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Rotational Motion

149749 Two particles of masses $m_{1}$ and $m_{2} (m_{1} > m_{2})$ , initially at rest, move towards each other under an inverse square law force of attraction. Pick out the correct statement about the centre of mass (CM) of the system

1 The CM moves towards

m_{1}

.

2 The CM moves towards

m_{2}

.

3 The CM remains at rest.

4 The motion of CM is accelerated.

Explanation:

C The correct option is the CM remains at rest

Now,
$\frac{r_{1}}{r_{2}} = \frac{m_{2}}{m_{1}}$
At the $m_{1}$ is near the centre of mass. As the two masses are attracted, $m_{2}$ is more accelerated then $m_{1}$ and at energy instant to the centre of mass remains at the same place, because the system is isolated.

AMU-2008

Rotational Motion

149755 A car of mass $M$ is tied to one end of a massless rope of length $10 m$ . The other end of the rope is in the hands of a man of mass $M$ . The entire system is on a smooth horizontal surface. The man is at $x = 0$ and the cart at $x = 10 m$ . If the man pulls the cart by the rope, the man and the cart will meet at the point

1 they will never meet

2

x = 10 m

3

x = 5 m

4

x = 0

Explanation:

C Since, there is no external force applied in on the system so the centre of mass will alwavs be at rest.

$x_{COM} = \frac{M_{man} \times 0 + M_{car} \times 10}{M_{man} + M_{car}}$
$x_{COM} = \frac{0 + 10 m}{2 m}$
$x_{COM} = 5$

AIPMT- 1997

Rotational Motion

149756 Three identical metal balls each of radius $r$ are placed touching each other on a horizontal surface such that an equilateral triangle is formed with centers of three balls joined. The centre of mass of the system is located at

1 horizontal surface

2 centre of one of the balls

3 line joining the centers of any two balls

4 point of intersection of the medians

Explanation:

D To the given data,
Three identical metal balls each of radius $r$ touching each other so the equilateral triangle therefore, centre of mass of the system at the which is the (point of intersection of the medians) of the triangle.

AIPMT-1999

Rotational Motion

149758 A solid sphere of radius $R$ is placed on a smooth horizontal surface. A horizontal force $F$ is applied at height $h$ from the lowest point. For the maximum acceleration of the centre of mass

1

h = R

2

h = 2 R

3

h = 0

4 the acceleration will be same and there is no relation between

h

and

R

Explanation:

D According to question surface is smooth so rolling motion is not possible. sphere will perform linear motion.
Hence,
The acceleration of the centre of the mass is $a = \frac{F}{m}$ , where $F$ is applied force.
Hence, the acceleration will be same and there is no relation between $h$ and $R$

AIPMT-2002

Rotational Motion

269276 When a force is applied on a body, Newton's second law is applicable to

1 center of mass

2 any part of the body

3 upper most part of body

4 lower most part of body

Rotational Motion

149749 Two particles of masses $m_{1}$ and $m_{2} (m_{1} > m_{2})$ , initially at rest, move towards each other under an inverse square law force of attraction. Pick out the correct statement about the centre of mass (CM) of the system

1 The CM moves towards

m_{1}

.

2 The CM moves towards

m_{2}

.

3 The CM remains at rest.

4 The motion of CM is accelerated.

Explanation:

C The correct option is the CM remains at rest

Now,
$\frac{r_{1}}{r_{2}} = \frac{m_{2}}{m_{1}}$
At the $m_{1}$ is near the centre of mass. As the two masses are attracted, $m_{2}$ is more accelerated then $m_{1}$ and at energy instant to the centre of mass remains at the same place, because the system is isolated.

AMU-2008

Rotational Motion

149755 A car of mass $M$ is tied to one end of a massless rope of length $10 m$ . The other end of the rope is in the hands of a man of mass $M$ . The entire system is on a smooth horizontal surface. The man is at $x = 0$ and the cart at $x = 10 m$ . If the man pulls the cart by the rope, the man and the cart will meet at the point

1 they will never meet

2

x = 10 m

3

x = 5 m

4

x = 0

Explanation:

C Since, there is no external force applied in on the system so the centre of mass will alwavs be at rest.

$x_{COM} = \frac{M_{man} \times 0 + M_{car} \times 10}{M_{man} + M_{car}}$
$x_{COM} = \frac{0 + 10 m}{2 m}$
$x_{COM} = 5$

AIPMT- 1997

Rotational Motion

149756 Three identical metal balls each of radius $r$ are placed touching each other on a horizontal surface such that an equilateral triangle is formed with centers of three balls joined. The centre of mass of the system is located at

1 horizontal surface

2 centre of one of the balls

3 line joining the centers of any two balls

4 point of intersection of the medians

Explanation:

D To the given data,
Three identical metal balls each of radius $r$ touching each other so the equilateral triangle therefore, centre of mass of the system at the which is the (point of intersection of the medians) of the triangle.

