QBManager

Rotational Motion

149709 In a two-particle system with particle masses $m_{1}$ and $m_{2}$ , the first particle is pushed towards the centre of mass through a distance $d$ , the distance through which second particle must be moved to keep the centre of mass at the same position is

1

\frac{m_{2} d}{m_{1}}

2 d

3

\frac{m_{1} d}{(m_{1} + m_{2})}

4

\frac{(m_{1} + m_{2}) d}{m_{1}}

5

\frac{m_{1} d}{m_{2}}

Explanation:

E
$x_{cm} = \frac{m_{1} (0) + m_{2} (l)}{m_{1} + m_{2}}$
$x_{cm} = \frac{m_{2} l}{m_{1} + m_{2}}$
$x_{cm} = \frac{m_{1} (d) + m_{2} (l - x)}{m_{1} + m_{2}}$
On equating equation (i) \& (ii), we get
$\frac{m_{1} d + m_{2} (l - x)}{m_{1} + m_{2}} = \frac{m_{2} l}{m_{1} + m_{2}}$
$m_{1} d + m_{2} l - m_{2} x = m_{2} l$
$m_{1} d = m_{2} x \Rightarrow x = \frac{m_{1} d}{m_{2}}$

Kerala CEE- 2013

Rotational Motion

149710 A system consists of 3 particles each of mass $m$ located at points $(1, 1) (2, 2)$ and $(3, 3)$ . The coordinates of the centre of mass are

1

(6, 6)

2

(3, 3)

3

(1, 1)

4

(2, 2)

5

(5, 5)

Explanation:

D Given, $m_{1} = m_{2} = m_{3} = m$
$x_{1} = 1, x_{2} = 2$ and $x_{3} = 3$
$y_{1} = 1, y_{2} = 2$ , and $y_{3} = 3$
$x_{cm} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}}$
$x_{cm} = \frac{{mx}_{1} + {mx}_{2} + {mx}_{3}}{m + m + m} = \frac{x_{1} + x_{2} + x_{3}}{3}$
$= \frac{1 + 2 + 3}{3} = \frac{6}{3} = 2$
$y_{cm} = \frac{{my}_{1} + {my}_{2} + {my}_{3}}{m + m + m} = \frac{y_{1} + y_{2} + y_{3}}{3}$
$= \frac{1 + 2 + 3}{3} = \frac{6}{3} = 2$
Hence, the coordinates of the centre of mass are $(2, 2)$ .

BITSAT-2010

Rotational Motion

149711 Three identical spheres, each of mass $3 kg$ are placed touching each other with their centres lying on a straight line. The centres of the sphere are marked at $P, Q$ and $R$ respectively. The distance of centre of mass of system from $P$ is

1

\frac{PQ + QR + PR}{3}

2

\frac{P Q + P R}{3}

3

\frac{PQ + QR + PR}{9}

4

\frac{P Q + P R}{9}

5

\frac{PQ + QR}{3}

Explanation:

B \lt smiles>[R]c1cccc2cccc([R])c12 \lt /smiles>
The centre of mass of the given system consisting of three identical spheres of mass $3 kg$ each is
$x_{c m} = \frac{M_{P} (x_{P}) + M_{Q} (x_{Q}) + M_{R} (x_{R})}{M_{P} + M_{Q} + M_{R}}$
$x_{cm} = \frac{M (0) + M (PQ) + M (PR)}{M + M + M}$
$= \frac{M (PQ + PR)}{3 M}$
$x_{cm} = \frac{PQ + PR}{3}$
Hence, COM of the given system is $\frac{P Q + P R}{3}$

Kerala CEE - 2008

Rotational Motion

149712 The distance between the centers of carbon and oxygen atoms in the carbon monoxide molecule is $1.130 \AA$ . Locate the centre of mass of the molecule relative to the carbon atom :

1

5.428 \AA

2

1.130 \AA

3

0.6457 \AA

4

0.3260 \AA

5 none of these

Explanation:

C $m_{1} = 16$ (Oxygen mass number)
$m_{2} = 12$ (carbon mass number)

$m_{2} x_{2} = m_{1} x_{1}$
$12 x_{2} = 16 (1.130 \times 10^{- 10} - x_{2})$
$12 x_{2} = 16 \times 1.130 \times 10^{- 10} - 16 x_{2}$
$x_{2} = \frac{16 \times 1.13 \times 10^{- 10}}{28}$
$x_{2} = 0.6457 \AA$
Hence, the centre of mass (COM) of molecule relative to the carbon atom is $0.6457^{\circ} A$ .

