Consider a two particle system with particles having masses m1 and m2. If the first particle is push

Q: Why do we take the displacement of $m_2$ as negative?

Since $m_1$ is moved towards the center of mass (let's say in the positive x direction), $m_2$ must be moved in the opposite direction (negative x direction) to keep the center of mass stationary.

Q: Does this principle apply if there are more than two particles?

Absolutely! The sum of the products of mass and displacement for all particles must equal zero if the center of mass is to remain stationary: $\sum m_i \Delta x_i = 0$.

Q: What if the particles were moving with constant velocities instead of being displaced?

The same logic applies to velocities. If the center of mass is stationary, its velocity is zero, meaning $m_1 v_1 + m_2 v_2 = 0$.

Q: Can we solve this using the concept of moments?

Yes! The center of mass is the point where the sum of mass moments is zero. If you change the moment of one mass by $m_1 d$, you must change the moment of the other by an equal and opposite amount $-m_2 d'$ to maintain balance.

Q: Why doesn't the initial position of the center of mass matter?

Because we are only concerned with the *change* in position. The shift formula $\Delta x_{CM} = \frac{m_1 \Delta x_1 + m_2 \Delta x_2}{m_1 + m_2}$ depends only on the displacements, not the absolute coordinates.

Question

Consider a two particle system with particles having masses $m_1$ and $m_2$. If the first particle is pushed towards the centre of mass through a distance $d$, by what distance should the second particle be moved, so as to keep the centre of mass at the same position ?

Options:

(A) $\frac{m_2}{m_1} d$

(B) $\frac{m_1}{m_1 + m_2} d$

(C) $\frac{m_1}{m_2} d$

(D) $d$

Sigma Percentile · Accepted Answer

### The Symmetrical Dance of Particles

Imagine a two-particle system. We have two masses, $m_1$ and $m_2$, placed on the x-axis. Their center of mass is located somewhere in between them, acting as the balancing point of the system.

Now, the question presents a scenario: $m_1$ is moved towards the center of mass by a distance $d$. If we want to keep the center of mass anchored at its original position, $m_2$ must also move. The question asks us to find this required displacement of $m_2$, which we will call $d'$.

### The Mathematical Anchor

Let's start with the fundamental formula for the position of the center of mass:

$$x_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$$

When $m_1$ is moved towards the center of mass by a distance $d$, its new position becomes $x_1 + d$. To keep the center of mass at the same place, $m_2$ must move towards the center of mass by a distance $d'$. So, the new position of $m_2$ will be $x_2 - d'$.

Since the center of mass has not changed, we can equate the old and new center of mass equations:

$$\frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} = \frac{m_1(x_1 + d) + m_2(x_2 - d')}{m_1 + m_2}$$

The denominator $m_1 + m_2$ cancels out from both sides. Expanding the numerator on the right side, we get:

$$m_1 x_1 + m_2 x_2 = m_1 x_1 + m_1 d + m_2 x_2 - m_2 d'$$

Notice how $m_1 x_1$ and $m_2 x_2$ beautifully cancel out from both sides! We are left with a very simple relation:

$$0 = m_1 d - m_2 d'$$

Solving for $d'$, we find:

$$d' = \frac{m_1}{m_2} d$$

### The Shift Formula Shortcut

We can also solve this problem much faster using the **shift formula**. The shift in the center of mass is directly related to the shifts of the individual particles:

$$\Delta x_{CM} = \frac{m_1 \Delta x_1 + m_2 \Delta x_2}{m_1 + m_2}$$

If the center of mass is stationary, its shift $\Delta x_{CM}$ is zero. This implies that the numerator must be zero:

$$m_1 \Delta x_1 + m_2 \Delta x_2 = 0$$

Let's substitute the values. The shift of $m_1$ is $d$ (positive, as it moves towards the CM), and the shift of $m_2$ is $-d'$ (negative, as it moves in the opposite direction to maintain balance).

$$m_1 (d) + m_2 (-d') = 0$$

$$m_1 d = m_2 d'$$

$$d' = \frac{m_1}{m_2} d$$

Both methods lead us to the same elegant conclusion. The displacement required for the second particle is inversely proportional to its mass relative to the first particle.

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Visualized Solution

$I ni t ia l S e t u p$

$C e n t r eo f M a ss F or m u l a$

$D i s pl a ce m e n t o f F i r s tP a r t i c l e$

$D i s pl a ce m e n t o f S eco n d P a r t i c l e$

$E q u a t in g C e n t r eo f M a ss$

$S o l v in g f or d^{'}$

$A l t er na t i v e M e t h o d : S hi f tF or m u l a$

$F ina l C a l c u l a t i o n$

The Sigma Insight: Centre of Mass

The Symmetrical Dance of Particles

The Mathematical Anchor

The Shift Formula Shortcut

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