KSEEB Class 11 Physics Solutions For Chapter 4 Motion In A Plane

KSEEB Class 11 Physics Solutions For Chapter 4 Motion in a Plane Very Short Answer Questions

Question 1. Write the equation for the horizontal – range covered by a projectile and specify
when it will be maximum.
Answer:

Range of a projectile (R) = \(\frac{u^2 \sin 2 \theta}{g}\)

When θ = 45° Range is maximum. (sin 90° = 1)

Maximum Range (R_max) = \(\frac{u^2}{g}\)

Read and Learn More KSEEB Class 11 Physics Solutions

Question 2. The vertical component of a vector is equal to its horizontal component. What is the angle made by the vector with the X-axis?
Answer:

Let R be a vector.

Vertical component = R sin θ;

Horizontal component = R cos θ

∴ R sin θ = R cos θ

So sin θ = cos θ ⇒ θ = 45°

Question 3. A vector V makes an angle θ with the horizontal. The vector is rotated through an angle α. Does this rotation change the vector V?
Answer:

Magnitude of vector = V ;

Let the initial angle with horizontal = θ

Angle rotated = α

So new angle with horizontal = θ + α

Now horizontal component, V_α = V cos (θ + α)

Vertical component, V_y = V sin (θ + α)

Magnitude of vector, V = \(\sqrt{V_x^2+V_y^2}=V\)

So rotating the vector does not change its magnitude.

KSEEB Class 11 Physics Chapter 4 Solutions PDF

Question 4. Two forces of magnitude 3 units and 5 units act at 60 with each other, What is the magnitude of their resultant?
Answer:

Given \(\bar{P}\) = 3 units, \(\bar{Q}\) = 5 units and θ = 60°

∴ R = \(\sqrt{P^2+Q^2+2 P Q \cos \theta}\)

= \(\sqrt{3^2+5^2+2(3)(5) \cos 60^{\circ}}\)

= \(\sqrt{9+25+30 \times \frac{1}{2}}=\sqrt{49}=7 \text { units. }\)

Question 5. A = \(\vec{i}\) + \(\vec{j}\). What is the angle between the vector and the X-axis?
Answer:

Given that, \(\vec{A}\) = \(\vec{i}\) + \(\vec{j}\)

|\(\vec{A}\)| = \(\sqrt{(1)^2+(1)^2}=\sqrt{2}\)

If ‘θ’ is the angle made by the vector with X-axis then, \(\cos \theta=\frac{A_x}{|\vec{A}|} \Rightarrow \cos \theta=\frac{1}{\sqrt{2}} \Rightarrow \theta=45^{\circ}\)

Question 6. When two right-angled vectors of magnitude 7 units and 24 units combine, what is the magnitude of their resultant?
Answer:

Given \(\vec{P}\) = 7 units; \(\vec{Q}\) = 24 units;  θ = 90°

∴ \(\sqrt{\mathrm{P}^2+\mathrm{Q}^2+2 \mathrm{PQ} \cos \theta}=\sqrt{\mathrm{P}^2+\mathrm{Q}^2}\)

∴ \(\mathrm{R}=\sqrt{\mathrm{P}^2+\mathrm{Q}^2}=\sqrt{7^2+24^2}=25\) units.

Karnataka 1st PUC Physics Chapter 4 Notes

Question 7. If \(\vec{P}\) = 2i + 4j + 14k and \(\vec{Q}\) = 4i + 4j + 10k, find the magnitude of \(\vec{P}\) + \(\vec{Q}\).
Answer:

Given \(\overline{\mathrm{P}}=2 \overline{\mathrm{i}}+4 \overline{\mathrm{j}}+14 \overline{\mathrm{k}}\) and

⇒ \(\bar{Q}=4 \overline{\mathrm{i}}+4 \overline{\mathrm{j}}+10 \overline{\mathrm{k}}\)

⇒ \(\overline{\mathrm{P}}+\overline{\mathrm{Q}}=6 \overline{\mathrm{i}}+8 \overline{\mathrm{j}}+24 \overline{\mathrm{k}}\)

Magnitude of \(|\overline{\mathrm{P}}+\overline{\mathrm{Q}}|=\sqrt{6^2+8^2+24^2}\)

= \(\sqrt{36+64+576}=\sqrt{676}=26\) units.

