Motion in a Plane — Page 6 | JEE Physics

Q51

The angle of projection for a projectile to have same horizontal range and maximum height is :

A

\tan ^{-1}\left(\dfrac{1}{2}\right)

B

\tan ^{-1}(2)

C

\tan ^{-1}\left(\dfrac{1}{4}\right)

D

\tan ^{-1}(4)

Correct Answer

Option D

Solution

To find the angle of projection for a projectile to have the same horizontal range and maximum height, we need to express both the range and the maximum height in terms of the projectile's initial velocity and the angle of projection, and then set them equal to each other.

The formula for the horizontal range $R$ of a projectile is given by: $R = \dfrac{v^2}{g} \sin 2\theta,$ where $v$ is the initial velocity of the projectile, $g$ is the acceleration due to gravity, and $\theta$ is the angle of projection.

The formula for the maximum height $H$ reached by the projectile is: $H = \dfrac{v^2}{2g} \sin^2\theta.$ To have the same numerical value for $R$ and $H$ , we set them equal to each other: $\dfrac{v^2}{g} \sin 2\theta = \dfrac{v^2}{2g} \sin^2\theta.$ Simplifying this equation, we get: $2 \sin 2\theta = \sin^2\theta.$ Using the double-angle formula, $\sin 2\theta = 2 \sin\theta \cos\theta$ , we can rewrite the equation as: $4 \sin\theta \cos\theta = \sin^2\theta.$ This can be simplified further to: $4 \sin\theta \cos\theta = (\sin\theta)^2.$ Dividing both sides by $\sin\theta$ (assuming $\sin\theta \neq 0$ ), we get: $4 \cos\theta = \sin\theta.$ Now, dividing both sides by $\cos\theta$ , we have: $4 = \tan\theta.$ So, the angle of projection $\theta$ is: $\theta = \tan^{-1}(4).$ Therefore, the correct answer is: Option D

\tan ^{-1}(4)

.

Q52

Two projectiles are fired with same initial speed from same point on ground at angles of

(45^\circ - \alpha)

and

(45^\circ + \alpha)

, respectively, with the horizontal direction. The ratio of their maximum heights attained is :

A

\dfrac{1+\sin\alpha}{1-\sin\alpha}

B

\dfrac{1+\sin2\alpha}{1-\sin2\alpha}

C

\dfrac{1-\tan\alpha}{1+\tan\alpha}

D

\dfrac{1-\sin2\alpha}{1+\sin2\alpha}

Correct Answer

Option D

Solution

\begin{aligned} & \mathrm{H}_{\mathrm{Max}}=\frac{(\mathrm{usin} \theta)^2}{2 \mathrm{~g}} \\ & \frac{\left(\mathrm{H}_{\max }\right)_1}{\left(\mathrm{H}_{\max }\right)_2}=\frac{\mathrm{u}^2 \sin ^2(45-\alpha)}{\mathrm{u}^2 \sin ^2(45+\alpha)} \\ & =\frac{\left(\frac{1}{\sqrt{2}} \cos \alpha-\frac{1}{\sqrt{2}} \sin \alpha\right)^2}{\left(\frac{1}{\sqrt{2}} \cos \alpha+\frac{1}{\sqrt{2}} \sin \alpha\right)^2} \\ & =\frac{1-\sin 2 \alpha}{1+\sin 2 \alpha} \end{aligned}

Q53

A ball of mass 100 g is projected with velocity

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

at

60^{\circ}

with horizontal. The decrease in kinetic energy of the ball during the motion from point of projection to highest point is

A 20 J

B 5 J

C 15 J

D zero

Correct Answer

Option C

Solution

\begin{aligned} & \mathrm{k}_{\mathrm{i}}=\frac{1}{2} \mathrm{mv}^2 \\ & \mathrm{k}_{\mathrm{f}}=\frac{1}{2} \mathrm{~m}\left(\mathrm{v} \cos 60^{\circ}\right)^2=\frac{1}{8} \mathrm{mv}^2 \\ & \Delta \mathrm{k}=\mathrm{k}_{\mathrm{i}}-\mathrm{k}_{\mathrm{f}}=\frac{3}{8} \mathrm{mv}^2=\frac{3}{8} \times 0.1 \times 400=15 \mathrm{~J} \end{aligned}

Q54

An object of mass ' m ' is projected from origin in a vertical xy plane at an angle

45^{\circ}

with the

\mathrm{x}-

axis with an initial velocity

\mathrm{v}_0

. The magnitude and direction of the angular momentum of the object with respect to origin, when it reaches at the maximum height, will be [ g is acceleration due to gravity]

