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PHXI04:MOTION IN A PLANE

361805 A group of particles are projected from a point on a smooth horizontal surface with same speed and different angles then all the particles after some arbitrary time will be on

1 Circular path

2 Parabolic path

3 Straight line path

4 Hyperbolic path

Explanation:

Assume that the particles are projected from the origin as shown.
For particle one we can write its $x$ and $y$ - coordinates $x = u \cos θ_{1} t, y = u \sin θ_{1} t$
$\sin^{2} θ_{1} + \cos^{2} θ_{1} = \frac{y^{2}}{u^{2} t^{2}} + \frac{x^{2}}{u^{2} t^{2}}$
$\Rightarrow x^{2} + y^{2} = u^{2} t^{2} (1)$
For second particle $x = u \cos θ_{2} t, y = u \sin θ_{2} t$
$\sin^{2} θ_{2} + \cos^{2} θ_{2} = \frac{y^{2}}{u^{2} t^{2}} + \frac{x^{2}}{u^{2} t^{2}}$
$\Rightarrow x^{2} + y^{2} = u^{2} t^{2} (2)$
From the above two equation we can conclude that all the particles will have same equation $x^{2} + y^{2} = u^{2} t^{2}$ by having $u t$ as the radius. All the particles lie on a circle of radius $u t$ .

PHXI04:MOTION IN A PLANE

361806 The $x$ and $y$ -coordinates of a particle moving in a plane are given by $x (t) = a \cos (p t)$ and $y (t) = b \sin (p t)$ , where $a, b (< a)$ and $p$ are positive constants of appropriate dimensions and $t$ is time. Then, which of the following is not true?

1 The path of the particle is an ellipse.

2 Velocity and acceleration of the particle are perpendicular to each other at

t = \frac{π}{2 p}

.

3 Acceleration of the particle is always. directed towards a fixed point.

4 Distance travelled by the particle in time interval between

t = 0

and

t = \frac{π}{2 p}

is a.

Explanation:

Given, $x = a \cos (p t), y = b \sin (p t)$
$∴ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ ,i.e. equation of ellipse
Now , $r = x \hat{i} + y \hat{j}$
$= a \cos (p t) \hat{i} + b \sin (p t) \hat{j}$
$v = \frac{d r}{d t} = - p a \sin (p t) \hat{i} + p b \cos (p t) \hat{j}$
$a = \frac{d v}{d t} = - p^{2} a \cos (p t) \hat{i} - p^{2} b \sin (p t) \hat{j}$
At $t = \frac{π}{2 p}$
$v = - p a \hat{i} a n d a = - p^{2} b \hat{j}$
$v$ perpendicular to $a$ .
Also, $a = - p^{2} b$ , i.e. directed towards a
fixed point as $p$ and $b$ are positive constants.
As, $v = \frac{d r}{d t} \Rightarrow Δ r = \int^{\frac{π}{2 p}} v d t$
$\Rightarrow Δ r = [a \cos (p t) \hat{i} + b \sin (p t) \hat{j}]_{0}^{\frac{π}{2 p}}$
$= - a \hat{i} + b \hat{j}$
So, $| r | = \sqrt{a^{2} + b^{2}}$
So option (4) is not true.

PHXI04:MOTION IN A PLANE

361807 A body lying initially at point (3, 7) starts moving with a constant acceleration of $4 \hat{i}$ . Its position after 3 $s$ is given by the coordinates:

1

(7, 18)

2

(7, 3)

3

(3, 7)

4

(21, 7)

Explanation:

Position vector of the body, $r = 3 \hat{i} + 7 \hat{j}$
$∵$ Acceleration of the body
$a = 4 \hat{i}$ and $t = 3 s$
Using $x = u_{0} t + \frac{1}{2} a t^{2} = 0 (3) + \frac{1}{2} (4) {(3)}^{2}$
$\Rightarrow x = 18$
$∴$ New Coordinates of the body are (21, 7)

