- The centre of mass of two particles lies on the line joining them. To verify this, let r1′ and r2′ be position vector of m1 and m2 relative to centre of mass. Then, m1(rcm+r1′)+m2(rcm+r′)=(m1+m2)rcm ⇒m1r1′+m2r2′=0 or m1r1′=−m2r2′ Thus, r1′ and r2′ are directed opposite to each other. This shows that r1′ and r2′ are at 180∘ with each other but their tails lie at a common point. Thus, C lies in the line joining the particles. - The centre of mass divides the distance between two particles in the inverse ratio of masses. m1r1′+m2r2′=0⇒∣r2′∣∣ri∣=m1m2 [taking modulus] Thus, the centre of mass is dividing the distance in inverse

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- The centre of mass of two particles lies on the line joining

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- The centre of mass of two particles lies on the line joining them. To verify this, let $r_{1}^{'}$ and $r_{2}^{'}$ be position vector of $m_{1}$ and $m_{2}$ relative to centre of mass. Then, $m_{1} (r_{cm} + r_{1}^{'}) + m_{2} (r_{cm} + r^{'}) = (m_{1} + m_{2}) r_{cm}$ $\Rightarrow m_{1} r_{1}^{'} + m_{2} r_{2}^{'} = 0$ or $m_{1} r_{1}^{'} = - m_{2} r_{2}^{'}$ Thus, $r_{1}^{'}$ and $r_{2}^{'}$ are directed opposite to each other. This shows that $r_{1}^{'}$ and $r_{2}^{'}$ are at $18 0^{\circ}$ with each other but their tails lie at a common point. Thus, $C$ lies in the line joining the particles. - The centre of mass divides the distance between two particles in the inverse ratio of masses. $m_{1} r_{1}^{'} + m_{2} r_{2}^{'} = 0 \Rightarrow \frac{∣ r _{i} ∣}{∣ r _{2}^{'} ∣} = \frac{m _{2}}{m _{1}}$ [taking modulus] Thus, the centre of mass is dividing the distance in inverse

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- The centre of mass of two particles lies on the line joining them. To verify this, let

r_{1}^{'}

and

r_{2}^{'}

be position vector of

m_{1}

and

m_{2}

relative to centre of mass. Then,

m_{1} (r_{cm} + r_{1}^{'}) + m_{2} (r_{cm} + r^{'}) = (m_{1} + m_{2}) r_{cm}

\Rightarrow m_{1} r_{1}^{'} + m_{2} r_{2}^{'} = 0

m_{1} r_{1}^{'} = - m_{2} r_{2}^{'}

Thus,

r_{1}^{'}

and

r_{2}^{'}

are directed opposite to each other. This shows that

r_{1}^{'}

and

r_{2}^{'}

are at

18 0^{\circ}

with each other but their tails lie at a common point. Thus,

C

lies in the line joining the particles. - The centre of mass divides the distance between two particles in the inverse ratio of masses.

m_{1} r_{1}^{'} + m_{2} r_{2}^{'} = 0 \Rightarrow \frac{∣ r _{i} ∣}{∣ r _{2}^{'} ∣} = \frac{m _{2}}{m _{1}}

[taking modulus] Thus, the centre of mass is dividing the distance in inverse

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Question Text	- The centre of mass of two particles lies on the line joining them. To verify this, let $r_{1}^{'}$ and $r_{2}^{'}$ be position vector of $m_{1}$ and $m_{2}$ relative to centre of mass. Then, $m_{1} (r_{cm} + r_{1}^{'}) + m_{2} (r_{cm} + r^{'}) = (m_{1} + m_{2}) r_{cm}$ $\Rightarrow m_{1} r_{1}^{'} + m_{2} r_{2}^{'} = 0$ or $m_{1} r_{1}^{'} = - m_{2} r_{2}^{'}$ Thus, $r_{1}^{'}$ and $r_{2}^{'}$ are directed opposite to each other. This shows that $r_{1}^{'}$ and $r_{2}^{'}$ are at $18 0^{\circ}$ with each other but their tails lie at a common point. Thus, $C$ lies in the line joining the particles. - The centre of mass divides the distance between two particles in the inverse ratio of masses. $m_{1} r_{1}^{'} + m_{2} r_{2}^{'} = 0 \Rightarrow \frac{∣ r _{i} ∣}{∣ r _{2}^{'} ∣} = \frac{m _{2}}{m _{1}}$ [taking modulus] Thus, the centre of mass is dividing the distance in inverse
Updated On	Dec 23, 2022
Topic	Mechanics
Subject	Physics
Class	Class 12
Answer Type	Video solution: 1
Upvotes	95
Avg. Video Duration	6 min

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