Question 8. Can a vector of magnitude zero have non-zero components?
Answer:

A vector with zero magnitude cannot have non-zero components, because the magnitude of a given vector \(\bar{V}=\sqrt{V_x^2+V_y^2}\) must be zero. This is possible only when \(V_x^2\) and \(V_y^2\) are zero.

Question 9. What is the acceleration of a projectile at the top of its trajectory?
Answer:

At Highest point acceleration, a = g. In projectile motion acceleration will always act towards the centre of the earth. It is irrespective of its position, whether it is at the highest point or somewhere.

Question 10. Can two vectors of unequal magnitude add up to give the zero vector? Can three unequal vectors add up to give the zero vector?
Answer:

No. Two unequal vectors can never give zero vector by addition. But three unequal vectors when added may give zero vector.

Motion In A Plane Solutions KSEEB Class 11 Physics Chapter 4 Short Answer Questions

Question 1. State parallelogram law of vectors. Derive an expression for the magnitude and direction of the resultant vector.
Answer:

Parallelogram Law: If two vectors are represented by the two adjacent sides of a parallelogram, then the diagonal passing through the intersection of given vectors represents their resultant both in direction and magnitude.

Proof: Let \(\overline{\mathrm{P}}\) and \(\overline{\mathrm{Q}}\) be two adjacent vectors ‘θ’ be the angle between them. Construct a parallelogram OACB as shown in the figure.

Extend the line OA and draw a normal D from C. The diagonal OC = the resultant \(\overline{\mathrm{R}}\) both in direction and magnitude

In figure OCD = right angle triangle ⇒ OC² = OD² + DC²

But OD = OA + AD = \(\overline{\mathrm{P}}\) + \(\overline{\mathrm{Q}}\) cos θ and CD = Q sin θ and \(\overrightarrow{\mathrm{OC}}=R\)

∴ \(R^2=[P+Q \cos \theta]^2+[Q \sin \theta]^2\)

⇒ \(\overline{\mathrm{R}}^2=\overline{\mathrm{P}}^2+\overline{\mathrm{Q}}^2 \cos ^2 \theta+2 \overline{\mathrm{PQ}} \cos \theta+\overline{\mathrm{Q}}^2 \sin ^2 \theta\)

∴ \(\overline{\mathrm{R}}^2=\overline{\mathrm{P}}^2+\overline{\mathrm{Q}}^2\left(\sin ^2 \theta+\cos ^2 \theta\right)+2 \overline{\mathrm{PQ}} \cos \theta\)

But \(\sin ^2 \theta+\cos ^2 \theta=1\)

∴ \(\mathrm{R}^2=[\mathrm{P}^2+\mathrm{Q}^2+2 \mathrm{PQ} \cos \theta\).

∴ Resultant vector, \(\overline{\mathrm{R}}=\sqrt{\mathrm{P}^2+\mathrm{Q}^2+2 \mathrm{PQ}}\)

The angle of resultant with adjacent side ‘α’

In triangle OCD, \(\tan \alpha=\frac{C D}{O D}\) (OD = OA + AD)

⇒ \(\text{Tan} \alpha=\frac{C D}{O A+A D}=\frac{Q \sin \theta}{P+Q \cos \theta}\)

(because A D = \(Q \cos \theta\))

Angle of \(\overline{\mathrm{R}}\) with adjacent side, \(\alpha=\tan ^{-1}\left[\frac{Q \sin \theta}{P+Q \cos \theta}\right]\)

Question 2. What is relative motion? Explain it.
Answer:

Relative velocity is the velocity of a body with respect to another moving body.