PHXI04:MOTION IN A PLANE

361808 A particle starts from the origin at $t = 0 s$ with a velocity of $10 \hat{j} m s^{- 1}$ and move in the $x$ - $y$ plane with a constant acceleration of $(8 \hat{i} + 2 \hat{j}) m s^{- 2}$ . At an instant when the $x$ -coordinate of the particle is $16 m$ , $y$ -coordinate of the particle is:

1

16 m

2

28 m

3

36 m

4

24 m

Explanation:

The $x$ and $y$ displacements of the particle are given as
$y = u_{x} t + \frac{1}{2} a_{x} t^{2} \Rightarrow 16 = 0 + \frac{1}{2} (8) t^{2} \Rightarrow t = 2 s$
$y = u_{y} t + \frac{1}{2} a_{y} t^{2} \Rightarrow y = 10 (2) + \frac{1}{2} (2) (2)^{2} = 24 m$

KCET - 2021

PHXI04:MOTION IN A PLANE

361805 A group of particles are projected from a point on a smooth horizontal surface with same speed and different angles then all the particles after some arbitrary time will be on

1 Circular path

2 Parabolic path

3 Straight line path

4 Hyperbolic path

Explanation:

Assume that the particles are projected from the origin as shown.
For particle one we can write its $x$ and $y$ - coordinates $x = u \cos θ_{1} t, y = u \sin θ_{1} t$
$\sin^{2} θ_{1} + \cos^{2} θ_{1} = \frac{y^{2}}{u^{2} t^{2}} + \frac{x^{2}}{u^{2} t^{2}}$
$\Rightarrow x^{2} + y^{2} = u^{2} t^{2} (1)$
For second particle $x = u \cos θ_{2} t, y = u \sin θ_{2} t$
$\sin^{2} θ_{2} + \cos^{2} θ_{2} = \frac{y^{2}}{u^{2} t^{2}} + \frac{x^{2}}{u^{2} t^{2}}$
$\Rightarrow x^{2} + y^{2} = u^{2} t^{2} (2)$
From the above two equation we can conclude that all the particles will have same equation $x^{2} + y^{2} = u^{2} t^{2}$ by having $u t$ as the radius. All the particles lie on a circle of radius $u t$ .

PHXI04:MOTION IN A PLANE

361806 The $x$ and $y$ -coordinates of a particle moving in a plane are given by $x (t) = a \cos (p t)$ and $y (t) = b \sin (p t)$ , where $a, b (< a)$ and $p$ are positive constants of appropriate dimensions and $t$ is time. Then, which of the following is not true?

1 The path of the particle is an ellipse.

2 Velocity and acceleration of the particle are perpendicular to each other at

t = \frac{π}{2 p}

.

3 Acceleration of the particle is always. directed towards a fixed point.

4 Distance travelled by the particle in time interval between

t = 0

and

t = \frac{π}{2 p}

is a.

Explanation:

Given, $x = a \cos (p t), y = b \sin (p t)$
$∴ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ ,i.e. equation of ellipse
Now , $r = x \hat{i} + y \hat{j}$
$= a \cos (p t) \hat{i} + b \sin (p t) \hat{j}$
$v = \frac{d r}{d t} = - p a \sin (p t) \hat{i} + p b \cos (p t) \hat{j}$
$a = \frac{d v}{d t} = - p^{2} a \cos (p t) \hat{i} - p^{2} b \sin (p t) \hat{j}$
At $t = \frac{π}{2 p}$
$v = - p a \hat{i} a n d a = - p^{2} b \hat{j}$
$v$ perpendicular to $a$ .
Also, $a = - p^{2} b$ , i.e. directed towards a
fixed point as $p$ and $b$ are positive constants.
As, $v = \frac{d r}{d t} \Rightarrow Δ r = \int^{\frac{π}{2 p}} v d t$
$\Rightarrow Δ r = [a \cos (p t) \hat{i} + b \sin (p t) \hat{j}]_{0}^{\frac{π}{2 p}}$
$= - a \hat{i} + b \hat{j}$
So, $| r | = \sqrt{a^{2} + b^{2}}$
So option (4) is not true.

