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PHXI05:LAWS OF MOTION

363549 Statement A :
The moment after a stone is released out of an accelerated train, there is no horizontal force or acceleration on the stone.
Statement B :
Force on a body at a given time is determined by the situation at the location of the body at that time.

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.

PHXI05:LAWS OF MOTION

363550 An object of mass $10 k g$ moves at a constant speed of $10 m / s .$ A constant force that acts for $4 s$ on the object, gives it a speed $2 m / s$ in opposite direction. The force acting on the object is

1

- 3 N

2

- 30 N

3

3 N

4

30 N

Explanation:

From the first equation of motion,
$v = u + a t$
$\Rightarrow v - u = a t$
$\Rightarrow a = \frac{v - u}{t}$
Given, $u = 10 m / s, v = - 2 m / s$
and $t = 4 s$ .
$a = \frac{- 2 - 10}{4} = - 3 m / s^{2}$
So, force acting on the object
$F = m a = 10 \times (- 3) = - 30 N$

PHXI05:LAWS OF MOTION

363551 Figure shows ( $x$ , $t$ ), ( $y$ , $t$ ) diagram of a particle moving in 2-dimensions.
If the particle has a mass of 500 $g$ , the force acting on the particle is
supporting img

1 0.5

N

along

x

- axis

2 0.5

N

along

y

- axis

3 0.1

N

along

y

- axis

4 0.1

N

along

x

- axis

Explanation:

Since the graphs between $x$ and $t$ is a straight line and passing through the origin.
At $x = 2, t = 2$
$∴ x = t$
At $t = 2, y = 4$
Since the graph between $y$ and $t$ is a parabola.
$∴ y = t^{2}$
$∴ v_{x} = \frac{d x}{d t} = 1$ and $a_{x} = \frac{d v_{x}}{d t} = 0$
and $v_{y} = \frac{d y}{d t} = 2 t$ and $a_{y} = 2 m s^{- 2}$
The force acting on the particle is
$F = m a_{y} = (0.5 k g) (2 m s^{- 2})$
$= 1 N$ along $y$ -axis

PHXI05:LAWS OF MOTION

363552 A body of mass 5 $k g$ starts from the origin with an initial velocity $\vec{u} = 30 \hat{i} + 40 \hat{j} m s^{- 1} .$ If a constant force $\vec{F} = - (\hat{i} + 5 \hat{j}) N$ acts on the body, the time in which the $y$ -component of the velocity becomes zero is

1 5 seconds

2 20 seconds

3 40 seconds

4 80 seconds

Explanation:

$u_{y} = 40 m / s, F_{y} = - 5 N, m = 5 k g$
So $a_{y} = \frac{F_{y}}{m} = - 1 m / s^{2}, (A s v = u + a t)$
$∴ v_{y} = 40 - 1 \times t = 0 \Rightarrow t = 40 \sec$

PHXI05:LAWS OF MOTION

363549 Statement A :
The moment after a stone is released out of an accelerated train, there is no horizontal force or acceleration on the stone.
Statement B :
Force on a body at a given time is determined by the situation at the location of the body at that time.

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.

PHXI05:LAWS OF MOTION

363550 An object of mass $10 k g$ moves at a constant speed of $10 m / s .$ A constant force that acts for $4 s$ on the object, gives it a speed $2 m / s$ in opposite direction. The force acting on the object is

1

- 3 N

2

- 30 N

3

3 N

4

30 N

Explanation:

From the first equation of motion,
$v = u + a t$
$\Rightarrow v - u = a t$
$\Rightarrow a = \frac{v - u}{t}$
Given, $u = 10 m / s, v = - 2 m / s$
and $t = 4 s$ .
$a = \frac{- 2 - 10}{4} = - 3 m / s^{2}$
So, force acting on the object
$F = m a = 10 \times (- 3) = - 30 N$

PHXI05:LAWS OF MOTION

363551 Figure shows ( $x$ , $t$ ), ( $y$ , $t$ ) diagram of a particle moving in 2-dimensions.
If the particle has a mass of 500 $g$ , the force acting on the particle is
supporting img

1 0.5

N

along

x

- axis

2 0.5

N

along

y

- axis

3 0.1

N

along

y

- axis

4 0.1

N

along

x

- axis

Explanation:

Since the graphs between $x$ and $t$ is a straight line and passing through the origin.
At $x = 2, t = 2$
$∴ x = t$
At $t = 2, y = 4$
Since the graph between $y$ and $t$ is a parabola.
$∴ y = t^{2}$
$∴ v_{x} = \frac{d x}{d t} = 1$ and $a_{x} = \frac{d v_{x}}{d t} = 0$
and $v_{y} = \frac{d y}{d t} = 2 t$ and $a_{y} = 2 m s^{- 2}$
The force acting on the particle is
$F = m a_{y} = (0.5 k g) (2 m s^{- 2})$
$= 1 N$ along $y$ -axis

PHXI05:LAWS OF MOTION

363552 A body of mass 5 $k g$ starts from the origin with an initial velocity $\vec{u} = 30 \hat{i} + 40 \hat{j} m s^{- 1} .$ If a constant force $\vec{F} = - (\hat{i} + 5 \hat{j}) N$ acts on the body, the time in which the $y$ -component of the velocity becomes zero is

1 5 seconds

2 20 seconds

3 40 seconds

4 80 seconds

Explanation:

$u_{y} = 40 m / s, F_{y} = - 5 N, m = 5 k g$
So $a_{y} = \frac{F_{y}}{m} = - 1 m / s^{2}, (A s v = u + a t)$
$∴ v_{y} = 40 - 1 \times t = 0 \Rightarrow t = 40 \sec$

PHXI05:LAWS OF MOTION

363549 Statement A :
The moment after a stone is released out of an accelerated train, there is no horizontal force or acceleration on the stone.
Statement B :
Force on a body at a given time is determined by the situation at the location of the body at that time.

