QBManager

PHXI05:LAWS OF MOTION

363493 Assertion :
Even though net external force on a body is zero, momentum need not be conserved.
Reason :
The internal interaction between particles of a body cancels out momentum of each other.

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:

Since ${\vec{F}}_{e x t} = 0 \Rightarrow \frac{d \vec{p}}{d t} = 0$
$\Rightarrow$ momentum gets conserved.
Also the internal interactions involve internal forces. Internal forces cancel each other not the linear momenta.

AIIMS - 2019

PHXI05:LAWS OF MOTION

363494 Statement A :
A body is momentarily at rest but still some force is acting on it at that time.
Statement B :
When a force acts on a body, it may not have some acceleration.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both Statements are correct.

4 Both Statements are incorrect.

Explanation:

A stationary body $(v = 0)$ may still have some acceleration, e.g. when a body is thrown in upward direction, it comes to rest at highest position, but at that time, it still have acceleration equal to acceleration due to gravity $g$ . Hence, gravitational force is acting at highest position and when a force acts on a body, then its accelerates. Therefore, Statement A is correct but Statement B is incorrect.

PHXI05:LAWS OF MOTION

363495 A gas molecule of mass $m$ strikes the wall of the container with a speed $v$ at an angle $θ$ with the normal to the wall at the point of collision. The impluse of the gas molecule has a magnitude

1

2 m V \cos θ

2

3 m V

3

Z e r o

4

m V

Explanation:

From adjoining figure the component of momentum along $x$ -axis (parallel to the wall of container) remains unchanged even after the collision.
$∴$ Impluse $=$ change in momentum of gas molecule along $y$ -axis,
$= 2 m v \cos θ$
supporting img

PHXI05:LAWS OF MOTION

363496 A body of mass $4 k g$ starts from rest, and a force is acting on the body, as shown in the figure. What is the speed of the body after $4 s$ ?
supporting img

1

15 m / s

2

20 m / s

3

5 m / s

4

25 m / s

Explanation:

Area under graph,
$\begin{aligned} Δ p & = \frac{1}{2} \times F \times t \\ = \frac{1}{2} \times 4 \times 10 = 20 k g m s^{- 1} \\ Δ p & = (m v - m u) \Rightarrow 20 = m (v - u) \\ 20 & = 4 (v - 0) v = 5 {m s}^{- 1} \end{aligned}$

PHXI05:LAWS OF MOTION

363493 Assertion :
Even though net external force on a body is zero, momentum need not be conserved.
Reason :
The internal interaction between particles of a body cancels out momentum of each other.

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:

Since ${\vec{F}}_{e x t} = 0 \Rightarrow \frac{d \vec{p}}{d t} = 0$
$\Rightarrow$ momentum gets conserved.
Also the internal interactions involve internal forces. Internal forces cancel each other not the linear momenta.

AIIMS - 2019

PHXI05:LAWS OF MOTION

363494 Statement A :
A body is momentarily at rest but still some force is acting on it at that time.
Statement B :
When a force acts on a body, it may not have some acceleration.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both Statements are correct.

4 Both Statements are incorrect.

Explanation:

A stationary body $(v = 0)$ may still have some acceleration, e.g. when a body is thrown in upward direction, it comes to rest at highest position, but at that time, it still have acceleration equal to acceleration due to gravity $g$ . Hence, gravitational force is acting at highest position and when a force acts on a body, then its accelerates. Therefore, Statement A is correct but Statement B is incorrect.

PHXI05:LAWS OF MOTION

363495 A gas molecule of mass $m$ strikes the wall of the container with a speed $v$ at an angle $θ$ with the normal to the wall at the point of collision. The impluse of the gas molecule has a magnitude

1

2 m V \cos θ

2

3 m V

3

Z e r o

4

m V

Explanation:

From adjoining figure the component of momentum along $x$ -axis (parallel to the wall of container) remains unchanged even after the collision.
$∴$ Impluse $=$ change in momentum of gas molecule along $y$ -axis,
$= 2 m v \cos θ$
supporting img

PHXI05:LAWS OF MOTION

363496 A body of mass $4 k g$ starts from rest, and a force is acting on the body, as shown in the figure. What is the speed of the body after $4 s$ ?
supporting img

1

15 m / s

2

20 m / s

3

5 m / s

4

25 m / s

Explanation:

Area under graph,
$\begin{aligned} Δ p & = \frac{1}{2} \times F \times t \\ = \frac{1}{2} \times 4 \times 10 = 20 k g m s^{- 1} \\ Δ p & = (m v - m u) \Rightarrow 20 = m (v - u) \\ 20 & = 4 (v - 0) v = 5 {m s}^{- 1} \end{aligned}$

PHXI05:LAWS OF MOTION

363493 Assertion :
Even though net external force on a body is zero, momentum need not be conserved.
Reason :
The internal interaction between particles of a body cancels out momentum of each other.

