Answer
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Hint: Let us consider a collection of 3 particles on the x axis. We can find the position of center of mass on the x axis by the formula to find the center of mass of a group of particles and the distance from origin will be exactly the same because we are calculating from the origin.
Formula used:
\[x=\dfrac{\sum\limits_{i=1}^{n}{{{m}_{i}}{{x}_{i}}}}{\sum\limits_{i=1}^{n}{{{m}_{i}}}}\]
Complete solution:
First of all let us understand what is the centre of mass. It is the point where the whole mass of the system is located.
In the question we have a system of 3 particles located at different positions on the x axis. If we represent it in a figure,
So now, let us find the position of the centre of mass by using the formula.
\[\begin{align}
& x=\dfrac{\sum\limits_{i=1}^{3}{{{m}_{i}}{{x}_{i}}}}{\sum\limits_{i=1}^{3}{{{m}_{i}}}} \\
& x=\dfrac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}+{{m}_{3}}{{x}_{3}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}} \\
\end{align}\]
Let, \[\begin{align}
& {{x}_{1}}=0cm \\
& {{x}_{2}}=40cm \\
& {{x}_{3}}=70cm \\
\end{align}\] \[\begin{align}
& {{m}_{1}}=300g \\
& {{m}_{2}}=500g \\
& {{m}_{3}}=400g \\
\end{align}\]
\[\Rightarrow x=\dfrac{300g\times 0cm+500g\times 40cm+400g\times 70cm}{300g+500g+400g}\]
\[x=\dfrac{20000+28000}{1200}\]
\[x=\dfrac{48000}{1200}=40cm\]
So, we got to the point where the centre of mass is located. Now, the distance from origin is exactly equal to \[40cm\] because we will be subtracting a 0 from it.
So the correct answer is option D.
Note: This question can also be solved by taking only 2 masses at a time. By doing this we will find the centre of mass of the 1st 2 particles and then equate the centre of mass of these particles with the 3rd one. After this we will get exactly the same answer which we got by using the formula.
Formula used:
\[x=\dfrac{\sum\limits_{i=1}^{n}{{{m}_{i}}{{x}_{i}}}}{\sum\limits_{i=1}^{n}{{{m}_{i}}}}\]
Complete solution:
First of all let us understand what is the centre of mass. It is the point where the whole mass of the system is located.
In the question we have a system of 3 particles located at different positions on the x axis. If we represent it in a figure,
So now, let us find the position of the centre of mass by using the formula.
\[\begin{align}
& x=\dfrac{\sum\limits_{i=1}^{3}{{{m}_{i}}{{x}_{i}}}}{\sum\limits_{i=1}^{3}{{{m}_{i}}}} \\
& x=\dfrac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}+{{m}_{3}}{{x}_{3}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}} \\
\end{align}\]
Let, \[\begin{align}
& {{x}_{1}}=0cm \\
& {{x}_{2}}=40cm \\
& {{x}_{3}}=70cm \\
\end{align}\] \[\begin{align}
& {{m}_{1}}=300g \\
& {{m}_{2}}=500g \\
& {{m}_{3}}=400g \\
\end{align}\]
\[\Rightarrow x=\dfrac{300g\times 0cm+500g\times 40cm+400g\times 70cm}{300g+500g+400g}\]
\[x=\dfrac{20000+28000}{1200}\]
\[x=\dfrac{48000}{1200}=40cm\]
So, we got to the point where the centre of mass is located. Now, the distance from origin is exactly equal to \[40cm\] because we will be subtracting a 0 from it.
So the correct answer is option D.
Note: This question can also be solved by taking only 2 masses at a time. By doing this we will find the centre of mass of the 1st 2 particles and then equate the centre of mass of these particles with the 3rd one. After this we will get exactly the same answer which we got by using the formula.
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