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Q7.33(a):
Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass : (a) Show $p_i = p'_i + m_i V$ where $p_i$ is the momentum of the ith particle (of mass $m_i$) and $p'_i = m_i v'_i$. Note $v'_i$ is the velocity of the ith particle relative to the centre of mass. Also, prove using the definition of the centre of mass $\sum p'_i = 0$.

Solution :

Step 1: Definition of Velocity Vectors

Let the position vector of the $i^{th}$ particle in a laboratory frame be $\vec{r}_i$ and the position of the Centre of Mass (CM) be $\vec{R}$. The position of the $i^{th}$ particle relative to the Centre of Mass is defined as:

$\vec{r}'_i = \vec{r}_i - \vec{R}$ [Definition of Relative Position Vector]

To find the corresponding velocities, we differentiate both sides with respect to time $t$:

$\frac{d\vec{r}'_i}{dt} = \frac{d\vec{r}_i}{dt} - \frac{d\vec{R}}{dt}$

$\vec{v}'_i = \vec{v}_i - \vec{V}$ [Where $\vec{v}_i$ is absolute velocity and $\vec{V}$ is the velocity of the CM]

Step 2: Derivation of Momentum Transformation

Given the velocity relationship $\vec{v}_i = \vec{v}'_i + \vec{V}$, we multiply both sides by the mass of the $i^{th}$ particle, $m_i$:

$m_i \vec{v}_i = m_i (\vec{v}'_i + \vec{V})$

$m_i \vec{v}_i = m_i \vec{v}'_i + m_i \vec{V}$

Substituting the definition of momentum $\vec{p}_i = m_i \vec{v}_i$ and $\vec{p}'_i = m_i \vec{v}'_i$:

$\vec{p}_i = \vec{p}'_i + m_i \vec{V}$

[This confirms the transformation of momentum into the CM frame components.]

Step 3: Proving the Summation of Relative Momenta is Zero

We analyze the sum of the relative momenta: $\sum \vec{p}'_i = \sum m_i \vec{v}'_i$.

Substitute $\vec{v}'_i = \vec{v}_i - \vec{V}$ into the expression:

$\sum \vec{p}'_i = \sum m_i (\vec{v}_i - \vec{V})$

$\sum \vec{p}'_i = \sum m_i \vec{v}_i - \sum m_i \vec{V}$

Since $\vec{V}$ is constant for the system at any instant, we factor it out:

$\sum \vec{p}'_i = \sum \vec{p}_i - \vec{V} \sum m_i$

Using the definition of total mass $M = \sum m_i$ and the definition of CM velocity $\vec{V} = \frac{\sum m_i \vec{v}_i}{M}$:

$\sum \vec{p}_i = M \vec{V}$

Substituting these back into the summation equation:

$\sum \vec{p}'_i = M \vec{V} - \vec{V} (M)$

$\sum \vec{p}'_i = 0$

[Per the definition of the Centre of Mass, the net momentum in the CM frame is identically zero.]

Visual Representation of Frames

CM (V) m_i v'_i

Final Solution: The relation $\vec{p}_i = \vec{p}'_i + m_i \vec{V}$ holds by definition of relative velocity, and the sum of relative momenta $\sum \vec{p}'_i = 0$ is proven by substituting the definition of the Centre of Mass, confirming the CM frame is the inertial frame where total momentum is zero.


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