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PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365724 Particles of masses $m, 2 m, 3 m, \dots . n m$ grams are placed on the same line at distance $l, 2 l, 3 l, \dots \dots . n l c m$ from a fixed point. The distance of centre of mass of the particles from the fixed point in centimetres is

1

\frac{1}{n + 1}

2

\frac{(2 n + 1) l}{3}

3

\frac{2 l}{n (n^{2} + 1)}

4

\frac{n (n^{2} + 1) l}{2}

Explanation:

$X_{C M} = \frac{m_{1} x_{1} + m_{2} x_{2} + \dots \dots}{m_{1} + m_{2} + \dots \dots}$
$= \frac{m l + 2 m \cdot 2 l + 3 m \cdot 3 l + \dots}{m + 2 m + 3 m + \dots}$
$= \frac{m l (1 + 4 + 9 + \dots \dots)}{m (1 + 2 + 3 + \dots .)} =$
$\frac{\frac{\ln (n + 1) (2 n + 1)}{6}}{\frac{n (n + 1)}{2}} = \frac{l (2 n + 1)}{3} .$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365725 In the $H C L$ molecule, the separation between the nuclei of the two atoms is about $1.27 \overset{o}{A} 1 \overset{o}{A} = 10^{- 10} m$ . Find the approximate location of the centre of mass of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucles.

1

r_{C M} = 2.22 \overset{o}{A}

2

r_{C M} = 1.24 \overset{o}{A}

3

r_{C M} = 3.24 \overset{o}{A}

4

r_{C M} = 0.24 \overset{o}{A}

Explanation:

Given, separation between the nuclei of
$H$ and $C l$ is $1.27 \overset{o}{A} = 1.27 \times 10^{- 10} m$
Let mass of hydrogen atom $= m$
$∴$ Mass of chlorine atom $= 35.5 m$
supporting img
Let mass of hydrogen atom be at origin, i.e., position vector of it, $r_{1} = 0$
$∴$ Position vector of chlorine atom
$r_{2} = 1.27 \times 10^{- 10} m$
$r_{C M} = \frac{m_{1} r_{1} + m_{2} r_{2}}{m_{1} + m_{2}}$
$= \frac{m \times 0 + 35.5 m \times 1.27 \times 10^{- 10}}{m + 35.5 m}$
$= \frac{35.5 \times 1.27 \times 10^{- 10}}{36.5} = 1.24 \overset{o}{A}$

NCERT Exemplar

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365726 A rigid body consists of a $3 k g$ mass located at $\vec{r} = (2 \hat{i} + 5 \hat{j}) m$ and a $2 k g$ mass located at $\vec{r} = (4 \hat{i} + 2 \hat{j}) m$ . The position of centre of mass is

1

(\frac{14}{5} \hat{j} + \frac{19}{5} \hat{i}) m

2

(\frac{14}{5} \hat{i} + \frac{19}{5} \hat{j}) m

3

(\frac{19}{5} \hat{i} + \frac{14}{5} \hat{j}) m

4

0

Explanation:

$L = \sqrt{4 + 9} = \sqrt{13}$
$x = \frac{3 \times 2 + 4 + 2}{3 + 2} = \frac{14}{5}$
$y = \frac{3 \times 5 + 2 \times 2}{5}$
$y = \frac{19}{5}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365727 A system consists of three particles, each of mass $m$ and located at $(1, 1), (2, 2)$ and $(3, 3)$ . The coordinates of the centre of mass are

1

(1, 1)

2

(2, 2)

3

(3, 3)

4

(6, 6)

Explanation:

The coordinates of $C . M$ of three particle are
$x = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}} &$
$y = \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3}}{m_{1} + m_{2} + m_{3}}$
here $m_{1} = m_{2} = m_{3} = m$
so $x = \frac{(x_{1} + x_{2} + x_{3}) m}{m + m + m} = 2,$
$y = \frac{(y_{1} + y_{2} + y_{3}) m}{m + m + m} = 2$
so coordinates of $C . M .$ of three particle are $(2, 2)$
$V = 300 m / s$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365724 Particles of masses $m, 2 m, 3 m, \dots . n m$ grams are placed on the same line at distance $l, 2 l, 3 l, \dots \dots . n l c m$ from a fixed point. The distance of centre of mass of the particles from the fixed point in centimetres is