PHXI04:MOTION IN A PLANE

361807 A body lying initially at point (3, 7) starts moving with a constant acceleration of $4 \hat{i}$ . Its position after 3 $s$ is given by the coordinates:

1

(7, 18)

2

(7, 3)

3

(3, 7)

4

(21, 7)

Explanation:

Position vector of the body, $r = 3 \hat{i} + 7 \hat{j}$
$∵$ Acceleration of the body
$a = 4 \hat{i}$ and $t = 3 s$
Using $x = u_{0} t + \frac{1}{2} a t^{2} = 0 (3) + \frac{1}{2} (4) {(3)}^{2}$
$\Rightarrow x = 18$
$∴$ New Coordinates of the body are (21, 7)

PHXI04:MOTION IN A PLANE

361808 A particle starts from the origin at $t = 0 s$ with a velocity of $10 \hat{j} m s^{- 1}$ and move in the $x$ - $y$ plane with a constant acceleration of $(8 \hat{i} + 2 \hat{j}) m s^{- 2}$ . At an instant when the $x$ -coordinate of the particle is $16 m$ , $y$ -coordinate of the particle is:

1

16 m

2

28 m

3

36 m

4

24 m

Explanation:

The $x$ and $y$ displacements of the particle are given as
$y = u_{x} t + \frac{1}{2} a_{x} t^{2} \Rightarrow 16 = 0 + \frac{1}{2} (8) t^{2} \Rightarrow t = 2 s$
$y = u_{y} t + \frac{1}{2} a_{y} t^{2} \Rightarrow y = 10 (2) + \frac{1}{2} (2) (2)^{2} = 24 m$

KCET - 2021

PHXI04:MOTION IN A PLANE

361805 A group of particles are projected from a point on a smooth horizontal surface with same speed and different angles then all the particles after some arbitrary time will be on

1 Circular path

2 Parabolic path

3 Straight line path

4 Hyperbolic path

Explanation:

Assume that the particles are projected from the origin as shown.
For particle one we can write its $x$ and $y$ - coordinates $x = u \cos θ_{1} t, y = u \sin θ_{1} t$
$\sin^{2} θ_{1} + \cos^{2} θ_{1} = \frac{y^{2}}{u^{2} t^{2}} + \frac{x^{2}}{u^{2} t^{2}}$
$\Rightarrow x^{2} + y^{2} = u^{2} t^{2} (1)$
For second particle $x = u \cos θ_{2} t, y = u \sin θ_{2} t$
$\sin^{2} θ_{2} + \cos^{2} θ_{2} = \frac{y^{2}}{u^{2} t^{2}} + \frac{x^{2}}{u^{2} t^{2}}$
$\Rightarrow x^{2} + y^{2} = u^{2} t^{2} (2)$
From the above two equation we can conclude that all the particles will have same equation $x^{2} + y^{2} = u^{2} t^{2}$ by having $u t$ as the radius. All the particles lie on a circle of radius $u t$ .

PHXI04:MOTION IN A PLANE

361806 The $x$ and $y$ -coordinates of a particle moving in a plane are given by $x (t) = a \cos (p t)$ and $y (t) = b \sin (p t)$ , where $a, b (< a)$ and $p$ are positive constants of appropriate dimensions and $t$ is time. Then, which of the following is not true?

1 The path of the particle is an ellipse.

2 Velocity and acceleration of the particle are perpendicular to each other at

t = \frac{π}{2 p}

.

3 Acceleration of the particle is always. directed towards a fixed point.

4 Distance travelled by the particle in time interval between

t = 0

and

t = \frac{π}{2 p}

is a.