A

\dfrac{m v_o{ }^3}{2 \sqrt{2} g}

along negative

z

-axis

B

\dfrac{m v_o^3}{2 \sqrt{2} g}

along positive

z

-axis

C

\dfrac{m v_o^3}{4 \sqrt{2} g}

along positive

z

-axis

D

\dfrac{m v_o^3}{4 \sqrt{2} g}

along negative z-axis

Correct Answer

Option D

Solution

\vec{v_0} = \frac{v_0}{\sqrt{2}}\,\hat{i} + \frac{v_0}{\sqrt{2}}\,\hat{j}.

At maximum height, the vertical component of the velocity becomes zero. Setting the vertical velocity to zero:

\frac{v_0}{\sqrt{2}} - g\,t = 0 \quad \Longrightarrow \quad t = \frac{v_0}{g\sqrt{2}}.

The coordinates at this time are:

x = \frac{v_0}{\sqrt{2}}\,t = \frac{v_0}{\sqrt{2}} \cdot \frac{v_0}{g\sqrt{2}} = \frac{v_0^2}{2g},

y = \frac{v_0}{\sqrt{2}}\,t - \frac{1}{2}g\,t^2 = \frac{v_0^2}{2g} - \frac{1}{2}g\,\frac{v_0^2}{2g^2} = \frac{v_0^2}{2g} - \frac{v_0^2}{4g} = \frac{v_0^2}{4g}.

Thus, the position vector at maximum height is:

\vec{r} = \frac{v_0^2}{2g}\,\hat{i} + \frac{v_0^2}{4g}\,\hat{j}.

At maximum height, the only nonzero component of velocity is the horizontal one:

\vec{v} = \frac{v_0}{\sqrt{2}}\,\hat{i}.

The angular momentum about the origin is given by:

\vec{L} = m\,\vec{r} \times \vec{v}.

Writing out the cross product:

\vec{r} \times \vec{v} = \left(\frac{v_0^2}{2g}\,\hat{i} + \frac{v_0^2}{4g}\,\hat{j}\right) \times \frac{v_0}{\sqrt{2}}\,\hat{i}.

Since the cross product of $\hat{i}$ with itself is zero and:

\hat{j} \times \hat{i} = -\hat{k},

we have:

\vec{r} \times \vec{v} = \frac{v_0^2}{4g} \cdot \frac{v_0}{\sqrt{2}}\, (-\hat{k}) = -\frac{v_0^3}{4\sqrt{2}g}\,\hat{k}.

Thus, the angular momentum is:

\vec{L} = -\frac{m\,v_0^3}{4\sqrt{2}g}\,\hat{k}.

This means the magnitude is:

\frac{m\,v_0^3}{4\sqrt{2}g},

and its direction is along the negative

\hat{k}

(or negative

z

-axis).

Q55

The position vector of a moving body at any instant of time is given as

\overrightarrow{\mathrm{r}}=\left(5 \mathrm{t}^2 \hat{i}-5 \mathrm{t} \hat{j}\right) \mathrm{m}

. The magnitude and direction of velocity at

t=2 s

is,

A

5 \sqrt{17} \mathrm{~m} / \mathrm{s}

, making an angle of

\tan ^{-1} 4

with - ve Y axis

B

5 \sqrt{15} \mathrm{~m} / \mathrm{s}

, making an angle of

\tan ^{-1} 4

with + ve

X

axis

C

5 \sqrt{17} \mathrm{~m} / \mathrm{s}

, making an angle of

\tan ^{-1} 4

with + ve

X

axis

D

5 \sqrt{15} \mathrm{~m} / \mathrm{s}

, making an angle of

\tan ^{-1} 4

with

-

ve

Y

axis

Correct Answer

Option A

Solution

\begin{aligned} & \overrightarrow{\mathrm{r}}=5 \mathrm{t}^2 \hat{\mathrm{i}}-5 \hat{\mathrm{t}} \\ & \overrightarrow{\mathrm{v}}=10 \mathrm{t} \hat{\mathrm{i}}-5 \hat{\mathrm{j}} \\ & \overrightarrow{\mathrm{v}}=20 \hat{\mathrm{i}}-5 \hat{\mathrm{j}} \quad \text{ at } \mathrm{t}=2 \mathrm{sec} \end{aligned}

\begin{aligned} & \tan \theta=\frac{20}{5}=4 \\ & \theta=\tan ^{-1} 4 \\ & \text{ From-veY-axis } \end{aligned}

Q56

Two balls with same mass and initial velocity, are projected at different angles in such a way that maximum height reached by first ball is 8 times higher than that of the second ball.