1

\frac{1}{n + 1}

2

\frac{(2 n + 1) l}{3}

3

\frac{2 l}{n (n^{2} + 1)}

4

\frac{n (n^{2} + 1) l}{2}

Explanation:

$X_{C M} = \frac{m_{1} x_{1} + m_{2} x_{2} + \dots \dots}{m_{1} + m_{2} + \dots \dots}$
$= \frac{m l + 2 m \cdot 2 l + 3 m \cdot 3 l + \dots}{m + 2 m + 3 m + \dots}$
$= \frac{m l (1 + 4 + 9 + \dots \dots)}{m (1 + 2 + 3 + \dots .)} =$
$\frac{\frac{\ln (n + 1) (2 n + 1)}{6}}{\frac{n (n + 1)}{2}} = \frac{l (2 n + 1)}{3} .$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365725 In the $H C L$ molecule, the separation between the nuclei of the two atoms is about $1.27 \overset{o}{A} 1 \overset{o}{A} = 10^{- 10} m$ . Find the approximate location of the centre of mass of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucles.

1

r_{C M} = 2.22 \overset{o}{A}

2

r_{C M} = 1.24 \overset{o}{A}

3

r_{C M} = 3.24 \overset{o}{A}

4

r_{C M} = 0.24 \overset{o}{A}

Explanation:

Given, separation between the nuclei of
$H$ and $C l$ is $1.27 \overset{o}{A} = 1.27 \times 10^{- 10} m$
Let mass of hydrogen atom $= m$
$∴$ Mass of chlorine atom $= 35.5 m$
supporting img
Let mass of hydrogen atom be at origin, i.e., position vector of it, $r_{1} = 0$
$∴$ Position vector of chlorine atom
$r_{2} = 1.27 \times 10^{- 10} m$
$r_{C M} = \frac{m_{1} r_{1} + m_{2} r_{2}}{m_{1} + m_{2}}$
$= \frac{m \times 0 + 35.5 m \times 1.27 \times 10^{- 10}}{m + 35.5 m}$
$= \frac{35.5 \times 1.27 \times 10^{- 10}}{36.5} = 1.24 \overset{o}{A}$

NCERT Exemplar

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365726 A rigid body consists of a $3 k g$ mass located at $\vec{r} = (2 \hat{i} + 5 \hat{j}) m$ and a $2 k g$ mass located at $\vec{r} = (4 \hat{i} + 2 \hat{j}) m$ . The position of centre of mass is

1

(\frac{14}{5} \hat{j} + \frac{19}{5} \hat{i}) m

2

(\frac{14}{5} \hat{i} + \frac{19}{5} \hat{j}) m

3

(\frac{19}{5} \hat{i} + \frac{14}{5} \hat{j}) m

4

0

Explanation:

$L = \sqrt{4 + 9} = \sqrt{13}$
$x = \frac{3 \times 2 + 4 + 2}{3 + 2} = \frac{14}{5}$
$y = \frac{3 \times 5 + 2 \times 2}{5}$
$y = \frac{19}{5}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365727 A system consists of three particles, each of mass $m$ and located at $(1, 1), (2, 2)$ and $(3, 3)$ . The coordinates of the centre of mass are

1

(1, 1)

2

(2, 2)

3

(3, 3)

4

(6, 6)

Explanation:

The coordinates of $C . M$ of three particle are
$x = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}} &$
$y = \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3}}{m_{1} + m_{2} + m_{3}}$
here $m_{1} = m_{2} = m_{3} = m$
so $x = \frac{(x_{1} + x_{2} + x_{3}) m}{m + m + m} = 2,$
$y = \frac{(y_{1} + y_{2} + y_{3}) m}{m + m + m} = 2$
so coordinates of $C . M .$ of three particle are $(2, 2)$
$V = 300 m / s$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365724 Particles of masses $m, 2 m, 3 m, \dots . n m$ grams are placed on the same line at distance $l, 2 l, 3 l, \dots \dots . n l c m$ from a fixed point. The distance of centre of mass of the particles from the fixed point in centimetres is

