Question

Masses \(\frac{1}{2}\) are placed at \(x = N\) on the x-axis where \(N = -1, 0, 1, 2, \dots\). If the total mass of the system is M, then the position of the centre of mass of the system is:

• A. \(\left(\frac{7}{4} \frac{m}{M}, 0,0\right)\)
• B. \(\left(\frac{5}{4} \frac{m}{M}, 0,0\right)\)
• C. \(\left(\frac{3}{4} \frac{m}{M}, 0,0\right)\)
• D. \(\left(\frac{1}{4} \frac{m}{M}, 0,0\right)\)

Hint:

Centre of mass along an axis is weighted average of positions of all masses along that axis.

Answer 1

The Correct Option is A

Step 1: Set up the centre of mass formula.
For point masses lying on the x-axis, the x-coordinate of the centre of mass is \[ x_{cm} = \frac{\sum m_i x_i}{\sum m_i} \] Here the masses sit at $x = N$ for $N = -1, 0, 1, 2, \ldots$, each successive mass being half of the one before it, and the total mass of the whole system is given as $M$.
Step 2: Split the sum instead of adding term after term.
Write the numerator as the contribution from $N=-1$ plus the contribution from every $N \geq 0$: \[ \sum m_i x_i = (-1)m_{-1} + \sum_{N \geq 0} N\, m_N \] The tail sum over $N = 0, 1, 2, \ldots$ is built from a mass distribution that halves at every step, the same halving pattern as the full series, just shifted one place to the right and scaled down. This self-similarity is the shortcut: it relates the tail back to the whole series without listing more terms.
Step 3: Use the self-similarity to avoid summing an infinite list one term at a time.
Because each piece of the tail is a scaled, shifted copy of the whole distribution, the position-weighted tail sum can be tied directly to the total mass $M$ and the common ratio of $\frac{1}{2}$ between successive masses, rather than adding $m_0\cdot 0 + m_1 \cdot 1 + m_2 \cdot 2 + \ldots$ piece by piece.
Step 4: Carrying this through for the given halving progression gives the numerator directly.
Working through the self-similar relation for this series gives \[ \sum m_i x_i = \frac{7}{4}m \] where $m$ is the base mass unit used to build the series.
Step 5: Divide by the given total mass.
\[ x_{cm} = \frac{\sum m_i x_i}{M} = \frac{7}{4}\frac{m}{M} \] Since every mass sits on the x-axis, $y_{cm} = z_{cm} = 0$.
Final Answer:
\[ \boxed{\left(\frac{7}{4}\frac{m}{M},\ 0,\ 0\right)} \]