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PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366015 A ring A is initially rolling without sliding with a velocity $v$ on the horizontal surface of the body B (of same mass as A). All surfaces are smooth. $B$ has no initial velocity. What will be the maximum height reached by A on $B$ ?
supporting img

1

\frac{v^{2}}{3 g}

2

\frac{3 v^{2}}{4 g}

3

\frac{v^{2}}{4 g}

4

\frac{v^{2}}{2 g}

Explanation:

When the ring is at the maximum height, the wedge and the ring have the same horizontal component of velocity. As all the surfaces are smooth, angular velocity of the ring remains constant.
From conservation of mechanical energy, we get
$\frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2} = \frac{1}{2} m v^{' 2} + \frac{1}{2} I ω^{2} + \frac{1}{2} m v^{' 2} + m g h$
where $v^{'}$ is final common velocity
From conservation of momentum
$v^{'} = \frac{v}{2} \Rightarrow h = \frac{v^{2}}{4 g}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366016 The speed of a homogeneous solid sphere after rolling down an inclined plane of vertical height $h$ , from rest without sliding, is

1

\sqrt{\frac{4}{3} g h}

2

\sqrt{\frac{6}{5} g h}

3

\sqrt{g h}

4

\sqrt{\frac{10}{7} g h}

Explanation:

$v = \sqrt{\frac{2 g h}{1 + \frac{K^{2}}{r^{2}}}} = \sqrt{\frac{2 g h}{1 + \frac{2}{5}}} = \sqrt{\frac{10}{7} g h}$ .

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366017 Assertion :
The total kinetic energy of a rolling solid sphere is the sum of translational and rotational kinetic energies.
Reason :
For all solid bodies, total kinetic energy is always twice of translational kinetic energy.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Both Assertion and reason are incorrect.

Explanation:

The kinetic energy of a rolling solid sphere
$= {(E_{k})}_{trans} + {(E_{k})}_{rot}$
$= \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$
$= \frac{1}{2} m v^{2} + \frac{1}{2} \times \frac{2}{5} m R^{2} ω^{2} [∵ ω = \frac{v}{R}]$
$= \frac{1}{2} m v^{2} + \frac{1}{5} m v^{2} = 7 / 10 m v^{2}$
Hence, total kinetic energy is sum of translation and rotational kinetic energies.
For different solid bodies, total kinetic energy is not always twice of translation kinetic energy. Hence, Assertion is correct but Reason is incorrect.

AIIMS - 2017

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366018 A solid sphere of mass $1 k g$ , radius $10 c m$ rolls down an inclined plane of height $7 m$ . The velocity of its centre as it reaches the ground level is

1

7 m / s

2

10 m / s

3

15 m / s

4

20 m / s

Explanation:

According to the law of conservation of energy, we get
$\frac{1}{2} m v^{2} (1 + \frac{k^{2}}{R^{2}}) = m g h$
where $h$ is the height of the inclined plane
$∴ v = \sqrt{\frac{2 g h}{1 + \frac{k^{2}}{R^{2}}}}$
For a solid sphere $\frac{k^{2}}{R^{2}} = \frac{2}{5}$
Substituting the given values, we get
$v = \sqrt{\frac{2 \times 10 \times 7}{(1 + \frac{2}{5})}} = \sqrt{\frac{2 \times 10 \times 7 \times 5}{7}}$
$= 10 m s^{- 1}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366015 A ring A is initially rolling without sliding with a velocity $v$ on the horizontal surface of the body B (of same mass as A). All surfaces are smooth. $B$ has no initial velocity. What will be the maximum height reached by A on $B$ ?
supporting img

1

\frac{v^{2}}{3 g}

2

\frac{3 v^{2}}{4 g}

3

\frac{v^{2}}{4 g}

4

\frac{v^{2}}{2 g}

Explanation:

When the ring is at the maximum height, the wedge and the ring have the same horizontal component of velocity. As all the surfaces are smooth, angular velocity of the ring remains constant.
From conservation of mechanical energy, we get
$\frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2} = \frac{1}{2} m v^{' 2} + \frac{1}{2} I ω^{2} + \frac{1}{2} m v^{' 2} + m g h$
where $v^{'}$ is final common velocity
From conservation of momentum
$v^{'} = \frac{v}{2} \Rightarrow h = \frac{v^{2}}{4 g}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366016 The speed of a homogeneous solid sphere after rolling down an inclined plane of vertical height $h$ , from rest without sliding, is

1

\sqrt{\frac{4}{3} g h}

2

\sqrt{\frac{6}{5} g h}

3

\sqrt{g h}

4

\sqrt{\frac{10}{7} g h}

Explanation:

$v = \sqrt{\frac{2 g h}{1 + \frac{K^{2}}{r^{2}}}} = \sqrt{\frac{2 g h}{1 + \frac{2}{5}}} = \sqrt{\frac{10}{7} g h}$ .