Relative velocity in two-dimensional motion

Let two bodies A and B are moving with velocities \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) and \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\) then relative velocity of A with respect to B is \(\overrightarrow{\mathrm{V}}_{\mathrm{AB}}=\overrightarrow{\mathrm{V}}_{\mathrm{A}}-\overrightarrow{\mathrm{V}}_{\mathrm{B}}\)

The relative velocity of B with respect to A is \(\overrightarrow{\mathrm{V}}_{\mathrm{BA}}=\overrightarrow{\mathrm{V}}_B-\overrightarrow{\mathrm{V}}_{\mathrm{A}}\)

Procedure To Find Resultant: To find relative velocity in two-dimensional motion use vectorial subtraction of V_A or V_B. Generally to find relative velocity one vector \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) or \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\) is reversed (as the case may be) and parallelogram is constructed. Now resultant of that parallelogram is equal to \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) – \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\) or \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\) – \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) (one vector is reversed VA is taken as –\(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) or \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\) is taken as –\(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\))

In the figure relative velocity of B with respect to A is V_BA = V_R = V_B -V_A

Question 3. Show that a boat must move at an angle with respect to river water in order to cross the river in minimum time.
Answer:

The motion of a boat in a river: Let a boat travel with a speed of V_bE in still water with respect to earth. It is used to cross a river which flows with a speed of V_WE with respect to earth. Let the width of the river be W.

We can cross the river in two different ways.

1. In shortest path
2. In the shortest time.

To cross the river in the shortest time:

Shortest time, t = \(\mathrm{t}=\frac{\text { width of river }}{\text { velocity of boat }}=\frac{W}{V_{\mathrm{bE}}}\)

To cross the river in the shortest time boat must be rowed along the width of the river i.e., the boat must be rowed perpendicular to the bank or 90° with the flow of water.

Because the width of the river is the shortest distance. So velocity must be taken in that direction to obtain the shortest time.

In this case, V_bE and V_WE are perpendicular and the boat will travel along AC. The distance BC is called drift. So to cross the river in the shortest time angle with the flow of water = 90°.

KSEEB Class 11 Physics Motion in a Plane Solutions

Question 4. Define unit vector, null vector and position vector.
Answer:

Unit Vector: A vector whose magnitude is one unit is called a unit vector.

Let \(\overline{\mathrm{a}}\) is a given vector then the unit vector of \(\bar{a}=\frac{\bar{a}}{|\bar{a}|}=\hat{a}\)

When \(\bar{a}\) ≠ 0 or \(\frac{\bar{a}}{|\bar{a}|}\) = unit vector of \(\bar{a}\). It is denoted by a \(\hat{a}\)

Null Vector: A vector whose magnitude is zero is called a null vector. But it has direction.

For a null vector, the origin and terminal point are the same.

Example: Let \(\overline{\mathrm{A}} \times \overline{\mathrm{B}}=\overline{0}\). Here magnitude of \(\overline{\mathrm{A}} \times \overline{\mathrm{B}}=\overline{0}\). But still, it has direction perpendicular to the plane of \(\overline{\mathrm{A}}\) and \(\overline{\mathrm{A}}\).

Position Vector: Any vector in space can be represented by the linear combination of \(\overline{\mathrm{i}}, \overline{\mathrm{j}}\) and \(\overline{\mathrm{k}} \text {. }\).

Let ‘O’ is the origin then \(\overline{\mathrm{OP}}\) is represented as \(\overline{\mathrm{OP}}=x \bar{i}+y \bar{j}+z \bar{k}\) where x, y and z are magnitudes of \(\overline{\mathrm{OP}}\) along \(\bar{i}, \bar{j}\) and \(\overline{\mathrm{k}}\) axis.

Magnitude of \(\overrightarrow{\mathrm{OP}}=\sqrt{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}\)

Question 5. If \(|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|\), prove that the angle between \(\vec{a}\) and \(\vec{b}\) is 90°.
Answer:

Let \(\vec{a}\), \(\vec{b}\) are the two vectors.

Sum of vectors = \(\bar{a}+\bar{b}=\sqrt{a^2+b^2+2 a b \cos \theta}\)

Difference of vectors = \(\bar{a}-\bar{b}=\sqrt{a^2+b^2-2 a b \cos \theta}\)

Given \(|\bar{a}+\bar{b}|=|\bar{a}-\bar{b}|\)

⇒ \(\sqrt{a^2+b^2+2 a b \cos \theta}\)

= \(\sqrt{a^2+b^2-2 a b \cos \theta}\) by squaring on both sides, \(\mathrm{a}^2+\mathrm{b}^2+2 \mathrm{ab} \cos \theta=\mathrm{a}^2+\mathrm{b}^2-2 \mathrm{ab} \cos \theta\)

∴ \(4 \mathrm{ab} \cos \theta=0\) or \(\theta=90^{\circ}\)

So if \(|\bar{a}+\bar{b}|=|\bar{a}-\bar{b}|\) then angle between \(\bar{a}\) and \(\bar{b}\) is \(90^{\circ}\).