AIPMT-1999

Rotational Motion

149758 A solid sphere of radius $R$ is placed on a smooth horizontal surface. A horizontal force $F$ is applied at height $h$ from the lowest point. For the maximum acceleration of the centre of mass

1

h = R

2

h = 2 R

3

h = 0

4 the acceleration will be same and there is no relation between

h

and

R

Explanation:

D According to question surface is smooth so rolling motion is not possible. sphere will perform linear motion.
Hence,
The acceleration of the centre of the mass is $a = \frac{F}{m}$ , where $F$ is applied force.
Hence, the acceleration will be same and there is no relation between $h$ and $R$

AIPMT-2002

Rotational Motion

269276 When a force is applied on a body, Newton's second law is applicable to

1 center of mass

2 any part of the body

3 upper most part of body

4 lower most part of body

Rotational Motion

149749 Two particles of masses $m_{1}$ and $m_{2} (m_{1} > m_{2})$ , initially at rest, move towards each other under an inverse square law force of attraction. Pick out the correct statement about the centre of mass (CM) of the system

1 The CM moves towards

m_{1}

.

2 The CM moves towards

m_{2}

.

3 The CM remains at rest.

4 The motion of CM is accelerated.

Explanation:

C The correct option is the CM remains at rest

Now,
$\frac{r_{1}}{r_{2}} = \frac{m_{2}}{m_{1}}$
At the $m_{1}$ is near the centre of mass. As the two masses are attracted, $m_{2}$ is more accelerated then $m_{1}$ and at energy instant to the centre of mass remains at the same place, because the system is isolated.

AMU-2008

Rotational Motion

149755 A car of mass $M$ is tied to one end of a massless rope of length $10 m$ . The other end of the rope is in the hands of a man of mass $M$ . The entire system is on a smooth horizontal surface. The man is at $x = 0$ and the cart at $x = 10 m$ . If the man pulls the cart by the rope, the man and the cart will meet at the point

1 they will never meet

2

x = 10 m

3

x = 5 m

4

x = 0

Explanation:

C Since, there is no external force applied in on the system so the centre of mass will alwavs be at rest.

$x_{COM} = \frac{M_{man} \times 0 + M_{car} \times 10}{M_{man} + M_{car}}$
$x_{COM} = \frac{0 + 10 m}{2 m}$
$x_{COM} = 5$

AIPMT- 1997

Rotational Motion

149756 Three identical metal balls each of radius $r$ are placed touching each other on a horizontal surface such that an equilateral triangle is formed with centers of three balls joined. The centre of mass of the system is located at

1 horizontal surface

2 centre of one of the balls

3 line joining the centers of any two balls

4 point of intersection of the medians

Explanation:

D To the given data,
Three identical metal balls each of radius $r$ touching each other so the equilateral triangle therefore, centre of mass of the system at the which is the (point of intersection of the medians) of the triangle.

AIPMT-1999

Rotational Motion

149758 A solid sphere of radius $R$ is placed on a smooth horizontal surface. A horizontal force $F$ is applied at height $h$ from the lowest point. For the maximum acceleration of the centre of mass

1

h = R

2

h = 2 R

3

h = 0

4 the acceleration will be same and there is no relation between

h

and

R

Explanation:

D According to question surface is smooth so rolling motion is not possible. sphere will perform linear motion.
Hence,
The acceleration of the centre of the mass is $a = \frac{F}{m}$ , where $F$ is applied force.
Hence, the acceleration will be same and there is no relation between $h$ and $R$

AIPMT-2002

Rotational Motion

269276 When a force is applied on a body, Newton's second law is applicable to

1 center of mass

2 any part of the body

3 upper most part of body

4 lower most part of body

Rotational Motion

149749 Two particles of masses $m_{1}$ and $m_{2} (m_{1} > m_{2})$ , initially at rest, move towards each other under an inverse square law force of attraction. Pick out the correct statement about the centre of mass (CM) of the system

1 The CM moves towards

m_{1}

.

2 The CM moves towards

m_{2}

.

3 The CM remains at rest.

4 The motion of CM is accelerated.

Explanation:

C The correct option is the CM remains at rest

Now,
$\frac{r_{1}}{r_{2}} = \frac{m_{2}}{m_{1}}$
At the $m_{1}$ is near the centre of mass. As the two masses are attracted, $m_{2}$ is more accelerated then $m_{1}$ and at energy instant to the centre of mass remains at the same place, because the system is isolated.