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366017 Assertion :
The total kinetic energy of a rolling solid sphere is the sum of translational and rotational kinetic energies.
Reason :
For all solid bodies, total kinetic energy is always twice of translational kinetic energy.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Both Assertion and reason are incorrect.

Explanation:

The kinetic energy of a rolling solid sphere
$= {(E_{k})}_{trans} + {(E_{k})}_{rot}$
$= \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$
$= \frac{1}{2} m v^{2} + \frac{1}{2} \times \frac{2}{5} m R^{2} ω^{2} [∵ ω = \frac{v}{R}]$
$= \frac{1}{2} m v^{2} + \frac{1}{5} m v^{2} = 7 / 10 m v^{2}$
Hence, total kinetic energy is sum of translation and rotational kinetic energies.
For different solid bodies, total kinetic energy is not always twice of translation kinetic energy. Hence, Assertion is correct but Reason is incorrect.

AIIMS - 2017

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366018 A solid sphere of mass $1 k g$ , radius $10 c m$ rolls down an inclined plane of height $7 m$ . The velocity of its centre as it reaches the ground level is

1

7 m / s

2

10 m / s

3

15 m / s

4

20 m / s

Explanation:

According to the law of conservation of energy, we get
$\frac{1}{2} m v^{2} (1 + \frac{k^{2}}{R^{2}}) = m g h$
where $h$ is the height of the inclined plane
$∴ v = \sqrt{\frac{2 g h}{1 + \frac{k^{2}}{R^{2}}}}$
For a solid sphere $\frac{k^{2}}{R^{2}} = \frac{2}{5}$
Substituting the given values, we get
$v = \sqrt{\frac{2 \times 10 \times 7}{(1 + \frac{2}{5})}} = \sqrt{\frac{2 \times 10 \times 7 \times 5}{7}}$
$= 10 m s^{- 1}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366015 A ring A is initially rolling without sliding with a velocity $v$ on the horizontal surface of the body B (of same mass as A). All surfaces are smooth. $B$ has no initial velocity. What will be the maximum height reached by A on $B$ ?
supporting img

1

\frac{v^{2}}{3 g}

2

\frac{3 v^{2}}{4 g}

3

\frac{v^{2}}{4 g}

4

\frac{v^{2}}{2 g}

Explanation:

When the ring is at the maximum height, the wedge and the ring have the same horizontal component of velocity. As all the surfaces are smooth, angular velocity of the ring remains constant.
From conservation of mechanical energy, we get
$\frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2} = \frac{1}{2} m v^{' 2} + \frac{1}{2} I ω^{2} + \frac{1}{2} m v^{' 2} + m g h$
where $v^{'}$ is final common velocity
From conservation of momentum
$v^{'} = \frac{v}{2} \Rightarrow h = \frac{v^{2}}{4 g}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366016 The speed of a homogeneous solid sphere after rolling down an inclined plane of vertical height $h$ , from rest without sliding, is

1

\sqrt{\frac{4}{3} g h}

2

\sqrt{\frac{6}{5} g h}

3

\sqrt{g h}

4

\sqrt{\frac{10}{7} g h}

Explanation:

$v = \sqrt{\frac{2 g h}{1 + \frac{K^{2}}{r^{2}}}} = \sqrt{\frac{2 g h}{1 + \frac{2}{5}}} = \sqrt{\frac{10}{7} g h}$ .

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366017 Assertion :
The total kinetic energy of a rolling solid sphere is the sum of translational and rotational kinetic energies.
Reason :
For all solid bodies, total kinetic energy is always twice of translational kinetic energy.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Both Assertion and reason are incorrect.

Explanation:

The kinetic energy of a rolling solid sphere
$= {(E_{k})}_{trans} + {(E_{k})}_{rot}$
$= \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$
$= \frac{1}{2} m v^{2} + \frac{1}{2} \times \frac{2}{5} m R^{2} ω^{2} [∵ ω = \frac{v}{R}]$
$= \frac{1}{2} m v^{2} + \frac{1}{5} m v^{2} = 7 / 10 m v^{2}$
Hence, total kinetic energy is sum of translation and rotational kinetic energies.
For different solid bodies, total kinetic energy is not always twice of translation kinetic energy. Hence, Assertion is correct but Reason is incorrect.