Question 6. Show that the trajectory of an object thrown at a certain angle with the horizontal is a parabola.
Answer:

Projectile: A body thrown into the air same angle as the horizontal, (other than 90°) its motion under the influence of gravity is called a projectile. The path followed by it is called a trajectory.

Let a body be projected from point O, with velocity ‘u’ at an angle θ with horizontal. The velocity ‘u’ can be resolved into two rectangular components u_x and u_y along X-axis and Y-axis.

u_x = u cos θ and u_y = u sin θ

After time t, Horizontal distance travelled x = u cos θ t……(1)

After a time sec; vertical displacement y = u sinθ • t – \(\frac{1}{2}\) gt²

y = \(u \sin \theta \cdot t-\frac{1}{2} g t^2\)

From (1) t = \(\frac{x}{u \cos \theta}\)

y = \(\frac{u \sin \theta \cdot x}{u \cos \theta}-\frac{1}{2} g \cdot \frac{x^2}{u^2 \cos ^2 \theta}\)

⇒ \(y=x \tan \theta-\frac{g x^2}{2 u^2 \cos ^2 \theta}\)

Let \(\tan \theta=\mathrm{A}\) and \(\frac{\mathrm{g}}{2 \mathrm{u}^2 \cos ^2 \theta}=\mathrm{B}\) then y = \(A x-B x^2\)

Let \(\tan \theta=\mathrm{A}\) and \(\frac{\mathrm{g}}{2 \mathrm{u}^2 \cos ^2 \theta}=\mathrm{B}\) then \(y=A x-B x^2\)

The above equation represents a “parabola”. Hence the path of a projectile is a parabola.

Question 7. Explain the terms average velocity and instantaneous velocity. When are they equal?
Answer:

Average Velocity: It is the ratio of total displacement to total time taken.

Average velocity = \(\frac{\text { total displacement }}{\text { total time taken }}\)

= \(\frac{x_2-x_1}{t_2-t_1}\)

Average velocity is independent of the path followed by the particle. It just deals with the initial and final positions of the body.

Instantaneous Velocity: The velocity of a body at any particular instant of time is defined as instantaneous velocity.

Mathematically \(\frac{\mathrm{dx}}{\mathrm{dt}}=\underset{\Delta t \rightarrow 0}{\text{Lt}} \frac{\Delta \mathrm{x}}{\Delta \mathrm{t}}\) is defined as instantaneous velocity.

For a body moving with uniform velocity its average velocity = Instantaneous velocity.

Question 8. Show that the maximum height and range of a projectile are \(\frac{\mathrm{U}^2 \sin ^2 \theta}{2 \mathrm{~g}}\) and \(\frac{\mathrm{U}^2 \sin 2 \theta}{\mathrm{g}}\) respectively where the terms have their regular meanings.
Answer:

Let a body be projected with an initial velocity ’u’ and with an angle θ to the horizontal.

Initial velocity along x direction, u_x = u cos θ

Initial velocity along y direction, u_y = u sin θ

Maximum Height: The vertical distance covered by the projectile until its vertical component becomes zero.

from equation v² – u² = 2as ……..(1)

at maximum height v = 0

u = u_y = u sin θ

a = -g

S = H

From equation (1)

0² – u² sin²θ = 2 (- g) H

– u² sin²θ = – 2 gH

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

KSEEB 1st PUC Physics Chapter 4 Important Questions

Horizontal Range: It is the distance covered by the projectile along the horizontal between the point of projection to the point on the ground, where the projectile returns again.

It is denoted by R. The horizontal distance covered by the projectile at the time of flight is called horizontal range.

Therefore, R = u cos θ x t.