Explanation:

Given, $x = a \cos (p t), y = b \sin (p t)$
$∴ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ ,i.e. equation of ellipse
Now , $r = x \hat{i} + y \hat{j}$
$= a \cos (p t) \hat{i} + b \sin (p t) \hat{j}$
$v = \frac{d r}{d t} = - p a \sin (p t) \hat{i} + p b \cos (p t) \hat{j}$
$a = \frac{d v}{d t} = - p^{2} a \cos (p t) \hat{i} - p^{2} b \sin (p t) \hat{j}$
At $t = \frac{π}{2 p}$
$v = - p a \hat{i} a n d a = - p^{2} b \hat{j}$
$v$ perpendicular to $a$ .
Also, $a = - p^{2} b$ , i.e. directed towards a
fixed point as $p$ and $b$ are positive constants.
As, $v = \frac{d r}{d t} \Rightarrow Δ r = \int^{\frac{π}{2 p}} v d t$
$\Rightarrow Δ r = [a \cos (p t) \hat{i} + b \sin (p t) \hat{j}]_{0}^{\frac{π}{2 p}}$
$= - a \hat{i} + b \hat{j}$
So, $| r | = \sqrt{a^{2} + b^{2}}$
So option (4) is not true.

PHXI04:MOTION IN A PLANE

361807 A body lying initially at point (3, 7) starts moving with a constant acceleration of $4 \hat{i}$ . Its position after 3 $s$ is given by the coordinates:

1

(7, 18)

2

(7, 3)

3

(3, 7)

4

(21, 7)

Explanation:

Position vector of the body, $r = 3 \hat{i} + 7 \hat{j}$
$∵$ Acceleration of the body
$a = 4 \hat{i}$ and $t = 3 s$
Using $x = u_{0} t + \frac{1}{2} a t^{2} = 0 (3) + \frac{1}{2} (4) {(3)}^{2}$
$\Rightarrow x = 18$
$∴$ New Coordinates of the body are (21, 7)

PHXI04:MOTION IN A PLANE

361808 A particle starts from the origin at $t = 0 s$ with a velocity of $10 \hat{j} m s^{- 1}$ and move in the $x$ - $y$ plane with a constant acceleration of $(8 \hat{i} + 2 \hat{j}) m s^{- 2}$ . At an instant when the $x$ -coordinate of the particle is $16 m$ , $y$ -coordinate of the particle is:

1

16 m

2

28 m

3

36 m

4

24 m

Explanation:

The $x$ and $y$ displacements of the particle are given as
$y = u_{x} t + \frac{1}{2} a_{x} t^{2} \Rightarrow 16 = 0 + \frac{1}{2} (8) t^{2} \Rightarrow t = 2 s$
$y = u_{y} t + \frac{1}{2} a_{y} t^{2} \Rightarrow y = 10 (2) + \frac{1}{2} (2) (2)^{2} = 24 m$

KCET - 2021

PHXI04:MOTION IN A PLANE

361805 A group of particles are projected from a point on a smooth horizontal surface with same speed and different angles then all the particles after some arbitrary time will be on

1 Circular path

2 Parabolic path

3 Straight line path

4 Hyperbolic path

Explanation:

Assume that the particles are projected from the origin as shown.
For particle one we can write its $x$ and $y$ - coordinates $x = u \cos θ_{1} t, y = u \sin θ_{1} t$
$\sin^{2} θ_{1} + \cos^{2} θ_{1} = \frac{y^{2}}{u^{2} t^{2}} + \frac{x^{2}}{u^{2} t^{2}}$
$\Rightarrow x^{2} + y^{2} = u^{2} t^{2} (1)$
For second particle $x = u \cos θ_{2} t, y = u \sin θ_{2} t$
$\sin^{2} θ_{2} + \cos^{2} θ_{2} = \frac{y^{2}}{u^{2} t^{2}} + \frac{x^{2}}{u^{2} t^{2}}$
$\Rightarrow x^{2} + y^{2} = u^{2} t^{2} (2)$
From the above two equation we can conclude that all the particles will have same equation $x^{2} + y^{2} = u^{2} t^{2}$ by having $u t$ as the radius. All the particles lie on a circle of radius $u t$ .