AIIMS - 2017

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366018 A solid sphere of mass $1 k g$ , radius $10 c m$ rolls down an inclined plane of height $7 m$ . The velocity of its centre as it reaches the ground level is

1

7 m / s

2

10 m / s

3

15 m / s

4

20 m / s

Explanation:

According to the law of conservation of energy, we get
$\frac{1}{2} m v^{2} (1 + \frac{k^{2}}{R^{2}}) = m g h$
where $h$ is the height of the inclined plane
$∴ v = \sqrt{\frac{2 g h}{1 + \frac{k^{2}}{R^{2}}}}$
For a solid sphere $\frac{k^{2}}{R^{2}} = \frac{2}{5}$
Substituting the given values, we get
$v = \sqrt{\frac{2 \times 10 \times 7}{(1 + \frac{2}{5})}} = \sqrt{\frac{2 \times 10 \times 7 \times 5}{7}}$
$= 10 m s^{- 1}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366015 A ring A is initially rolling without sliding with a velocity $v$ on the horizontal surface of the body B (of same mass as A). All surfaces are smooth. $B$ has no initial velocity. What will be the maximum height reached by A on $B$ ?
supporting img

1

\frac{v^{2}}{3 g}

2

\frac{3 v^{2}}{4 g}

3

\frac{v^{2}}{4 g}

4

\frac{v^{2}}{2 g}

Explanation:

When the ring is at the maximum height, the wedge and the ring have the same horizontal component of velocity. As all the surfaces are smooth, angular velocity of the ring remains constant.
From conservation of mechanical energy, we get
$\frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2} = \frac{1}{2} m v^{' 2} + \frac{1}{2} I ω^{2} + \frac{1}{2} m v^{' 2} + m g h$
where $v^{'}$ is final common velocity
From conservation of momentum
$v^{'} = \frac{v}{2} \Rightarrow h = \frac{v^{2}}{4 g}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366016 The speed of a homogeneous solid sphere after rolling down an inclined plane of vertical height $h$ , from rest without sliding, is

1

\sqrt{\frac{4}{3} g h}

2

\sqrt{\frac{6}{5} g h}

3

\sqrt{g h}

4

\sqrt{\frac{10}{7} g h}

Explanation:

$v = \sqrt{\frac{2 g h}{1 + \frac{K^{2}}{r^{2}}}} = \sqrt{\frac{2 g h}{1 + \frac{2}{5}}} = \sqrt{\frac{10}{7} g h}$ .

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366017 Assertion :
The total kinetic energy of a rolling solid sphere is the sum of translational and rotational kinetic energies.
Reason :
For all solid bodies, total kinetic energy is always twice of translational kinetic energy.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Both Assertion and reason are incorrect.

Explanation:

The kinetic energy of a rolling solid sphere
$= {(E_{k})}_{trans} + {(E_{k})}_{rot}$
$= \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$
$= \frac{1}{2} m v^{2} + \frac{1}{2} \times \frac{2}{5} m R^{2} ω^{2} [∵ ω = \frac{v}{R}]$
$= \frac{1}{2} m v^{2} + \frac{1}{5} m v^{2} = 7 / 10 m v^{2}$
Hence, total kinetic energy is sum of translation and rotational kinetic energies.
For different solid bodies, total kinetic energy is not always twice of translation kinetic energy. Hence, Assertion is correct but Reason is incorrect.

AIIMS - 2017

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366018 A solid sphere of mass $1 k g$ , radius $10 c m$ rolls down an inclined plane of height $7 m$ . The velocity of its centre as it reaches the ground level is

1

7 m / s

2

10 m / s

3

15 m / s

4

20 m / s

Explanation:

According to the law of conservation of energy, we get
$\frac{1}{2} m v^{2} (1 + \frac{k^{2}}{R^{2}}) = m g h$
where $h$ is the height of the inclined plane
$∴ v = \sqrt{\frac{2 g h}{1 + \frac{k^{2}}{R^{2}}}}$
For a solid sphere $\frac{k^{2}}{R^{2}} = \frac{2}{5}$
Substituting the given values, we get
$v = \sqrt{\frac{2 \times 10 \times 7}{(1 + \frac{2}{5})}} = \sqrt{\frac{2 \times 10 \times 7 \times 5}{7}}$
$= 10 m s^{- 1}$