R = \(\mathrm{u} \cos \theta \times \frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}=\frac{\mathrm{u}^2(2 \sin \theta \cos \theta)}{\mathrm{g}}\),

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

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

Angle Of Projection For Maximum Range: For a given velocity of projection, the horizontal range will be maximum, when sin 2θ = 1.

∴ Angle of projection for maximum range is 2θ = 90° or θ = 45°

∴ \(R_{\max }=\frac{u^2}{g}\)

Question 9. If the trajectory of a body IN parabolic In one reference frame, can It be parabolic in another reference frame that moves at constant velocity with respect to the first reference frame? If the trajectory can be other than parabolic, what else can it be?
Answer:

Yes. According to Newton’s first law, a body at rest or a body moving with uniform velocity is treated as the same. Both of them belong to an inertial frame of reference.

If a frame (say 1) is moving with uniform velocity with respect to others, then that second frame must be at rest or it maintains a constant velocity concerning the first.

So both frames are inertial frames. So if the trajectory of a body in one frame is a parabola, then the trajectory of that body in another frame is also a parabola.

Question 10. A force 2i + j – k Newton acts on a body which is initially at rest. At the end of 20 seconds, the velocity of the body is 4i + 2j – 2k ms^-1. What is the mass of the body?
Answer:

Force, F = 2i + j – k

time, t = 20

Initial velocity, u = 0

Final velocity, v = 4i + 2j – 2k = 2(2i + j – k)

Acceleration, \(\mathrm{a}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}\)

= \(\frac{2(2 i+j-k)-0}{20}=\frac{2 i+j-k}{10}\)

Mass of the body, m = \(\frac{F}{a}\)

= \(\frac{2 i+j-k}{(2 i+j-k) / 10}=10 \mathrm{~kg}\)

Chapter 4 Motion In-Plane Problems In KSEEB Physics Class 11

Question 1. Ship A is 10 km due west of ship B. Ship A is heading directly north at a speed of 30 km/h, while ship B is heading in a direction 60° west of north at a speed of 20 km/h.

1. Determine the magnitude of the velocity of ship B relative to ship A.
2. What will be their distance of closest approach?

Answer:

The velocity of A 30 kmph due North

∴ V_A=30 \(\hat{j}\)

Velocity of B = 20 kmph 60° west of North

∴ \(V_B=-20 \sin 60^{\circ}+20 \cos 60^{\circ}=10 \sqrt{3} \hat{\mathrm{i}}+10 \hat{\mathrm{j}}\)

Velocity of B with respect to A = V_B – V_A

= \(-10 \sqrt{3} \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-30 \hat{\mathrm{j}}=-10 \sqrt{3} \hat{\mathrm{i}}-20 \hat{\mathrm{j}}\)

∴ \(\mathrm{V}_R=\left|\mathrm{V}_B-\mathrm{V}_A\right|=\sqrt{100 \times 3+400}\)

= \(10 \sqrt{7} \mathrm{kmph}\)

Shortest Distance: In Δ le ANB shortest distance, AN = AB sin θ

But distance, AB = 10 km

∴ AN = \(10 \times \frac{20}{10 \sqrt{7}}=\frac{20}{\sqrt{7}}=7.56 \mathrm{~km}\)

Karnataka Board Class 11 Physics Chapter 4 MCQs

Question 2. If θ is the angle of projection, R is the range, h is the maximum height, and T is the time of flight, then show that

1. tan θ = 4h/R and
2. h = gT²/8

Answer:

1. Given angle of projection = η,

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

h = \(\mathrm{h}_{\max }=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}} \text {. }\)

Time of flight \(\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}\)

\(\frac{\mathrm{h}}{\mathrm{R}}=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}} \times \frac{\mathrm{g}}{\mathrm{u}^2 \sin 2 \theta}\)

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

∴ \(\frac{h}{R}=\frac{\tan \theta}{4} \Rightarrow \tan \theta=\frac{4 h}{R}\)

2. \(\mathrm{h}=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}\)

But \(\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}} \Rightarrow \frac{\mathrm{u} \sin \theta}{\mathrm{g}}=\mathrm{T}/2\).

multiply with (g/g)

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

= \(\left(\frac{T}{2}\right)^2 \times \frac{g}{2}=\frac{g T^2}{8}\)

∴ \(\mathrm{h}=\frac{\mathrm{gT}^2}{8}\) is proved.