Kerala CEE 2006

Rotational Motion

149709 In a two-particle system with particle masses $m_{1}$ and $m_{2}$ , the first particle is pushed towards the centre of mass through a distance $d$ , the distance through which second particle must be moved to keep the centre of mass at the same position is

1

\frac{m_{2} d}{m_{1}}

2 d

3

\frac{m_{1} d}{(m_{1} + m_{2})}

4

\frac{(m_{1} + m_{2}) d}{m_{1}}

5

\frac{m_{1} d}{m_{2}}

Explanation:

E
$x_{cm} = \frac{m_{1} (0) + m_{2} (l)}{m_{1} + m_{2}}$
$x_{cm} = \frac{m_{2} l}{m_{1} + m_{2}}$
$x_{cm} = \frac{m_{1} (d) + m_{2} (l - x)}{m_{1} + m_{2}}$
On equating equation (i) \& (ii), we get
$\frac{m_{1} d + m_{2} (l - x)}{m_{1} + m_{2}} = \frac{m_{2} l}{m_{1} + m_{2}}$
$m_{1} d + m_{2} l - m_{2} x = m_{2} l$
$m_{1} d = m_{2} x \Rightarrow x = \frac{m_{1} d}{m_{2}}$

Kerala CEE- 2013

Rotational Motion

149710 A system consists of 3 particles each of mass $m$ located at points $(1, 1) (2, 2)$ and $(3, 3)$ . The coordinates of the centre of mass are

1

(6, 6)

2

(3, 3)

3

(1, 1)

4

(2, 2)

5

(5, 5)

Explanation:

D Given, $m_{1} = m_{2} = m_{3} = m$
$x_{1} = 1, x_{2} = 2$ and $x_{3} = 3$
$y_{1} = 1, y_{2} = 2$ , and $y_{3} = 3$
$x_{cm} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}}$
$x_{cm} = \frac{{mx}_{1} + {mx}_{2} + {mx}_{3}}{m + m + m} = \frac{x_{1} + x_{2} + x_{3}}{3}$
$= \frac{1 + 2 + 3}{3} = \frac{6}{3} = 2$
$y_{cm} = \frac{{my}_{1} + {my}_{2} + {my}_{3}}{m + m + m} = \frac{y_{1} + y_{2} + y_{3}}{3}$
$= \frac{1 + 2 + 3}{3} = \frac{6}{3} = 2$
Hence, the coordinates of the centre of mass are $(2, 2)$ .

BITSAT-2010

Rotational Motion

149711 Three identical spheres, each of mass $3 kg$ are placed touching each other with their centres lying on a straight line. The centres of the sphere are marked at $P, Q$ and $R$ respectively. The distance of centre of mass of system from $P$ is

1

\frac{PQ + QR + PR}{3}

2

\frac{P Q + P R}{3}

3

\frac{PQ + QR + PR}{9}

4

\frac{P Q + P R}{9}

5

\frac{PQ + QR}{3}

Explanation:

B \lt smiles>[R]c1cccc2cccc([R])c12 \lt /smiles>
The centre of mass of the given system consisting of three identical spheres of mass $3 kg$ each is
$x_{c m} = \frac{M_{P} (x_{P}) + M_{Q} (x_{Q}) + M_{R} (x_{R})}{M_{P} + M_{Q} + M_{R}}$
$x_{cm} = \frac{M (0) + M (PQ) + M (PR)}{M + M + M}$
$= \frac{M (PQ + PR)}{3 M}$
$x_{cm} = \frac{PQ + PR}{3}$
Hence, COM of the given system is $\frac{P Q + P R}{3}$

Kerala CEE - 2008

Rotational Motion

149712 The distance between the centers of carbon and oxygen atoms in the carbon monoxide molecule is $1.130 \AA$ . Locate the centre of mass of the molecule relative to the carbon atom :

1

5.428 \AA

2

1.130 \AA

3

0.6457 \AA

4

0.3260 \AA

5 none of these

Explanation:

C $m_{1} = 16$ (Oxygen mass number)
$m_{2} = 12$ (carbon mass number)

$m_{2} x_{2} = m_{1} x_{1}$
$12 x_{2} = 16 (1.130 \times 10^{- 10} - x_{2})$
$12 x_{2} = 16 \times 1.130 \times 10^{- 10} - 16 x_{2}$
$x_{2} = \frac{16 \times 1.13 \times 10^{- 10}}{28}$
$x_{2} = 0.6457 \AA$
Hence, the centre of mass (COM) of molecule relative to the carbon atom is $0.6457^{\circ} A$ .

