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PHXI05:LAWS OF MOTION

363545 A force $F_{1}$ of $500 N$ is required to push a car of mass $1000 k g$ slowly at constant speed on a levelled road. If a force $F_{2}$ of $1000 N$ is applied, then the acceleration of the car will be

1 zero

2

1.5 m s^{- 2}

3

1 m s^{- 2}

4

0.5 m s^{- 2}

Explanation:

We know that a force of $500 N$ does not
produce any acceleration as it pushes the car
slowly at constant speed. Out of $1000 N$ force,
only $500 N$ produces
acceleration, thus acceleration produced,
$a = \frac{force}{mass} = \frac{500}{1000} = 0.5 m s^{- 2}$

PHXI05:LAWS OF MOTION

363546 A body of mass $2 k g$ has an initial velocity of $3 m s^{- 1}$ along $O E$ and it is subjected to a force of $4 N$ in a direction perpendicular to $O E$ . The distance of body from $O$ after $4 s$ will be.
supporting img

1

12 m

2

20 m

3

8 m

4

48 m

Explanation:

The acceleration of the body perpendicular
to $O E$ is a
supporting img

$a = \frac{F}{m} = \frac{4}{2} = 2 m s^{- 2}$
Displacement along $O E$ ,
$s_{1} = v t = 3 \times 4 = 12 m$
Displacement perpendicular to $O E$ , $i . e .$
along $O A$
$s_{2} = \frac{1}{2} a t^{2} = \frac{1}{2} \times 2 \times (4)^{2} = 16 m$
The resultant displacement
$s = \sqrt{s_{1}^{2} + s_{2}^{2}}$
$= \sqrt{144 + 256} = 20 m$

PHXI05:LAWS OF MOTION

363547 The position vector of a particle related to time $t$ is given by $\vec{r} = (10 t \hat{i} + 15 t^{2} \hat{j} + 7 \hat{k}) m$ .The direction of net force experienced by the particle is

1 positive

y

-axis

2 positive

x

-axis

3 positive

z

-axis

4 in

x - y

plane.

Explanation:

Velocity of the particle,
$\vec{v} = \frac{d \vec{r}}{d t} = \frac{d}{d t} (10 t \hat{i} + 15 t^{2} \hat{j} + 7 \hat{k})$
or $\vec{v} = 10 \hat{i} + 30 t \hat{j}$
Acceleration of the particle, $\vec{a} = \frac{d \vec{v}}{d t} = 30 \hat{j}$
The particle has acceleration only along positive
$y$ -axis. According to Newton's second law, net force must also acts along positive $y$ -axis.

JEE - 2023

PHXI05:LAWS OF MOTION

363548 A cricket player catches a ball of mass $120 g$ moving with $25 m / s$ speed. If the catching process is completed in $0.1 s$ then the magnitude of force exerted by the ball on the hand of player will be (in $S I$ unit)

1 30

2 24

3 12

4 25

Explanation:

Force exerted by the ball on the hand of player,
$F = \frac{m | v - u |}{t} = \frac{120 \times 10^{- 3} \times 25}{0.1}$
$= 300 \times 10^{- 1} = 30 N$

JEE - 2024

PHXI05:LAWS OF MOTION

363545 A force $F_{1}$ of $500 N$ is required to push a car of mass $1000 k g$ slowly at constant speed on a levelled road. If a force $F_{2}$ of $1000 N$ is applied, then the acceleration of the car will be

1 zero

2

1.5 m s^{- 2}

3

1 m s^{- 2}

4

0.5 m s^{- 2}

Explanation:

We know that a force of $500 N$ does not
produce any acceleration as it pushes the car
slowly at constant speed. Out of $1000 N$ force,
only $500 N$ produces
acceleration, thus acceleration produced,
$a = \frac{force}{mass} = \frac{500}{1000} = 0.5 m s^{- 2}$

PHXI05:LAWS OF MOTION

363546 A body of mass $2 k g$ has an initial velocity of $3 m s^{- 1}$ along $O E$ and it is subjected to a force of $4 N$ in a direction perpendicular to $O E$ . The distance of body from $O$ after $4 s$ will be.
supporting img

1

12 m

2

20 m

3

8 m

4

48 m

Explanation:

The acceleration of the body perpendicular
to $O E$ is a
supporting img

$a = \frac{F}{m} = \frac{4}{2} = 2 m s^{- 2}$
Displacement along $O E$ ,
$s_{1} = v t = 3 \times 4 = 12 m$
Displacement perpendicular to $O E$ , $i . e .$
along $O A$
$s_{2} = \frac{1}{2} a t^{2} = \frac{1}{2} \times 2 \times (4)^{2} = 16 m$
The resultant displacement
$s = \sqrt{s_{1}^{2} + s_{2}^{2}}$
$= \sqrt{144 + 256} = 20 m$

PHXI05:LAWS OF MOTION

363547 The position vector of a particle related to time $t$ is given by $\vec{r} = (10 t \hat{i} + 15 t^{2} \hat{j} + 7 \hat{k}) m$ .The direction of net force experienced by the particle is

1 positive

y

-axis

2 positive

x

-axis

3 positive

z

-axis

4 in

x - y

plane.

Explanation:

Velocity of the particle,
$\vec{v} = \frac{d \vec{r}}{d t} = \frac{d}{d t} (10 t \hat{i} + 15 t^{2} \hat{j} + 7 \hat{k})$
or $\vec{v} = 10 \hat{i} + 30 t \hat{j}$
Acceleration of the particle, $\vec{a} = \frac{d \vec{v}}{d t} = 30 \hat{j}$
The particle has acceleration only along positive
$y$ -axis. According to Newton's second law, net force must also acts along positive $y$ -axis.

JEE - 2023

PHXI05:LAWS OF MOTION

363548 A cricket player catches a ball of mass $120 g$ moving with $25 m / s$ speed. If the catching process is completed in $0.1 s$ then the magnitude of force exerted by the ball on the hand of player will be (in $S I$ unit)

1 30

2 24

3 12

4 25

Explanation:

Force exerted by the ball on the hand of player,
$F = \frac{m | v - u |}{t} = \frac{120 \times 10^{- 3} \times 25}{0.1}$
$= 300 \times 10^{- 1} = 30 N$

JEE - 2024

PHXI05:LAWS OF MOTION

363545 A force $F_{1}$ of $500 N$ is required to push a car of mass $1000 k g$ slowly at constant speed on a levelled road. If a force $F_{2}$ of $1000 N$ is applied, then the acceleration of the car will be

1 zero

2

1.5 m s^{- 2}

3

1 m s^{- 2}

4

0.5 m s^{- 2}

Explanation:

We know that a force of $500 N$ does not
produce any acceleration as it pushes the car
slowly at constant speed. Out of $1000 N$ force,
only $500 N$ produces
acceleration, thus acceleration produced,
$a = \frac{force}{mass} = \frac{500}{1000} = 0.5 m s^{- 2}$

PHXI05:LAWS OF MOTION

363546 A body of mass $2 k g$ has an initial velocity of $3 m s^{- 1}$ along $O E$ and it is subjected to a force of $4 N$ in a direction perpendicular to $O E$ . The distance of body from $O$ after $4 s$ will be.
supporting img

1

12 m

2

20 m

3

8 m

4

48 m

Explanation:

The acceleration of the body perpendicular
to $O E$ is a
supporting img

$a = \frac{F}{m} = \frac{4}{2} = 2 m s^{- 2}$
Displacement along $O E$ ,
$s_{1} = v t = 3 \times 4 = 12 m$
Displacement perpendicular to $O E$ , $i . e .$
along $O A$
$s_{2} = \frac{1}{2} a t^{2} = \frac{1}{2} \times 2 \times (4)^{2} = 16 m$
The resultant displacement
$s = \sqrt{s_{1}^{2} + s_{2}^{2}}$
$= \sqrt{144 + 256} = 20 m$

PHXI05:LAWS OF MOTION

363547 The position vector of a particle related to time $t$ is given by $\vec{r} = (10 t \hat{i} + 15 t^{2} \hat{j} + 7 \hat{k}) m$ .The direction of net force experienced by the particle is

1 positive

y

-axis

2 positive

x

-axis

3 positive

z

-axis

4 in

x - y

plane.