AMU-2008

Rotational Motion

149755 A car of mass $M$ is tied to one end of a massless rope of length $10 m$ . The other end of the rope is in the hands of a man of mass $M$ . The entire system is on a smooth horizontal surface. The man is at $x = 0$ and the cart at $x = 10 m$ . If the man pulls the cart by the rope, the man and the cart will meet at the point

1 they will never meet

2

x = 10 m

3

x = 5 m

4

x = 0

Explanation:

C Since, there is no external force applied in on the system so the centre of mass will alwavs be at rest.

$x_{COM} = \frac{M_{man} \times 0 + M_{car} \times 10}{M_{man} + M_{car}}$
$x_{COM} = \frac{0 + 10 m}{2 m}$
$x_{COM} = 5$

AIPMT- 1997

Rotational Motion

149756 Three identical metal balls each of radius $r$ are placed touching each other on a horizontal surface such that an equilateral triangle is formed with centers of three balls joined. The centre of mass of the system is located at

1 horizontal surface

2 centre of one of the balls

3 line joining the centers of any two balls

4 point of intersection of the medians

Explanation:

D To the given data,
Three identical metal balls each of radius $r$ touching each other so the equilateral triangle therefore, centre of mass of the system at the which is the (point of intersection of the medians) of the triangle.

AIPMT-1999

Rotational Motion

149758 A solid sphere of radius $R$ is placed on a smooth horizontal surface. A horizontal force $F$ is applied at height $h$ from the lowest point. For the maximum acceleration of the centre of mass

1

h = R

2

h = 2 R

3

h = 0

4 the acceleration will be same and there is no relation between

h

and

R

Explanation:

D According to question surface is smooth so rolling motion is not possible. sphere will perform linear motion.
Hence,
The acceleration of the centre of the mass is $a = \frac{F}{m}$ , where $F$ is applied force.
Hence, the acceleration will be same and there is no relation between $h$ and $R$

AIPMT-2002

Rotational Motion

269276 When a force is applied on a body, Newton's second law is applicable to

1 center of mass

2 any part of the body

3 upper most part of body

4 lower most part of body

Rotational Motion

149749 Two particles of masses $m_{1}$ and $m_{2} (m_{1} > m_{2})$ , initially at rest, move towards each other under an inverse square law force of attraction. Pick out the correct statement about the centre of mass (CM) of the system

1 The CM moves towards

m_{1}

.

2 The CM moves towards

m_{2}

.

3 The CM remains at rest.

4 The motion of CM is accelerated.

Explanation:

C The correct option is the CM remains at rest

Now,
$\frac{r_{1}}{r_{2}} = \frac{m_{2}}{m_{1}}$
At the $m_{1}$ is near the centre of mass. As the two masses are attracted, $m_{2}$ is more accelerated then $m_{1}$ and at energy instant to the centre of mass remains at the same place, because the system is isolated.

AMU-2008

Rotational Motion

149755 A car of mass $M$ is tied to one end of a massless rope of length $10 m$ . The other end of the rope is in the hands of a man of mass $M$ . The entire system is on a smooth horizontal surface. The man is at $x = 0$ and the cart at $x = 10 m$ . If the man pulls the cart by the rope, the man and the cart will meet at the point

1 they will never meet

2

x = 10 m

3

x = 5 m

4

x = 0

Explanation:

C Since, there is no external force applied in on the system so the centre of mass will alwavs be at rest.

$x_{COM} = \frac{M_{man} \times 0 + M_{car} \times 10}{M_{man} + M_{car}}$
$x_{COM} = \frac{0 + 10 m}{2 m}$
$x_{COM} = 5$

AIPMT- 1997

Rotational Motion

149756 Three identical metal balls each of radius $r$ are placed touching each other on a horizontal surface such that an equilateral triangle is formed with centers of three balls joined. The centre of mass of the system is located at

1 horizontal surface

2 centre of one of the balls

3 line joining the centers of any two balls

4 point of intersection of the medians

Explanation:

D To the given data,
Three identical metal balls each of radius $r$ touching each other so the equilateral triangle therefore, centre of mass of the system at the which is the (point of intersection of the medians) of the triangle.

AIPMT-1999

Rotational Motion

149758 A solid sphere of radius $R$ is placed on a smooth horizontal surface. A horizontal force $F$ is applied at height $h$ from the lowest point. For the maximum acceleration of the centre of mass

1

h = R

2

h = 2 R

3

h = 0

4 the acceleration will be same and there is no relation between

h

and

R

Explanation:

D According to question surface is smooth so rolling motion is not possible. sphere will perform linear motion.
Hence,
The acceleration of the centre of the mass is $a = \frac{F}{m}$ , where $F$ is applied force.
Hence, the acceleration will be same and there is no relation between $h$ and $R$

AIPMT-2002

Rotational Motion

269276 When a force is applied on a body, Newton's second law is applicable to

1 center of mass

2 any part of the body

3 upper most part of body

4 lower most part of body