Motion in a Plane — Page 5 | JEE Physics

Q41

A child stands on the edge of the cliff

10 \mathrm{~m}

above the ground and throws a stone horizontally with an initial speed of

5 \mathrm{~ms}^{-1}

. Neglecting the air resistance, the speed with which the stone hits the ground will be

\mathrm{ms}^{-1}

(given,

g=10 \mathrm{~ms}^{-2}

).

A 20

B 25

C 30

D 15

Correct Answer

Option D

Solution

\begin{aligned} & \mathrm{v}_{\mathrm{y}}=\sqrt{2 g h}=\sqrt{200} \\\\ & v_{n e t}=\sqrt{25+200}=15 \mathrm{~m} / \mathrm{s} \end{aligned}

Q42

Two objects are projected with same velocity 'u' however at different angles

\alpha

and

\beta

with the horizontal. If

\alpha+\beta=90^\circ

, the ratio of horizontal range of the first object to the 2nd object will be :

A 1 : 1

B 2 : 1

C 1 : 2

D 4 : 1

Correct Answer

Option A

Solution

\text{Range}=\frac{u^2 \sin 2 \theta}{g}

Range for projection angle " $\alpha$ "

\mathrm{R}_1=\frac{\mathrm{u}^2 \sin 2 \alpha}{\mathrm{g}}

Range for projection angle " $\beta$ "

\begin{aligned} & \mathrm{R}_2=\frac{\mathrm{u}^2 \sin 2 \beta}{\mathrm{g}} \\\\ & \alpha+\beta=90^{\circ}(\text{ Given }) \\\\ & \Rightarrow \beta=90^{\circ}-\alpha \\\\ & \mathrm{R}_2=\frac{\mathrm{u}^2 \sin 2\left(90^{\circ}-\alpha\right)}{\mathrm{g}} \\\\ & \mathrm{R}_2=\frac{\mathrm{u}^2 \sin \left(180^{\circ}-2 \alpha\right)}{\mathrm{g}} \\\\ & \mathrm{R}_2=\frac{\mathrm{u}^2 \sin 2 \alpha}{\mathrm{g}} \\\\ & \Rightarrow \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\left(\frac{\mathrm{u}^2 \sin 2 \alpha}{\mathrm{g}}\right)}{\left(\frac{\mathrm{u}^2 \sin 2 \alpha}{\mathrm{g}}\right)}=\frac{1}{1} \end{aligned}

Q43

A projectile is projected at

30^{\circ}

from horizontal with initial velocity

40 \mathrm{~ms}^{-1}

. The velocity of the projectile at

\mathrm{t}=2 \mathrm{~s}

from the start will be : (Given

g=10 \mathrm{~m} / \mathrm{s}^{2}

)

A

20 \sqrt{3} \mathrm{~ms}^{-1}

B Zero

C

20 \mathrm{~ms}^{-1}

D

40 \sqrt{3} \mathrm{~ms}^{-1}

Correct Answer

Option A

Solution

To find the velocity of the projectile at t = 2 s, we need to find the horizontal and vertical components of the velocity at that time.

The initial horizontal component of the velocity is constant and is given by:

v_{0x} = v_0 \cos\theta = 40\,\mathrm{ms}^{-1} \cos(30^{\circ}) = 40\,\mathrm{ms}^{-1} \cdot \frac{\sqrt{3}}{2} = 20\sqrt{3}\,\mathrm{ms}^{-1}

The initial vertical component of the velocity is:

v_{0y} = v_0 \sin\theta = 40\,\mathrm{ms}^{-1} \sin(30^{\circ}) = 40\,\mathrm{ms}^{-1} \cdot \frac{1}{2} = 20\,\mathrm{ms}^{-1}

To find the vertical component of the velocity at t = 2 s, we use the equation:

v_y = v_{0y} - gt = 20\,\mathrm{ms}^{-1} - (10\,\mathrm{ms}^{-2})(2\,\mathrm{s}) = 20\,\mathrm{ms}^{-1} - 20\,\mathrm{ms}^{-1} = 0\,\mathrm{ms}^{-1}

At t = 2 s, the horizontal component of the velocity is still

20\sqrt{3}\,\mathrm{ms}^{-1}

, and the vertical component is 0. The overall velocity at t = 2 s is:

\vec{v} = 20\sqrt{3}\,\mathrm{ms}^{-1} \hat{i} + 0\,\mathrm{ms}^{-1} \hat{j} = 20\sqrt{3}\,\mathrm{ms}^{-1}

So the correct answer is:

20\sqrt{3}\,\mathrm{ms}^{-1}

Q44

Two projectiles are projected at

30^{\circ}

and

60^{\circ}

with the horizontal with the same speed. The ratio of the maximum height attained by the two projectiles respectively is:

A

1: \sqrt{3}

B

\sqrt{3}: 1

C 1 : 3

D

2: \sqrt{3}

Correct Answer

Option C

Solution

Let the initial speed of both projectiles be v.

The maximum height attained by a projectile can be calculated using the formula:

H = \frac{v^2 \sin^2 \theta}{2g}

where H is the maximum height, v is the initial speed, θ is the angle of projection, and g is the acceleration due to gravity.

For the projectile projected at 30°, the maximum height is:

H_1 = \frac{v^2 \sin^2 30^{\circ}}{2g} = \frac{v^2 \times \frac{1}{4}}{2g} = \frac{v^2}{8g}

For the projectile projected at 60°, the maximum height is:

H_2 = \frac{v^2 \sin^2 60^{\circ}}{2g} = \frac{v^2 \times \frac{3}{4}}{2g} = \frac{3v^2}{8g}

Now, let's find the ratio of the maximum heights:

\frac{H_1}{H_2} = \frac{\frac{v^2}{8g}}{\frac{3v^2}{8g}} = \frac{v^2}{3v^2}

The v² terms cancel out, and we get:

\frac{H_1}{H_2} = \frac{1}{3}

Therefore, the ratio of the maximum heights attained by the two projectiles is 1 : 3

Q45

The range of the projectile projected at an angle of 15

^\circ

with horizontal is 50 m. If the projectile is projected with same velocity at an angle of 45

^\circ

with horizontal, then its range will be

A 50

\sqrt2

m

B 100 m

C 100

\sqrt2

m

D 50 m

Correct Answer

Option B

Solution

The range $R$ of a projectile launched with an initial speed $v$ and at an angle $\theta$ to the horizontal is given by: $R = \dfrac{v^2}{g} \sin(2\theta)$ where $g$ is the acceleration due to gravity.

From this equation, we can see that the range is dependent on the sine of twice the launch angle.

Given that the range at $15^\circ$ is $50$ m, if we launch the projectile at $45^\circ$ with the same velocity, we can compare the ranges by comparing $\sin(2 \times 15^\circ)$ and $\sin(2 \times 45^\circ)$ : $\sin(30^\circ) = \dfrac{1}{2}$ $\sin(90^\circ) = 1$ Therefore, the range at $45^\circ$ will be twice the range at $15^\circ$ , because $\sin(90^\circ)$ is twice as large as $\sin(30^\circ)$ .

So, the range when the projectile is launched at $45^\circ$ will be $2 \times 50$ m = $100$ m.

So, $100$ m is the correct answer.

Q46

The trajectory of projectile, projected from the ground is given by

y=x-\dfrac{x^{2}}{20}

. Where

x

and

y

are measured in meter. The maximum height attained by the projectile will be.

A 10 m

B 5 m

C 200 m

D 10

\sqrt2

m

Correct Answer

Option B

Solution

The equation of the trajectory given is $y = x - \dfrac{x^2}{20}$ .

This is a parabola, and it represents the path of the projectile.

The maximum height of the projectile corresponds to the vertex of the parabola.

The x-coordinate of the vertex for a parabola given by $y = ax^2 + bx + c$ is $-b/2a$ .

In this case, $a = -1/20$ and $b = 1$ , so the x-coordinate of the vertex is: $x_{\text{vertex}} = -\dfrac{b}{2a} = -\dfrac{1}{2 \times (-1/20)} = 10$ Substituting this into the equation of the trajectory gives the y-coordinate of the vertex, which is the maximum height: $y_{\text{max}} = 10 - \dfrac{10^2}{20} = 10 - 5 = 5$ So, the maximum height attained by the projectile is 5 m.

Q47

Two projectiles A and B are thrown with initial velocities of

40 \mathrm{~m} / \mathrm{s}

and

60 \mathrm{~m} / \mathrm{s}

at angles

30^{\circ}

and

60^{\circ}

with the horizontal respectively. The ratio of their ranges respectively is

\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)

A

4: 9

B

2: \sqrt{3}

C

\sqrt{3}: 2

D

1: 1

Correct Answer

Option A

Solution

The range of a projectile launched with an initial velocity $v$ at an angle $\theta$ with respect to the horizontal is given by: $R = \dfrac{v^2 \sin(2\theta)}{g}$ , where $g$ is the acceleration due to gravity.

Let's calculate the ranges of projectiles A and B: For projectile A, $v = 40 \, \text{m/s}$ and $\theta = 30^\circ$ , so: $R_A = \dfrac{(40)^2 \sin(2 \times 30)}{10} = 4 \times 40 = 160 \, \text{m}$ .

For projectile B, $v = 60 \, \text{m/s}$ and $\theta = 60^\circ$ , so: $R_B = \dfrac{(60)^2 \sin(2 \times 60)}{10} = 6 \times 60 = 360 \, \text{m}$ .

Therefore, the ratio of their ranges is $R_A : R_B = 160 : 360 = 4 : 9$ .