1

\frac{1}{n + 1}

2

\frac{(2 n + 1) l}{3}

3

\frac{2 l}{n (n^{2} + 1)}

4

\frac{n (n^{2} + 1) l}{2}

Explanation:

$X_{C M} = \frac{m_{1} x_{1} + m_{2} x_{2} + \dots \dots}{m_{1} + m_{2} + \dots \dots}$
$= \frac{m l + 2 m \cdot 2 l + 3 m \cdot 3 l + \dots}{m + 2 m + 3 m + \dots}$
$= \frac{m l (1 + 4 + 9 + \dots \dots)}{m (1 + 2 + 3 + \dots .)} =$
$\frac{\frac{\ln (n + 1) (2 n + 1)}{6}}{\frac{n (n + 1)}{2}} = \frac{l (2 n + 1)}{3} .$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365725 In the $H C L$ molecule, the separation between the nuclei of the two atoms is about $1.27 \overset{o}{A} 1 \overset{o}{A} = 10^{- 10} m$ . Find the approximate location of the centre of mass of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucles.

1

r_{C M} = 2.22 \overset{o}{A}

2

r_{C M} = 1.24 \overset{o}{A}

3

r_{C M} = 3.24 \overset{o}{A}

4

r_{C M} = 0.24 \overset{o}{A}

Explanation:

Given, separation between the nuclei of
$H$ and $C l$ is $1.27 \overset{o}{A} = 1.27 \times 10^{- 10} m$
Let mass of hydrogen atom $= m$
$∴$ Mass of chlorine atom $= 35.5 m$
supporting img
Let mass of hydrogen atom be at origin, i.e., position vector of it, $r_{1} = 0$
$∴$ Position vector of chlorine atom
$r_{2} = 1.27 \times 10^{- 10} m$
$r_{C M} = \frac{m_{1} r_{1} + m_{2} r_{2}}{m_{1} + m_{2}}$
$= \frac{m \times 0 + 35.5 m \times 1.27 \times 10^{- 10}}{m + 35.5 m}$
$= \frac{35.5 \times 1.27 \times 10^{- 10}}{36.5} = 1.24 \overset{o}{A}$

NCERT Exemplar

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365726 A rigid body consists of a $3 k g$ mass located at $\vec{r} = (2 \hat{i} + 5 \hat{j}) m$ and a $2 k g$ mass located at $\vec{r} = (4 \hat{i} + 2 \hat{j}) m$ . The position of centre of mass is

1

(\frac{14}{5} \hat{j} + \frac{19}{5} \hat{i}) m

2

(\frac{14}{5} \hat{i} + \frac{19}{5} \hat{j}) m

3

(\frac{19}{5} \hat{i} + \frac{14}{5} \hat{j}) m

4

0

Explanation:

$L = \sqrt{4 + 9} = \sqrt{13}$
$x = \frac{3 \times 2 + 4 + 2}{3 + 2} = \frac{14}{5}$
$y = \frac{3 \times 5 + 2 \times 2}{5}$
$y = \frac{19}{5}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365727 A system consists of three particles, each of mass $m$ and located at $(1, 1), (2, 2)$ and $(3, 3)$ . The coordinates of the centre of mass are

1

(1, 1)

2

(2, 2)

3

(3, 3)

4

(6, 6)

Explanation:

The coordinates of $C . M$ of three particle are
$x = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}} &$
$y = \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3}}{m_{1} + m_{2} + m_{3}}$
here $m_{1} = m_{2} = m_{3} = m$
so $x = \frac{(x_{1} + x_{2} + x_{3}) m}{m + m + m} = 2,$
$y = \frac{(y_{1} + y_{2} + y_{3}) m}{m + m + m} = 2$
so coordinates of $C . M .$ of three particle are $(2, 2)$
$V = 300 m / s$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365724 Particles of masses $m, 2 m, 3 m, \dots . n m$ grams are placed on the same line at distance $l, 2 l, 3 l, \dots \dots . n l c m$ from a fixed point. The distance of centre of mass of the particles from the fixed point in centimetres is