Kerala CEE 2006

Rotational Motion

149709 In a two-particle system with particle masses $m_{1}$ and $m_{2}$ , the first particle is pushed towards the centre of mass through a distance $d$ , the distance through which second particle must be moved to keep the centre of mass at the same position is

1

\frac{m_{2} d}{m_{1}}

2 d

3

\frac{m_{1} d}{(m_{1} + m_{2})}

4

\frac{(m_{1} + m_{2}) d}{m_{1}}

5

\frac{m_{1} d}{m_{2}}

Explanation:

E
$x_{cm} = \frac{m_{1} (0) + m_{2} (l)}{m_{1} + m_{2}}$
$x_{cm} = \frac{m_{2} l}{m_{1} + m_{2}}$
$x_{cm} = \frac{m_{1} (d) + m_{2} (l - x)}{m_{1} + m_{2}}$
On equating equation (i) \& (ii), we get
$\frac{m_{1} d + m_{2} (l - x)}{m_{1} + m_{2}} = \frac{m_{2} l}{m_{1} + m_{2}}$
$m_{1} d + m_{2} l - m_{2} x = m_{2} l$
$m_{1} d = m_{2} x \Rightarrow x = \frac{m_{1} d}{m_{2}}$

Kerala CEE- 2013

Rotational Motion

149710 A system consists of 3 particles each of mass $m$ located at points $(1, 1) (2, 2)$ and $(3, 3)$ . The coordinates of the centre of mass are

1

(6, 6)

2

(3, 3)

3

(1, 1)

4

(2, 2)

5

(5, 5)

Explanation:

D Given, $m_{1} = m_{2} = m_{3} = m$
$x_{1} = 1, x_{2} = 2$ and $x_{3} = 3$
$y_{1} = 1, y_{2} = 2$ , and $y_{3} = 3$
$x_{cm} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}}$
$x_{cm} = \frac{{mx}_{1} + {mx}_{2} + {mx}_{3}}{m + m + m} = \frac{x_{1} + x_{2} + x_{3}}{3}$
$= \frac{1 + 2 + 3}{3} = \frac{6}{3} = 2$
$y_{cm} = \frac{{my}_{1} + {my}_{2} + {my}_{3}}{m + m + m} = \frac{y_{1} + y_{2} + y_{3}}{3}$
$= \frac{1 + 2 + 3}{3} = \frac{6}{3} = 2$
Hence, the coordinates of the centre of mass are $(2, 2)$ .

BITSAT-2010

Rotational Motion

149711 Three identical spheres, each of mass $3 kg$ are placed touching each other with their centres lying on a straight line. The centres of the sphere are marked at $P, Q$ and $R$ respectively. The distance of centre of mass of system from $P$ is

1

\frac{PQ + QR + PR}{3}

2

\frac{P Q + P R}{3}

3

\frac{PQ + QR + PR}{9}

4

\frac{P Q + P R}{9}

5

\frac{PQ + QR}{3}

Explanation:

B \lt smiles>[R]c1cccc2cccc([R])c12 \lt /smiles>
The centre of mass of the given system consisting of three identical spheres of mass $3 kg$ each is
$x_{c m} = \frac{M_{P} (x_{P}) + M_{Q} (x_{Q}) + M_{R} (x_{R})}{M_{P} + M_{Q} + M_{R}}$
$x_{cm} = \frac{M (0) + M (PQ) + M (PR)}{M + M + M}$
$= \frac{M (PQ + PR)}{3 M}$
$x_{cm} = \frac{PQ + PR}{3}$
Hence, COM of the given system is $\frac{P Q + P R}{3}$

Kerala CEE - 2008

Rotational Motion

149712 The distance between the centers of carbon and oxygen atoms in the carbon monoxide molecule is $1.130 \AA$ . Locate the centre of mass of the molecule relative to the carbon atom :

1

5.428 \AA

2

1.130 \AA

3

0.6457 \AA

4

0.3260 \AA

5 none of these

Explanation:

C $m_{1} = 16$ (Oxygen mass number)
$m_{2} = 12$ (carbon mass number)

$m_{2} x_{2} = m_{1} x_{1}$
$12 x_{2} = 16 (1.130 \times 10^{- 10} - x_{2})$
$12 x_{2} = 16 \times 1.130 \times 10^{- 10} - 16 x_{2}$
$x_{2} = \frac{16 \times 1.13 \times 10^{- 10}}{28}$
$x_{2} = 0.6457 \AA$
Hence, the centre of mass (COM) of molecule relative to the carbon atom is $0.6457^{\circ} A$ .

Kerala CEE 2006

Rotational Motion

149709 In a two-particle system with particle masses $m_{1}$ and $m_{2}$ , the first particle is pushed towards the centre of mass through a distance $d$ , the distance through which second particle must be moved to keep the centre of mass at the same position is

1

\frac{m_{2} d}{m_{1}}

2 d

3

\frac{m_{1} d}{(m_{1} + m_{2})}

4

\frac{(m_{1} + m_{2}) d}{m_{1}}

5

\frac{m_{1} d}{m_{2}}

Explanation:

E
$x_{cm} = \frac{m_{1} (0) + m_{2} (l)}{m_{1} + m_{2}}$
$x_{cm} = \frac{m_{2} l}{m_{1} + m_{2}}$
$x_{cm} = \frac{m_{1} (d) + m_{2} (l - x)}{m_{1} + m_{2}}$
On equating equation (i) \& (ii), we get
$\frac{m_{1} d + m_{2} (l - x)}{m_{1} + m_{2}} = \frac{m_{2} l}{m_{1} + m_{2}}$
$m_{1} d + m_{2} l - m_{2} x = m_{2} l$
$m_{1} d = m_{2} x \Rightarrow x = \frac{m_{1} d}{m_{2}}$

Kerala CEE- 2013

Rotational Motion

149710 A system consists of 3 particles each of mass $m$ located at points $(1, 1) (2, 2)$ and $(3, 3)$ . The coordinates of the centre of mass are

1

(6, 6)

2

(3, 3)

3

(1, 1)

4

(2, 2)

5

(5, 5)

Explanation:

D Given, $m_{1} = m_{2} = m_{3} = m$
$x_{1} = 1, x_{2} = 2$ and $x_{3} = 3$
$y_{1} = 1, y_{2} = 2$ , and $y_{3} = 3$
$x_{cm} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}}$
$x_{cm} = \frac{{mx}_{1} + {mx}_{2} + {mx}_{3}}{m + m + m} = \frac{x_{1} + x_{2} + x_{3}}{3}$
$= \frac{1 + 2 + 3}{3} = \frac{6}{3} = 2$
$y_{cm} = \frac{{my}_{1} + {my}_{2} + {my}_{3}}{m + m + m} = \frac{y_{1} + y_{2} + y_{3}}{3}$
$= \frac{1 + 2 + 3}{3} = \frac{6}{3} = 2$
Hence, the coordinates of the centre of mass are $(2, 2)$ .

BITSAT-2010

Rotational Motion

149711 Three identical spheres, each of mass $3 kg$ are placed touching each other with their centres lying on a straight line. The centres of the sphere are marked at $P, Q$ and $R$ respectively. The distance of centre of mass of system from $P$ is

1

\frac{PQ + QR + PR}{3}

2

\frac{P Q + P R}{3}

3

\frac{PQ + QR + PR}{9}

4

\frac{P Q + P R}{9}

5

\frac{PQ + QR}{3}

Explanation:

B \lt smiles>[R]c1cccc2cccc([R])c12 \lt /smiles>
The centre of mass of the given system consisting of three identical spheres of mass $3 kg$ each is
$x_{c m} = \frac{M_{P} (x_{P}) + M_{Q} (x_{Q}) + M_{R} (x_{R})}{M_{P} + M_{Q} + M_{R}}$
$x_{cm} = \frac{M (0) + M (PQ) + M (PR)}{M + M + M}$
$= \frac{M (PQ + PR)}{3 M}$
$x_{cm} = \frac{PQ + PR}{3}$
Hence, COM of the given system is $\frac{P Q + P R}{3}$

Kerala CEE - 2008

Rotational Motion

149712 The distance between the centers of carbon and oxygen atoms in the carbon monoxide molecule is $1.130 \AA$ . Locate the centre of mass of the molecule relative to the carbon atom :

1

5.428 \AA

2

1.130 \AA

3

0.6457 \AA

4

0.3260 \AA

5 none of these

Explanation:

C $m_{1} = 16$ (Oxygen mass number)
$m_{2} = 12$ (carbon mass number)

$m_{2} x_{2} = m_{1} x_{1}$
$12 x_{2} = 16 (1.130 \times 10^{- 10} - x_{2})$
$12 x_{2} = 16 \times 1.130 \times 10^{- 10} - 16 x_{2}$
$x_{2} = \frac{16 \times 1.13 \times 10^{- 10}}{28}$
$x_{2} = 0.6457 \AA$
Hence, the centre of mass (COM) of molecule relative to the carbon atom is $0.6457^{\circ} A$ .

Kerala CEE 2006