Explanation:

Velocity of the particle,
$\vec{v} = \frac{d \vec{r}}{d t} = \frac{d}{d t} (10 t \hat{i} + 15 t^{2} \hat{j} + 7 \hat{k})$
or $\vec{v} = 10 \hat{i} + 30 t \hat{j}$
Acceleration of the particle, $\vec{a} = \frac{d \vec{v}}{d t} = 30 \hat{j}$
The particle has acceleration only along positive
$y$ -axis. According to Newton's second law, net force must also acts along positive $y$ -axis.

JEE - 2023

PHXI05:LAWS OF MOTION

363548 A cricket player catches a ball of mass $120 g$ moving with $25 m / s$ speed. If the catching process is completed in $0.1 s$ then the magnitude of force exerted by the ball on the hand of player will be (in $S I$ unit)

1 30

2 24

3 12

4 25

Explanation:

Force exerted by the ball on the hand of player,
$F = \frac{m | v - u |}{t} = \frac{120 \times 10^{- 3} \times 25}{0.1}$
$= 300 \times 10^{- 1} = 30 N$

JEE - 2024

PHXI05:LAWS OF MOTION

363545 A force $F_{1}$ of $500 N$ is required to push a car of mass $1000 k g$ slowly at constant speed on a levelled road. If a force $F_{2}$ of $1000 N$ is applied, then the acceleration of the car will be

1 zero

2

1.5 m s^{- 2}

3

1 m s^{- 2}

4

0.5 m s^{- 2}

Explanation:

We know that a force of $500 N$ does not
produce any acceleration as it pushes the car
slowly at constant speed. Out of $1000 N$ force,
only $500 N$ produces
acceleration, thus acceleration produced,
$a = \frac{force}{mass} = \frac{500}{1000} = 0.5 m s^{- 2}$

PHXI05:LAWS OF MOTION

363546 A body of mass $2 k g$ has an initial velocity of $3 m s^{- 1}$ along $O E$ and it is subjected to a force of $4 N$ in a direction perpendicular to $O E$ . The distance of body from $O$ after $4 s$ will be.
supporting img

1

12 m

2

20 m

3

8 m

4

48 m

Explanation:

The acceleration of the body perpendicular
to $O E$ is a
supporting img

$a = \frac{F}{m} = \frac{4}{2} = 2 m s^{- 2}$
Displacement along $O E$ ,
$s_{1} = v t = 3 \times 4 = 12 m$
Displacement perpendicular to $O E$ , $i . e .$
along $O A$
$s_{2} = \frac{1}{2} a t^{2} = \frac{1}{2} \times 2 \times (4)^{2} = 16 m$
The resultant displacement
$s = \sqrt{s_{1}^{2} + s_{2}^{2}}$
$= \sqrt{144 + 256} = 20 m$

PHXI05:LAWS OF MOTION

363547 The position vector of a particle related to time $t$ is given by $\vec{r} = (10 t \hat{i} + 15 t^{2} \hat{j} + 7 \hat{k}) m$ .The direction of net force experienced by the particle is

1 positive

y

-axis

2 positive

x

-axis

3 positive

z

-axis

4 in

x - y

plane.

Explanation:

Velocity of the particle,
$\vec{v} = \frac{d \vec{r}}{d t} = \frac{d}{d t} (10 t \hat{i} + 15 t^{2} \hat{j} + 7 \hat{k})$
or $\vec{v} = 10 \hat{i} + 30 t \hat{j}$
Acceleration of the particle, $\vec{a} = \frac{d \vec{v}}{d t} = 30 \hat{j}$
The particle has acceleration only along positive
$y$ -axis. According to Newton's second law, net force must also acts along positive $y$ -axis.

JEE - 2023

PHXI05:LAWS OF MOTION

363548 A cricket player catches a ball of mass $120 g$ moving with $25 m / s$ speed. If the catching process is completed in $0.1 s$ then the magnitude of force exerted by the ball on the hand of player will be (in $S I$ unit)

1 30

2 24

3 12

4 25

Explanation:

Force exerted by the ball on the hand of player,
$F = \frac{m | v - u |}{t} = \frac{120 \times 10^{- 3} \times 25}{0.1}$
$= 300 \times 10^{- 1} = 30 N$

JEE - 2024

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