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.

PHXI05:LAWS OF MOTION

363550 An object of mass $10 k g$ moves at a constant speed of $10 m / s .$ A constant force that acts for $4 s$ on the object, gives it a speed $2 m / s$ in opposite direction. The force acting on the object is

1

- 3 N

2

- 30 N

3

3 N

4

30 N

Explanation:

From the first equation of motion,
$v = u + a t$
$\Rightarrow v - u = a t$
$\Rightarrow a = \frac{v - u}{t}$
Given, $u = 10 m / s, v = - 2 m / s$
and $t = 4 s$ .
$a = \frac{- 2 - 10}{4} = - 3 m / s^{2}$
So, force acting on the object
$F = m a = 10 \times (- 3) = - 30 N$

PHXI05:LAWS OF MOTION

363551 Figure shows ( $x$ , $t$ ), ( $y$ , $t$ ) diagram of a particle moving in 2-dimensions.
If the particle has a mass of 500 $g$ , the force acting on the particle is
supporting img

1 0.5

N

along

x

- axis

2 0.5

N

along

y

- axis

3 0.1

N

along

y

- axis

4 0.1

N

along

x

- axis

Explanation:

Since the graphs between $x$ and $t$ is a straight line and passing through the origin.
At $x = 2, t = 2$
$∴ x = t$
At $t = 2, y = 4$
Since the graph between $y$ and $t$ is a parabola.
$∴ y = t^{2}$
$∴ v_{x} = \frac{d x}{d t} = 1$ and $a_{x} = \frac{d v_{x}}{d t} = 0$
and $v_{y} = \frac{d y}{d t} = 2 t$ and $a_{y} = 2 m s^{- 2}$
The force acting on the particle is
$F = m a_{y} = (0.5 k g) (2 m s^{- 2})$
$= 1 N$ along $y$ -axis

PHXI05:LAWS OF MOTION

363552 A body of mass 5 $k g$ starts from the origin with an initial velocity $\vec{u} = 30 \hat{i} + 40 \hat{j} m s^{- 1} .$ If a constant force $\vec{F} = - (\hat{i} + 5 \hat{j}) N$ acts on the body, the time in which the $y$ -component of the velocity becomes zero is

1 5 seconds

2 20 seconds

3 40 seconds

4 80 seconds

Explanation:

$u_{y} = 40 m / s, F_{y} = - 5 N, m = 5 k g$
So $a_{y} = \frac{F_{y}}{m} = - 1 m / s^{2}, (A s v = u + a t)$
$∴ v_{y} = 40 - 1 \times t = 0 \Rightarrow t = 40 \sec$

PHXI05:LAWS OF MOTION

363549 Statement A :
The moment after a stone is released out of an accelerated train, there is no horizontal force or acceleration on the stone.
Statement B :
Force on a body at a given time is determined by the situation at the location of the body at that time.

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.

PHXI05:LAWS OF MOTION

363550 An object of mass $10 k g$ moves at a constant speed of $10 m / s .$ A constant force that acts for $4 s$ on the object, gives it a speed $2 m / s$ in opposite direction. The force acting on the object is

1

- 3 N

2

- 30 N

3

3 N

4

30 N

Explanation:

From the first equation of motion,
$v = u + a t$
$\Rightarrow v - u = a t$
$\Rightarrow a = \frac{v - u}{t}$
Given, $u = 10 m / s, v = - 2 m / s$
and $t = 4 s$ .
$a = \frac{- 2 - 10}{4} = - 3 m / s^{2}$
So, force acting on the object
$F = m a = 10 \times (- 3) = - 30 N$

PHXI05:LAWS OF MOTION

363551 Figure shows ( $x$ , $t$ ), ( $y$ , $t$ ) diagram of a particle moving in 2-dimensions.
If the particle has a mass of 500 $g$ , the force acting on the particle is
supporting img

1 0.5

N

along

x

- axis

2 0.5

N

along

y

- axis

3 0.1

N

along

y

- axis

4 0.1

N

along

x

- axis

Explanation:

Since the graphs between $x$ and $t$ is a straight line and passing through the origin.
At $x = 2, t = 2$
$∴ x = t$
At $t = 2, y = 4$
Since the graph between $y$ and $t$ is a parabola.
$∴ y = t^{2}$
$∴ v_{x} = \frac{d x}{d t} = 1$ and $a_{x} = \frac{d v_{x}}{d t} = 0$
and $v_{y} = \frac{d y}{d t} = 2 t$ and $a_{y} = 2 m s^{- 2}$
The force acting on the particle is
$F = m a_{y} = (0.5 k g) (2 m s^{- 2})$
$= 1 N$ along $y$ -axis

PHXI05:LAWS OF MOTION

363552 A body of mass 5 $k g$ starts from the origin with an initial velocity $\vec{u} = 30 \hat{i} + 40 \hat{j} m s^{- 1} .$ If a constant force $\vec{F} = - (\hat{i} + 5 \hat{j}) N$ acts on the body, the time in which the $y$ -component of the velocity becomes zero is

1 5 seconds

2 20 seconds

3 40 seconds

4 80 seconds

Explanation:

$u_{y} = 40 m / s, F_{y} = - 5 N, m = 5 k g$
So $a_{y} = \frac{F_{y}}{m} = - 1 m / s^{2}, (A s v = u + a t)$
$∴ v_{y} = 40 - 1 \times t = 0 \Rightarrow t = 40 \sec$