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:

Since ${\vec{F}}_{e x t} = 0 \Rightarrow \frac{d \vec{p}}{d t} = 0$
$\Rightarrow$ momentum gets conserved.
Also the internal interactions involve internal forces. Internal forces cancel each other not the linear momenta.

AIIMS - 2019

PHXI05:LAWS OF MOTION

363494 Statement A :
A body is momentarily at rest but still some force is acting on it at that time.
Statement B :
When a force acts on a body, it may not have some acceleration.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both Statements are correct.

4 Both Statements are incorrect.

Explanation:

A stationary body $(v = 0)$ may still have some acceleration, e.g. when a body is thrown in upward direction, it comes to rest at highest position, but at that time, it still have acceleration equal to acceleration due to gravity $g$ . Hence, gravitational force is acting at highest position and when a force acts on a body, then its accelerates. Therefore, Statement A is correct but Statement B is incorrect.

PHXI05:LAWS OF MOTION

363495 A gas molecule of mass $m$ strikes the wall of the container with a speed $v$ at an angle $θ$ with the normal to the wall at the point of collision. The impluse of the gas molecule has a magnitude

1

2 m V \cos θ

2

3 m V

3

Z e r o

4

m V

Explanation:

From adjoining figure the component of momentum along $x$ -axis (parallel to the wall of container) remains unchanged even after the collision.
$∴$ Impluse $=$ change in momentum of gas molecule along $y$ -axis,
$= 2 m v \cos θ$
supporting img

PHXI05:LAWS OF MOTION

363496 A body of mass $4 k g$ starts from rest, and a force is acting on the body, as shown in the figure. What is the speed of the body after $4 s$ ?
supporting img

1

15 m / s

2

20 m / s

3

5 m / s

4

25 m / s

Explanation:

Area under graph,
$\begin{aligned} Δ p & = \frac{1}{2} \times F \times t \\ = \frac{1}{2} \times 4 \times 10 = 20 k g m s^{- 1} \\ Δ p & = (m v - m u) \Rightarrow 20 = m (v - u) \\ 20 & = 4 (v - 0) v = 5 {m s}^{- 1} \end{aligned}$

PHXI05:LAWS OF MOTION

363493 Assertion :
Even though net external force on a body is zero, momentum need not be conserved.
Reason :
The internal interaction between particles of a body cancels out momentum of each other.

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:

Since ${\vec{F}}_{e x t} = 0 \Rightarrow \frac{d \vec{p}}{d t} = 0$
$\Rightarrow$ momentum gets conserved.
Also the internal interactions involve internal forces. Internal forces cancel each other not the linear momenta.

AIIMS - 2019

PHXI05:LAWS OF MOTION

363494 Statement A :
A body is momentarily at rest but still some force is acting on it at that time.
Statement B :
When a force acts on a body, it may not have some acceleration.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both Statements are correct.

4 Both Statements are incorrect.

Explanation:

A stationary body $(v = 0)$ may still have some acceleration, e.g. when a body is thrown in upward direction, it comes to rest at highest position, but at that time, it still have acceleration equal to acceleration due to gravity $g$ . Hence, gravitational force is acting at highest position and when a force acts on a body, then its accelerates. Therefore, Statement A is correct but Statement B is incorrect.

PHXI05:LAWS OF MOTION

363495 A gas molecule of mass $m$ strikes the wall of the container with a speed $v$ at an angle $θ$ with the normal to the wall at the point of collision. The impluse of the gas molecule has a magnitude

1

2 m V \cos θ

2

3 m V

3

Z e r o

4

m V

Explanation:

From adjoining figure the component of momentum along $x$ -axis (parallel to the wall of container) remains unchanged even after the collision.
$∴$ Impluse $=$ change in momentum of gas molecule along $y$ -axis,
$= 2 m v \cos θ$
supporting img

PHXI05:LAWS OF MOTION

363496 A body of mass $4 k g$ starts from rest, and a force is acting on the body, as shown in the figure. What is the speed of the body after $4 s$ ?
supporting img

1

15 m / s

2

20 m / s

3

5 m / s

4

25 m / s

Explanation:

Area under graph,
$\begin{aligned} Δ p & = \frac{1}{2} \times F \times t \\ = \frac{1}{2} \times 4 \times 10 = 20 k g m s^{- 1} \\ Δ p & = (m v - m u) \Rightarrow 20 = m (v - u) \\ 20 & = 4 (v - 0) v = 5 {m s}^{- 1} \end{aligned}$

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: