Motion in a Plane — Page 4 | JEE Physics

Q31

A projectile is launched at an angle '

\alpha

' with the horizontal with a velocity 20 ms

-

1. After 10 s, its inclination with horizontal is '

\beta

'. The value of tan

\beta

will be : (g = 10 ms

-

2).

A tan

\alpha

+ 5sec

\alpha

B tan

\alpha

-

5sec

\alpha

C 2tan

\alpha

-

5sec

\alpha

D 2tan

\alpha

+

5sec

\alpha

Correct Answer

Option B

Solution

At $t=0$ , the motion of projectile is given as $\tan \alpha=\dfrac{u_{y}}{u_{x}}$ where, $u_{y}$ is the vertical component of initial velocity and $u_{x}$ is the horizontal component of initial velocity.

At $t=10 \mathrm{~s}$ , the motion of projectile is given as $\tan \beta=\dfrac{v_{y}}{v_{x}}$ .........(ii) where, $v_{y}$ is the vertical component of final velocity after $t=10 \mathrm{~s}$ and $v_{x}$ is the horizontal component of final velocity.

From Eqs. (i) and (ii), we get $\dfrac{\tan \beta}{\tan \alpha}=\dfrac{v_y}{u_y} \quad\left(\right.$ as $v_x=u_x$ ) ....(iii) Using the following equation in vertical direction, we get

\begin{aligned} &v_{y}=u_{y}-g t \\\\ &v_{y}=u_{y}-100 \end{aligned}

Using Eq, (iii)}

\frac{\tan \beta}{\tan \alpha}=\frac{u_{y}-100}{u_{y}}

\begin{aligned} & \frac{\tan \beta}{\tan \alpha}=1-\frac{100}{u_{y}}=1-\frac{100}{20 \sin \alpha} \quad\left(\because u_{y}=20 \sin \alpha\right) \end{aligned}

=1-\frac{5}{\sin \alpha}

$\Rightarrow \tan \beta=\tan \alpha\left(1-\dfrac{5}{\sin \alpha}\right)=\tan \alpha-5 \sec \alpha$

Q32

A girl standing on road holds her umbrella at 45

^\circ

with the vertical to keep the rain away. If she starts running without umbrella with a speed of 15

\sqrt2

kmh

-

1, the rain drops hit her head vertically. The speed of rain drops with respect to the moving girl is :

A 30 kmh

-

1

B

{{25} \over {\sqrt 2 }}

kmh

-

1

C

{{30} \over {\sqrt 2 }}

kmh

-

1

D 25 kmh

-

1

Correct Answer

Option C

Solution

From graph,

{v_{RG}} = 15\sqrt 2 \tan 45^\circ

= 15\sqrt 2

= {{30} \over {\sqrt 2 }}

Q33

Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Two identical balls A and B thrown with same velocity 'u' at two different angles with horizontal attained the same range R. IF A and B reached the maximum height h1 and h2 respectively, then

R = 4\sqrt {{h_1}{h_2}}

Reason R : Product of said heights.

{h_1}{h_2} = \left( {{{{u^2}{{\sin }^2}\theta } \over {2g}}} \right)\,.\,\left( {{{{u^2}{{\cos }^2}\theta } \over {2g}}} \right)

Choose the correct answer :

A Both A and R are true and R is the correct explanation of A.

B Both A and R are true but R is NOT the correct explanation of A.

C A is true but R is false.

D A is false but R is true.

Correct Answer

Option A

Solution

When two projectiles are thrown with the same initial velocity 'u' but at complementary angles (say, $\theta$ and $(90^\circ - \theta)$ ) with the horizontal, they attain the same range R.

The formula for the range R of a projectile is:

R = \frac{u^2 \sin 2\theta}{g}

For complementary angles, $2 \theta$ and $180^\circ - 2 \theta$ (which simplifies to the same value for the sine function), the ranges are equal.

The maximum height $h_1$ for angle $\theta$ is given by:

h_1 = \frac{u^2 \sin^2 \theta}{2g}

And the maximum height $h_2$ for angle $(90^\circ - \theta)$ is given by:

h_2 = \frac{u^2 \cos^2 \theta}{2g}

Now, multiplying these heights:

h_1 h_2 = \left( \frac{u^2 \sin^2 \theta}{2g} \right) \cdot \left( \frac{u^2 \cos^2 \theta}{2g} \right)

This simplifies to:

h_1 h_2 = \frac{u^4 \sin^2 \theta \cos^2 \theta}{4g^2}

Since $\sin^2 \theta \cos^2 \theta = \left( \dfrac{\sin 2 \theta}{2} \right)^2 = \dfrac{1}{4} \sin^2 2 \theta$ :

h_1 h_2 = \frac{u^4 \sin^2 2 \theta}{16g^2}

Using the range formula $R = \dfrac{u^2 \sin 2 \theta}{g}$ , we get:

R^2 = \left( \frac{u^2 \sin 2 \theta}{g} \right)^2

Thus, we have:

4h_1 h_2 = \frac{u^4 \sin^2 2 \theta}{4g^2} = \frac{R^2}{4}

This simplifies to:

R = 4\sqrt{h_1 h_2}

Both the assertion and the reason are correct, and the reason correctly explains the assertion.

The correct answer is: Option A : Both A and R are true and R is the correct explanation of A.

Q34

A projectile is projected with velocity of 25 m/s at an angle

\theta

with the horizontal. After t seconds its inclination with horizontal becomes zero. If R represents horizontal range of the projectile, the value of

\theta

will be : [use g = 10 m/s2]

A

{1 \over 2}{\sin ^{ - 1}}\left( {{{5{t^2}} \over {4R}}} \right)

B

{1 \over 2}{\sin ^{ - 1}}\left( {{{4R} \over {5{t^2}}}} \right)

C

{\tan ^{ - 1}}\left( {{{4{t^2}} \over {5R}}} \right)

D

{\cot ^{ - 1}}\left( {{R \over {20{t^2}}}} \right)

Correct Answer

Option D

Solution

t = {{25\sin \theta } \over g}

and,

R = {{{{(25)}^2}(2\sin \theta \cos \theta )} \over g}

\Rightarrow R = {{25 \times 25 \times 2} \over g} \times {{gt} \over {25}} \times \cos \theta

\Rightarrow R = 50t\cos \theta

$\therefore$

tan\theta = {{gt} \over {25}} \times {{50t} \over R}

= {{20{t^2}} \over R}

\Rightarrow \theta = {\cot ^{ - 1}}\left( {{R \over {20{t^2}}}} \right)

Q35

At t = 0, truck, starting from rest, moves in the positive x-direction at uniform acceleration of 5 ms

-

2. At t = 20 s, a ball is released from the top of the truck. The ball strikes the ground in 1 s after the release. The velocity of the ball, when it strikes the ground, will be : (Given g = 10 ms

-

2)

A

100\widehat i - 10\widehat j

B

10\widehat i - 100\widehat j

C

100\widehat i

D

- 10\widehat j

Correct Answer

Option A

Solution

At t = 20 s, velocity of truck, v = 0 + 5 $\times$ 20 = 100 m/s At 20 sec a ball is dropped from the truck, so velocity of ball will be same as truck.

Velocity of truck at x-direction = 100 m/s and in y-direction = 0.

$\therefore$ Velocity of ball vx = 100 m/s, vy = 0 Now ball will show projectile motion where vertically downward acceleration g = 10 m/s act on the ball.

As horizontally no acceleration acting on the ball so horizontal velocity 100 m/s will remain unchanged.

Velocity of the ball when it reach the ground along y-direction after 1 sec.

{v_y} = 0 - 10 \times 1

$\Rightarrow$

{v_y} = - 10

m/s $\therefore$ Velocity of ball

(\overrightarrow v ) = 100\widehat i - 10\widehat j

Q36

Two projectiles P1 and P2 thrown with speed in the ratio

\sqrt3

:

\sqrt2

, attain the same height during their motion. If P2 is thrown at an angle of 60

^\circ

with the horizontal, the angle of projection of P1 with horizontal will be :

A 15

^\circ

B 30

^\circ

C 45

^\circ

D 60

^\circ

Correct Answer

Option C

Solution

We know, Maximum height of a projectile

(H) = {{{u^2}{{\sin }^2}\theta } \over {2g}}

Given, Ratio of initial velocity of two projectile

{{{u_1}} \over {{u_2}}} = {{\sqrt 3 } \over {\sqrt 2 }}

Both projectile reach the same maximum height. $\therefore$

{H_1} = {H_2}

\Rightarrow {{u_1^2{{\sin }^2}{\theta _1}} \over {2g}} = {{u_2^2{{\sin }^2}{\theta _2}} \over {2g}}

\Rightarrow u_1^2{\sin ^2}{\theta _1} = u_2^2{\sin ^2}{\theta _2}

\Rightarrow {\left( {{{{u_1}} \over {{u_2}}}} \right)^2} = {\left( {{{\sin {\theta _2}} \over {\sin {\theta _1}}}} \right)^2}

\Rightarrow {\left( {{{\sqrt 3 } \over {\sqrt 2 }}} \right)^2} = {\left( {{{\sin 60^\circ } \over {\sin {\theta _1}}}} \right)^2}

\Rightarrow {3 \over 2} = {3 \over {4{{\sin }^2}{\theta _1}}}

\Rightarrow \sin _{{\theta _1}}^2 = {1 \over 2}

\Rightarrow \sin {\theta _1} = {1 \over {\sqrt 2 }} = \sin 45^\circ

\Rightarrow {\theta _1} = 45^\circ

Q37

A ball is projected from the ground with a speed 15 ms

-

1 at an angle

\theta

with horizontal so that its range and maximum height are equal, then 'tan

\theta

' will be equal to :

A

{1 \over 4}

B

{1 \over 2}

C 2

D 4

Correct Answer

Option D

Solution

To solve this problem, we will use the equations for the range and maximum height of a projectile. The range

R

and maximum height

H

of a projectile launched with speed

u

at an angle $\theta$ can be expressed as follows: Range:

R = \frac{u^2 \sin(2\theta)}{g}

Maximum height:

H = \frac{u^2 \sin^2(\theta)}{2g}

We are given that the range and maximum height are equal. So, we set these equations equal to each other:

\frac{u^2 \sin(2\theta)}{g} = \frac{u^2 \sin^2(\theta)}{2g}

We can cancel out the common terms

u^2

and

g

on both sides:

\sin(2\theta) = \frac{1}{2} \sin^2(\theta)

Using the double angle identity for sine,

\sin(2\theta) = 2 \sin(\theta) \cos(\theta)

, we substitute it into the equation:

2 \sin(\theta) \cos(\theta) = \frac{1}{2} \sin^2(\theta)

We can simplify this by dividing both sides by

\sin(\theta)

(assuming

\theta \neq 0

):

2 \cos(\theta) = \frac{1}{2} \sin(\theta)

Rearranging to get all terms on one side gives us:

4 \cos(\theta) = \sin(\theta)

Dividing both sides by

\cos(\theta)

, we get:

4 = \tan(\theta)

So, we find that:

\tan(\theta) = 4

Thus, the correct answer is: Option D: 4

Q38

Two projectiles thrown at

30^{\circ}

and

45^{\circ}

with the horizontal respectively, reach the maximum height in same time. The ratio of their initial velocities is :

A

1: \sqrt{2}

B

2: 1

C

\sqrt{2}: 1

D

1: 2

Correct Answer

Option C

Solution

To solve this problem, we need to understand the relationship between the angles of projection, the initial velocities, and the time taken to reach maximum height for each projectile.

The formula to calculate the time to reach the maximum height is given by:

t = \frac{u \sin \theta}{g}

where:

t

is the time to reach maximum height,

u

is the initial velocity, $\theta$ is the angle of projection, and

g

is the acceleration due to gravity.

Given that the projectiles reach the maximum height in the same time, we can set up the following equation:

\frac{u_1 \sin 30^\circ}{g} = \frac{u_2 \sin 45^\circ}{g}

Since

\sin 30^\circ = \frac{1}{2}

and

\sin 45^\circ = \frac{\sqrt{2}}{2}

, the equation simplifies to:

\frac{u_1 \cdot \frac{1}{2}}{g} = \frac{u_2 \cdot \frac{\sqrt{2}}{2}}{g}

Canceling out the common terms (i.e.,

g

and

\frac{1}{2}

), we get:

u_1 = u_2 \sqrt{2}

Hence, the ratio of their initial velocities is:

\frac{u_1}{u_2} = \sqrt{2}

Therefore, the correct answer is Option C:

\sqrt{2}:1

.

Q39

Two projectiles are thrown with same initial velocity making an angle of

45^{\circ}

and

30^{\circ}

with the horizontal respectively. The ratio of their respective ranges will be :

A

1: \sqrt{2}

B

\sqrt{2}: 1

C

2: \sqrt{3}

D

\sqrt{3}: 2

Correct Answer

Option C

Solution

Here's how to determine the ratio of the ranges for the two projectiles: Understanding the Concepts Projectile Motion: Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity.

The object is called a projectile, and its path is called its trajectory.

Range: The range of a projectile is the horizontal distance it covers from its launch point to the point where it returns to the same vertical height.

Key Formula The formula for the range (R) of a projectile is:

R = \frac{u^2 \sin 2\theta}{g}

where: R = Range u = Initial velocity (same for both projectiles) θ = Angle of projection g = Acceleration due to gravity (constant) Calculations Projectile 1 (45° angle): Let the range of the projectile launched at 45° be R1.

R_1 = \frac{u^2 \sin (2 \times 45^{\circ})}{g} = \frac{u^2 \sin 90^{\circ}}{g} = \frac{u^2}{g}

Projectile 2 (30° angle): Let the range of the projectile launched at 30° be R2.

R_2 = \frac{u^2 \sin (2 \times 30^{\circ})}{g} = \frac{u^2 \sin 60^{\circ}}{g} = \frac{u^2 \sqrt{3}}{2g}

Ratio of Ranges (R1 : R2): Divide the range of projectile 1 by the range of projectile 2:

\frac{R_1}{R_2} = \frac{\frac{u^2}{g}}{\frac{u^2 \sqrt{3}}{2g}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}

To simplify the ratio, multiply both numerator and denominator by √3:

\frac{R_1}{R_2} = \frac{2\sqrt{3} \times \sqrt{3}}{3 \times \sqrt{3}} = \frac{6}{3\sqrt{3}} = \frac{2}{\sqrt{3}}

Therefore, the ratio of the ranges of the two projectiles is 2 : √3, which corresponds to Option C.

Q40

At time

t=0

a particle starts travelling from a height

7 \hat{z} \mathrm{~cm}

in a plane keeping z coordinate constant. At any instant of time it's position along the

\hat{x}

and

\hat{y}

directions are defined as

3 \mathrm{t}

and

5 \mathrm{t}^{3}

respectively. At t = 1s acceleration of the particle will be

A

-30 \hat{y}

B

30 \hat{y}

C

3 \hat{x}+15 \hat{y}

D

3 \hat{x}+15 \hat{y}+7 \hat{z}

Correct Answer

Option B

Solution

x = 3t \Rightarrow {a_x} = 0

y = 5{t^3} \Rightarrow {a_y} = 30t

\overrightarrow a (t = 1) = 30\widehat y