Q48

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R Assertion A : When a body is projected at an angle

45^{\circ}

, it's range is maximum. Reason R : For maximum range, the value of

\sin 2 \theta

should be equal to one. In the light of the above statements, choose the correct answer from the options given below:

A Both

\mathbf{A}

and

\mathbf{R}

are correct and

\mathbf{R}

is the correct explanation of

\mathbf{A}

B

\mathbf{A}

is true but

\mathbf{R}

is false

C

\mathbf{A}

is false but

\mathbf{R}

is true

D Both

\mathbf{A}

and

\mathbf{R}

are correct but

\mathbf{R}

is NOT the correct explanation of

\mathbf{A}

Correct Answer

Option A

Solution

Assertion A: When a body is projected at an angle of $45^{\circ}$ , its range is maximum.

This is true, and it's a well-established fact in physics.

The maximum range of a projectile, assuming no air resistance and flat terrain, is achieved at an angle of $45^{\circ}$ .

Reason R: For maximum range, the value of $\sin 2\theta$ should be equal to one.

This is also true.

The range of a projectile, again assuming no air resistance and flat terrain, can be calculated using the formula $R = (v^{2}/g) \cdot \sin(2\theta)$ , where $v$ is the initial velocity of the projectile, $g$ is the acceleration due to gravity, and $\theta$ is the launch angle.

For the range to be maximized, $\sin(2\theta)$ must be maximized, and the maximum value of $\sin(2\theta)$ is 1.

This occurs when $2\theta = 90$ degrees, or $\theta = 45$ degrees, which corresponds to the assertion.

Q49

Position of an ant (

\mathrm{S}

in metres) moving in

\mathrm{Y}

-

\mathrm{Z}

plane is given by

S=2 t^2 \hat{j}+5 \hat{k}

(where

t

is in second). The magnitude and direction of velocity of the ant at

\mathrm{t}=1 \mathrm{~s}

will be :

A

16 \mathrm{~m} / \mathrm{s}

in

y

-direction

B

4 \mathrm{~m} / \mathrm{s}

in

x

-direction

C

9 \mathrm{~m} / \mathrm{s}

in

\mathrm{z}

-direction

D

4 \mathrm{~m} / \mathrm{s}

in

y

-direction

Correct Answer

Option D

Solution

The position of an ant, denoted as $\mathrm{S}$ in meters, moving in the $\mathrm{Y}$ - $\mathrm{Z}$ plane is given by $S = 2t^2 \hat{j} + 5 \hat{k}$ , where $t$ is in seconds.

To determine the magnitude and direction of the ant's velocity at $t = 1$ second, we need to differentiate the position function with respect to time.

The velocity $\overrightarrow{\mathrm{v}}$ is given by:

\overrightarrow{\mathrm{v}} = \frac{d\mathrm{S}}{dt} = \frac{d}{dt} (2t^2 \hat{j} + 5 \hat{k})

On differentiating, we get:

\overrightarrow{\mathrm{v}} = 4t \hat{j}

At $t = 1$ second:

\overrightarrow{\mathrm{v}} = 4 \cdot 1 \hat{j} = 4 \hat{j}

Therefore, the magnitude of the velocity is $4 \mathrm{~m/s}$ and it is directed along the $y$ -axis.

Q50

The co-ordinates of a particle moving in

x

-

y

plane are given by :

x=2+4 \mathrm{t}, y=3 \mathrm{t}+8 \mathrm{t}^2

. The motion of the particle is :

A uniform motion along a straight line.

B non-uniformly accelerated.

C uniformly accelerated having motion along a straight line.

D uniformly accelerated having motion along a parabolic path.

Correct Answer

Option D

Solution

To determine the nature of the motion of the particle given by its coordinates in the

x

-

y

plane, we analyze the given equations for

x

and

y

in terms of time

t

:

x = 2 + 4t

y = 3t + 8t^2

Firstly, the equation for

x

is of the form

x = x_0 + vt

, where

x_0 = 2

is the initial position and

v = 4

is the constant velocity along the

x

-axis. This suggests a uniform motion along the

x

-axis because the velocity remains constant with time. Secondly, the equation for

y

is a second-degree polynomial in

t

, which indicates a parabolic path. The presence of the

t^2

term (

8t^2

) signifies acceleration since the position along the

y

-axis is changing at a rate that itself changes over time. The equation for

y

can show two types of motion depending on the terms: If it was of the form

y = y_0 + vt

, it would indicate uniform motion. If it was of the form

y = y_0 + vt + \frac{1}{2}at^2

, where

a

would represent acceleration, it would indicate uniformly accelerated motion. The presence of the

8t^2

term here plays a similar role, indicating that the motion is uniformly accelerated in the

y

-direction due to the constant acceleration implied by this term. Since the motion in the

y

direction is determined by a quadratic equation, and the path of the particle depends on both the

x

and

y

coordinates, the motion of the particle is not along a straight line but rather follows a parabolic path due to the quadratic (second-degree) dependence on time in the

y

-coordinate. Additionally, the acceleration is not changing with time, as deduced from the constant coefficient of the

t^2

term in the

y

equation, indicating uniform acceleration.

Therefore, the motion is uniformly accelerated and follows a parabolic path.

Hence, the correct answer is: Option D: uniformly accelerated having motion along a parabolic path.