PHXI04:MOTION IN A PLANE

361806 The $x$ and $y$ -coordinates of a particle moving in a plane are given by $x (t) = a \cos (p t)$ and $y (t) = b \sin (p t)$ , where $a, b (< a)$ and $p$ are positive constants of appropriate dimensions and $t$ is time. Then, which of the following is not true?

1 The path of the particle is an ellipse.

2 Velocity and acceleration of the particle are perpendicular to each other at

t = \frac{π}{2 p}

.

3 Acceleration of the particle is always. directed towards a fixed point.

4 Distance travelled by the particle in time interval between

t = 0

and

t = \frac{π}{2 p}

is a.

Explanation:

Given, $x = a \cos (p t), y = b \sin (p t)$
$∴ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ ,i.e. equation of ellipse
Now , $r = x \hat{i} + y \hat{j}$
$= a \cos (p t) \hat{i} + b \sin (p t) \hat{j}$
$v = \frac{d r}{d t} = - p a \sin (p t) \hat{i} + p b \cos (p t) \hat{j}$
$a = \frac{d v}{d t} = - p^{2} a \cos (p t) \hat{i} - p^{2} b \sin (p t) \hat{j}$
At $t = \frac{π}{2 p}$
$v = - p a \hat{i} a n d a = - p^{2} b \hat{j}$
$v$ perpendicular to $a$ .
Also, $a = - p^{2} b$ , i.e. directed towards a
fixed point as $p$ and $b$ are positive constants.
As, $v = \frac{d r}{d t} \Rightarrow Δ r = \int^{\frac{π}{2 p}} v d t$
$\Rightarrow Δ r = [a \cos (p t) \hat{i} + b \sin (p t) \hat{j}]_{0}^{\frac{π}{2 p}}$
$= - a \hat{i} + b \hat{j}$
So, $| r | = \sqrt{a^{2} + b^{2}}$
So option (4) is not true.

PHXI04:MOTION IN A PLANE

361807 A body lying initially at point (3, 7) starts moving with a constant acceleration of $4 \hat{i}$ . Its position after 3 $s$ is given by the coordinates:

1

(7, 18)

2

(7, 3)

3

(3, 7)

4

(21, 7)

Explanation:

Position vector of the body, $r = 3 \hat{i} + 7 \hat{j}$
$∵$ Acceleration of the body
$a = 4 \hat{i}$ and $t = 3 s$
Using $x = u_{0} t + \frac{1}{2} a t^{2} = 0 (3) + \frac{1}{2} (4) {(3)}^{2}$
$\Rightarrow x = 18$
$∴$ New Coordinates of the body are (21, 7)

PHXI04:MOTION IN A PLANE

361808 A particle starts from the origin at $t = 0 s$ with a velocity of $10 \hat{j} m s^{- 1}$ and move in the $x$ - $y$ plane with a constant acceleration of $(8 \hat{i} + 2 \hat{j}) m s^{- 2}$ . At an instant when the $x$ -coordinate of the particle is $16 m$ , $y$ -coordinate of the particle is:

1

16 m

2

28 m

3

36 m

4

24 m

Explanation:

The $x$ and $y$ displacements of the particle are given as
$y = u_{x} t + \frac{1}{2} a_{x} t^{2} \Rightarrow 16 = 0 + \frac{1}{2} (8) t^{2} \Rightarrow t = 2 s$
$y = u_{y} t + \frac{1}{2} a_{y} t^{2} \Rightarrow y = 10 (2) + \frac{1}{2} (2) (2)^{2} = 24 m$

KCET - 2021