T_1

and

T_2

are the total flying times of first and second ball, respectively, then the ratio of

T_1

and

T_2

is

A 2 : 1

B

\sqrt{2} : 1

C 2

\sqrt{2} : 1

D 4 : 1

Correct Answer

Option C

Solution

Given that the maximum height reached by the first ball is 8 times that of the second ball: $\left(H_{\max}\right)_1 = 8 \times \left(H_{\max}\right)_2$ We can relate this to the initial velocities and angles using the formula for maximum height: $\dfrac{u^2 \sin^2 \theta_1}{2g} = 8 \times \dfrac{u^2 \sin^2 \theta_2}{2g}$ Simplifying, we find: $\sin \theta_1 = 2 \sqrt{2} \sin \theta_2$ Now, the ratio of the total flying times $T_1$ and $T_2$ is given by: $\dfrac{T_1}{T_2} = \dfrac{2u \sin \theta_1 / g}{2u \sin \theta_2 / g} = \dfrac{\sin \theta_1}{\sin \theta_2} = 2 \sqrt{2}$ Thus, the ratio of $T_1$ to $T_2$ is $2 \sqrt{2} : 1$ .

Q57

A river is flowing from west to east direction with speed of

9 \mathrm{~km} \mathrm{~h}^{-1}

. If a boat capable of moving at a maximum speed of

27 \mathrm{~km} \mathrm{~h}^{-1}

in still water, crosses the river in half a minute, while moving with maximum speed at an angle of

150^{\circ}

to direction of river flow, then the width of the river is :

A 112.5 m

B 75 m

C 300 m

D

112.5 \times \sqrt{3} \mathrm{~m}

Correct Answer

Option A

Solution

Step 1: Find the boat’s speed across the river The boat's maximum speed is $27~\mathrm{km/h}$ , and it makes an angle of $150^\circ$ with the direction of the river.

The part of the boat’s speed that helps it cross the river straight (perpendicular to the river) is found by: $27 \times \cos 60^\circ$ .

Since $\cos 60^\circ = \dfrac{1}{2}$ , this gives: $27 \times \dfrac{1}{2} = \dfrac{27}{2}~\mathrm{km/h}$ .

Step 2: Change speed to meters per second $\dfrac{27}{2}~\mathrm{km/h} = \dfrac{27}{2} \times \dfrac{1000}{3600} = \dfrac{27}{2} \times \dfrac{5}{18}~\mathrm{m/s}$ .

Step 3: Use the time the boat takes The boat takes $30$ seconds to cross.

Distance = speed × time So the width of the river is:

\mathrm{S} = \frac{27}{2} \times \frac{5}{18} \times 30~\mathrm{m}

Step 4: Do the calculation Calculate the value: $\dfrac{27}{2} \times \dfrac{5}{18} \times 30 = 112.5~\mathrm{m}$ So, the width of the river is $112.5$ meters.

Q58

Two projectiles are fired from ground with same initial speeds from same point at angles

\left(45^{\circ}+\right.

\alpha)

and

\left(45^{\circ}-\alpha\right)

with horizontal direction. The ratio of their times of flights is

A

\dfrac{1+\tan \alpha}{1-\tan \alpha}

B

\dfrac{1+\sin 2 \alpha}{1-\sin 2 \alpha}

C

\dfrac{1-\tan \alpha}{1+\tan \alpha}

D 1

Correct Answer

Option A

Solution

To find the ratio of the times of flight for two projectiles fired from the ground with the same initial speed but at different angles, we follow these steps: Projectile Angles: The first projectile is launched at an angle $\theta_1 = 45^\circ + \alpha$ .

The second projectile is launched at an angle $\theta_2 = 45^\circ - \alpha$ .

Time of Flight Formula: The time of flight $T$ for a projectile is given by: $T = \dfrac{2v \sin \theta}{g}$ where $v$ is the initial speed and $g$ is the acceleration due to gravity.