Question 3. A projectile is fired at an angle of 60° to the horizontal with an initial velocity of 800 m/s:

1. Find the time of flight of the projectile before it hits the ground.
2. Find the distance it travels before it hits the ground (range).
3. Find the time of flight for the projectile to reach its maximum height.

Answer:

Angle of projection, θ = 60°,

Initial velocity, u = 800 m/s

1. Time of flight, \(\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}\)

= \(\frac{2 \times 800}{9.8} \times \frac{\sqrt{3}}{2}=\frac{800 \times 1.732}{9.8}=141.4 \mathrm{sec}\)

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

= \(\frac{800 \times 800 \times \sin \left(2 \times 60^{\circ}\right)^{\prime}}{9.8}\)

R = \(\frac{800 \times 800}{9.8} \frac{\sqrt{3}}{2}\)

= \(\frac{800 \times 800}{9.8} \times 0.8660=56.56 \mathrm{~km}\)

3. Time of flight to reach maximum height = \(\frac{T}{2}\)

\(\mathrm{T}_1=\frac{2 \mathrm{u} \sin \theta}{2 \mathrm{~g}}=\frac{\mathrm{u} \sin \theta}{\mathrm{g}}=\frac{800 \sin 60^{\circ}}{9.8}\)

= \(\frac{800 \times 0.8660}{9.80}=70.7 \mathrm{sec}\)

Motion In A Plane Problems In KSEEB Physics

Question 4. For a particle projected slantwise from the ground, the magnitude of its position vector with respect to the point of projection, when it is at the highest point of the path is found to be √2 times the maximum height reached by it. Show that the angle of projection is tan^-1 √2.
Answer:

The position vector of h (max point) from 0, is \(\frac{\mathrm{R}}{2}=\frac{1}{2} \frac{\mathrm{u}^2 \sin 2 \theta}{\mathrm{g}}\)

But R/2 = √2 h_max(given)

∴ \(\frac{1}{2} \frac{u^2 \sin 2 \theta}{g}=\frac{\sqrt{2} \mathrm{u}^2 \sin ^2 \theta}{2 g} \)

∴ \(2 \sin \theta \cos \theta=\sqrt{2} \sin \theta \cdot \sin \dot{ }\)

⇒ \(\frac{2}{\sqrt{2}}=\frac{\sin \theta}{\cos \theta} \Rightarrow \tan \theta=\sqrt{2}\)

∴ Angle of projection, \(\theta=\tan ^{-1} \sqrt{2}\)

Question 5. An object is launched from a cliff 20 m above the ground at an angle of 30° above the horizontal with an initial speed of 30 m/s. How far horizontally does the object travel before landing on the ground? (g = 10 m/s²).
Answer:

Height of cliff = 20m

The angle of projection, θ = 30°

Velocity of projection, u = 30 m/s

Total horizontal distance travelled = OC = OB’+ B’C

But OB’ = AB = Range R = \(\frac{u^2 \sin 2 \theta}{\mathrm{g}}\)

∴ R = \(\frac{30 \times 30 \sin 60^{\circ}}{10}=90 \frac{\sqrt{3}}{2}=45 \sqrt{3} \rightarrow(1)\)

2. Distance B’C = Range of a horizontal projectile.

∴ Range = u cos θ t

u \(\cos \theta=30 \cdot \frac{\sqrt{3}}{2}=15 \sqrt{3}\)

Time taken to reach the ground, t =?

Given \(S_y=20, \quad u_y=u \sin \theta=30 \sin 30^{\circ}\)

= \(15 \mathrm{~m} / \mathrm{s}\)

from \(\mathrm{s}=\mathrm{ut}+\frac{1}{2} a t^2\)

∴ \(\mathrm{s}_{\mathrm{y}}=20=15 \mathrm{t}+\frac{10}{2} \mathrm{t}^2 \Rightarrow 5 \mathrm{t}^2+15 \mathrm{t}-20=0\)

or \(t^2+3 t-4=0\) or \((t+4)(t-1)=0\)

∴ \(\mathrm{t}=-4\) or \(\mathrm{t}=1\)

∴ \(\mathrm{t}\) is Not -ve use t=1

∴ Range = \(\mathrm{u} \cdot \cos \theta \cdot \mathrm{t}=15 \sqrt{3} \times 1 \rightarrow(2)\)

Total distance travelled before reaching the equation (1) + (2) ground = 45√3 + 15√3 =60√3 m.

