2}}{{z}_{2}}+{{m}_{3}}{{z}_{3}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}\Rightarrow 3=\dfrac{6(3)+3(-2)+5z}{6+3+5}\Rightarrow 2=\dfrac{12+5z}{14}\Rightarrow z=\dfrac{16}{5} \\
\end{align}$
Therefore, the point at which a particle of mass $5kg$ needs to be placed so that the centre of mass of the entire system lies at the centre of mass of the first system is $\left( \dfrac{11}{5},\dfrac{7}{5},\dfrac{16}{5} \right)$.

So, the correct answer is “Option D”.

Note:
As clear from the above solution, there are many points and masses involved here. Students need to be careful while naming each point such that they are not repeating any variables twice. Also, the final answer can be cross-checked twice or thrice to opt the correct answer from the given options since minute mistakes can cause serious variations in the answer.