1

\frac{1}{n + 1}

2

\frac{(2 n + 1) l}{3}

3

\frac{2 l}{n (n^{2} + 1)}

4

\frac{n (n^{2} + 1) l}{2}

Explanation:

$X_{C M} = \frac{m_{1} x_{1} + m_{2} x_{2} + \dots \dots}{m_{1} + m_{2} + \dots \dots}$
$= \frac{m l + 2 m \cdot 2 l + 3 m \cdot 3 l + \dots}{m + 2 m + 3 m + \dots}$
$= \frac{m l (1 + 4 + 9 + \dots \dots)}{m (1 + 2 + 3 + \dots .)} =$
$\frac{\frac{\ln (n + 1) (2 n + 1)}{6}}{\frac{n (n + 1)}{2}} = \frac{l (2 n + 1)}{3} .$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365725 In the $H C L$ molecule, the separation between the nuclei of the two atoms is about $1.27 \overset{o}{A} 1 \overset{o}{A} = 10^{- 10} m$ . Find the approximate location of the centre of mass of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucles.

1

r_{C M} = 2.22 \overset{o}{A}

2

r_{C M} = 1.24 \overset{o}{A}

3

r_{C M} = 3.24 \overset{o}{A}

4

r_{C M} = 0.24 \overset{o}{A}

Explanation:

Given, separation between the nuclei of
$H$ and $C l$ is $1.27 \overset{o}{A} = 1.27 \times 10^{- 10} m$
Let mass of hydrogen atom $= m$
$∴$ Mass of chlorine atom $= 35.5 m$
supporting img
Let mass of hydrogen atom be at origin, i.e., position vector of it, $r_{1} = 0$
$∴$ Position vector of chlorine atom
$r_{2} = 1.27 \times 10^{- 10} m$
$r_{C M} = \frac{m_{1} r_{1} + m_{2} r_{2}}{m_{1} + m_{2}}$
$= \frac{m \times 0 + 35.5 m \times 1.27 \times 10^{- 10}}{m + 35.5 m}$
$= \frac{35.5 \times 1.27 \times 10^{- 10}}{36.5} = 1.24 \overset{o}{A}$

NCERT Exemplar

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365726 A rigid body consists of a $3 k g$ mass located at $\vec{r} = (2 \hat{i} + 5 \hat{j}) m$ and a $2 k g$ mass located at $\vec{r} = (4 \hat{i} + 2 \hat{j}) m$ . The position of centre of mass is

1

(\frac{14}{5} \hat{j} + \frac{19}{5} \hat{i}) m

2

(\frac{14}{5} \hat{i} + \frac{19}{5} \hat{j}) m

3

(\frac{19}{5} \hat{i} + \frac{14}{5} \hat{j}) m

4

0

Explanation:

$L = \sqrt{4 + 9} = \sqrt{13}$
$x = \frac{3 \times 2 + 4 + 2}{3 + 2} = \frac{14}{5}$
$y = \frac{3 \times 5 + 2 \times 2}{5}$
$y = \frac{19}{5}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365727 A system consists of three particles, each of mass $m$ and located at $(1, 1), (2, 2)$ and $(3, 3)$ . The coordinates of the centre of mass are

1

(1, 1)

2

(2, 2)

3

(3, 3)

4

(6, 6)

Explanation:

The coordinates of $C . M$ of three particle are
$x = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}} &$
$y = \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3}}{m_{1} + m_{2} + m_{3}}$
here $m_{1} = m_{2} = m_{3} = m$
so $x = \frac{(x_{1} + x_{2} + x_{3}) m}{m + m + m} = 2,$
$y = \frac{(y_{1} + y_{2} + y_{3}) m}{m + m + m} = 2$
so coordinates of $C . M .$ of three particle are $(2, 2)$
$V = 300 m / s$