Question 6. ‘O’ is a point on the ground chosen as origin. A body first suffers a displacement of 10√2 mm North-East, next 10 m North and finally 10√2 North-West. How far it is from its origin?
Answer:

1. 10√2 m North-East
2. 10 m North
3. 10√2 m North West

From the figure total displacement from the origin, ‘O’ is OC

But OC = OA’ + A’B’ + B’C = 10 + 10 + 10 = 30 m.

Question 7. From a point on the ground, a particle is projected with initial velocity u, such that its horizontal range is maximum. Find the magnitude of average velocity during its ascent.
Answer:

Velocity of projection = u.

Range is maximum ⇒ θ = 45°

During time of ascent ⇒ when h = h_max ⇒ u_x = v_x= u · cos θ

Average velocity, \(V_A=\sqrt{V_x^2+V_y^2}\)

V_x = Average velocity along X-axis = \(\frac{\mathrm{u} \cdot \cos \theta+\mathrm{u} \cos \theta}{2}\)

⇒ \(\mathrm{v}_{\mathrm{x}}=\mathrm{u} \cdot \cos \theta=\mathrm{u} \cdot \cos 45^{\circ}=\mathrm{u} / \sqrt{2} \longrightarrow(1)\)

Average velocity along \(\mathrm{Y} \text {-axis }=\frac{\mathrm{u}_{\mathrm{y}}+\mathrm{v}_{\mathrm{y}}}{2}\)

⇒ \(\mathrm{u}_{\mathrm{y}}=\mathrm{u} \sin \theta=\mathrm{u} / \sqrt{2}, \mathrm{v}_{\mathrm{y}}=0\left(\text { at }_{\mathrm{h}_{\max }}\right)\)

∴ \(\mathrm{v}_{\mathrm{y}}=\frac{\mathrm{u} / \sqrt{2}}{2}=\frac{\mathrm{u}}{2 \sqrt{2}} \longrightarrow(2)\)

Average velocity during the time of ascent

= \(\sqrt{\left(\frac{u}{\sqrt{2}}\right)^2+\left(\frac{u}{2 \sqrt{2}}\right)^2}\)

Average velocity

= \(\sqrt{\frac{u^2}{2}+\frac{u^2}{4 \times 2}}=\sqrt{\frac{4 u^2+u^2}{8}}=\frac{u \sqrt{5}}{2 \sqrt{2}}\)

Question 8. A particle is projected from the ground with some initial velocity making an angle of 45° with the horizontal. It reaches a height of 7.5 m above the ground while it travels a horizontal distance of 10 m from the point of projection. Find the initial speed of projection (g = 10 m/s²).
Answer:

Angle of projection = 45°

Vertical height, h_y = 7.5 m

Horizontal distance, h_x = 10 m

From figure \(\tan \theta=\frac{y}{x}=\frac{7.5}{10}=3 / 4, \cos \theta=\frac{4}{5}\)

⇒ \(\mathrm{s}_{\mathrm{x}}=\mathrm{u} \cdot \cos \theta \cdot \mathrm{t} \Rightarrow 10=\mathrm{u} \cdot \frac{1}{\sqrt{2}} \cdot \mathrm{t}\)

t = \(\frac{10 \sqrt{2}}{\mathrm{u}}\)…..(1)

⇒ \(\mathrm{s}_{\mathrm{y}}=7.5=(\mathrm{u} \sin \theta) \mathrm{t}-\frac{1}{2} g \mathrm{t}^2\)

7.5 = \(\frac{\mathrm{u}}{\sqrt{2}} \frac{10 \sqrt{2}}{\mathrm{u}}-\frac{1}{2} \times \frac{10 \times 10 \times 10 \times 2}{\mathrm{u}^2}\)

7.5 = \(10-\frac{1000}{\mathrm{u}^2} \Rightarrow-2.5=\frac{-1000}{\mathrm{u}^2}\)

∴ \(\mathrm{u}^2=\frac{1000}{2.5}=400\)

∴ \(u^2=400 \Rightarrow u=20 \mathrm{~m} / \mathrm{s}\)

Question 9. The wind is blowing from the south at 5 ms^-1. To a cyclist, It appears to be blowing from the cast at 5 ms^-1. Show that the velocity of the cyclist is ms^-1 towards the north-cast.
Answer:

The direction of wind is South to North 5 m/s.