Ratio of Times of Flight: The ratio of the times of flights $\dfrac{T_1}{T_2}$ can be determined as follows: $\dfrac{T_1}{T_2} = \dfrac{\sin(45^\circ + \alpha)}{\sin(45^\circ - \alpha)}$ Applying Trigonometric Identities: Use the sine addition and subtraction formulas: $\sin(45^\circ + \alpha) = \dfrac{1}{\sqrt{2}}(\cos \alpha + \sin \alpha)$ $\sin(45^\circ - \alpha) = \dfrac{1}{\sqrt{2}}(\cos \alpha - \sin \alpha)$ Simplifying the Ratio: Substitute the expressions for $\sin(45^\circ + \alpha)$ and $\sin(45^\circ - \alpha)$ into the ratio: $\dfrac{T_1}{T_2} = \dfrac{\dfrac{1}{\sqrt{2}} (\cos \alpha + \sin \alpha)}{\dfrac{1}{\sqrt{2}} (\cos \alpha - \sin \alpha)} = \dfrac{\cos \alpha + \sin \alpha}{\cos \alpha - \sin \alpha}$ Factor out the trigonometric functions to express in terms of tangent: $\dfrac{T_1}{T_2} = \dfrac{1 + \tan \alpha}{1 - \tan \alpha}$ Therefore, the ratio of the times of flights for the two projectiles is $\dfrac{1 + \tan \alpha}{1 - \tan \alpha}$ .

Q59

A helicopter flying horizontally with a speed of 360 km/h at an altitude of 2 km, drops an object at an instant. The object hits the ground at a point O, 20 s after it is dropped. Displacement of 'O' from the position of helicopter where the object was released is : (use acceleration due to gravity g = 10 m/s2 and neglect air resistance)

A 7.2 km

B 2

\sqrt{5}

km

C 2

\sqrt{2}

km

D 4 km

Correct Answer

Option C

Solution

\begin{aligned} & u=360 \times \frac{5}{18}=100 \mathrm{~m} / \mathrm{s} \\ & \mathrm{x}=\mathrm{u} \times \mathrm{t}=2 \times 10^3 \mathrm{~m} \\ & \mathrm{t}=\sqrt{\frac{2 \mathrm{H}}{\mathrm{~g}}} \Rightarrow \mathrm{H}=\frac{\mathrm{t}^2 \mathrm{~g}}{2} \\ & \mathrm{H}=\frac{400 \times 10}{2} \\ & \mathrm{H}=2000 \mathrm{~m} \\ & \mathrm{D}=\sqrt{\mathrm{x}^2+\mathrm{H}^2} \\ & \mathrm{D}=2 \sqrt{2} \mathrm{~km} \end{aligned}

Q60

A particle is projected with velocity

u

so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as

\dfrac{n u^2}{25 g}

, where value of

n

is: (Given, '

g

' is the acceleration due to gravity.)

A 6

B 12

C 18

D 24

Correct Answer

Option D

Solution

Given that the horizontal range of a projectile is three times its maximum height, we need to determine the value of $n$ in the expression for the range $\dfrac{n u^2}{25 g}$ .

Let's start by setting up the equations for the range and maximum height of a projectile.

The formula for the range $R$ is: $R = \dfrac{u^2 \sin 2\theta}{g}$ The formula for the maximum height $H_{\max}$ is: $H_{\max} = \dfrac{u^2 \sin^2 \theta}{2g}$ According to the problem, the range $R$ is three times the maximum height $H_{\max}$ : $R = 3H_{\max}$ Substituting the formulas for $R$ and $H_{\max}$ into this condition gives: $\dfrac{u^2 \sin 2\theta}{g} = 3 \cdot \dfrac{u^2 \sin^2 \theta}{2g}$ Simplifying this equation, we get: $2 \sin \theta \cos \theta = \dfrac{3}{2} \sin^2 \theta$ Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$ , we can rewrite: $2 \sin \theta \cos \theta = \sin 2\theta$ This gives us: $\sin 2\theta = \dfrac{3}{2} \sin^2 \theta$ Dividing both sides by $\sin \theta$ , assuming $\sin \theta \neq 0$ , we have: $2 \cos \theta = \dfrac{3}{2} \sin \theta$ Rearranging the terms to solve for $\tan \theta$ , we get: $\tan \theta = \dfrac{4}{3}$ Thus, the angle $\theta$ where $\tan \theta = \dfrac{4}{3}$ corresponds to $\theta = 53^\circ$ .

Now, substituting back into the range formula: $R = \dfrac{u^2 \cdot 2 \cdot \dfrac{3}{5} \cdot \dfrac{4}{5}}{g}$ The calculation yields: $R = \dfrac{24 u^2}{25 g}$ Therefore, the value of $n$ in the expression $\dfrac{n u^2}{25 g}$ is $n = 24$ .