The apparent direction is from East to West at 5 m/s.

This is relative velocity.

To find the velocity of the cyclist reverse the direction of the resultant vector OB and find the resultant

∴ Velocity of cyclist = \(\sqrt{5^2+5^2+0}\)

= \(5 \sqrt{2} \mathrm{~m}\)

Question 10. A person walking at 4 m/s finds raindrops falling slantwise into his face at a speed of 4 m/s at an angle of 30° with the vertical. Show that the actual speech of the raindrops is 4 m/s.
Answer:

Velocity of man = 4 m/sec

The apparent velocity of raindrop = 4 m/sec with θ = 30° with vertical.

This is relative velocity V_B.

1. Velocity of man = 4 \(\hat{i}\)

Velocity of rain = 4 m/s, 30° with vertical

∴ It is represented by = -4 sin 30 i – 4.cos 30° \(\hat{j}\)

= \(-4 \cdot \frac{1}{2} \hat{\mathrm{i}}+4 \frac{\sqrt{3}}{2} \text { say } V_R=-2 \hat{\mathrm{i}}+2 \sqrt{3} \hat{\mathrm{j}}\)

But V_R = V_B – V_A where V_B is the actual velocity of rain

⇒ \(V_B-V_A+V_A=V_B\)

= \(-2 \hat{i}+2 \sqrt{3} \hat{j}+4 \hat{\mathrm{i}}=2 \hat{\mathrm{i}}+2 \sqrt{3} \hat{\mathrm{j}}\)

∴ \(\left|V_B\right|=\sqrt{4+4 \times 3}=\sqrt{16}=4 \mathrm{~m} / \mathrm{s}\)

Question 11. The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 ms_^-1 can go without hitting the ceiling of the hall?
Solution:

Here, u = 40 ms^-1; H = 25m, R = ?

Let θ be the angle of projection with the horizontal direction to have the maximum range, with maximum height = 25 m.

Maximum height, \(\mathrm{H} \Rightarrow 25=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}\)

= \(\frac{(40)^2 \sin ^2 \theta}{2 \times 9.8}\)

or \(\sin \theta=\left(\frac{25 \times 2 \times 9.8}{40^2}\right)=0.5534\)

= \(\sin 33.6^{\circ}\) or \(\theta=33.6^{\circ}\)

Horizontal range, \(\mathrm{R}=\frac{\mathrm{u}^2 \sin 2 \theta}{\mathrm{g}}\)

= \(\frac{(40)^2 \times \sin \left(2 \times 33.6^{\circ}\right)}{9.8}\)

= \(\frac{1600 \times 0.9219}{9.8}=150.5 \mathrm{~m}\)

Question 12. A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 seconds, what is the magnitude and direction of acceleration of the stone?
Solution:

Here, r = 80 cm = 0.8 m; v = 14/25 s^-1.

∴ \(\omega=2 \pi v=2 \times \frac{22}{7} \times \frac{14}{25}=\frac{88}{25} \text { rad. s } \mathrm{s}^{-1} \text {. }\)

The centripetal acceleration, \(\mathrm{a}=\omega^2 \mathrm{r}=\left(\frac{88}{25}\right)^2 \times 0.80=9.90 \mathrm{~m} / \mathrm{s}^2\)

The direction of centripetal acceleration is along the string directed towards the centre of the circular path.

Class 11 Physics Motion in a Plane KSEEB Guide

Question 13. An aircraft executes a horizontal loop of radius 1 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.
Solution:

Here, r = 1 km = 1000 m;

v = 900 km h^-1 = 900 x (1000m) x (60 x 60s)^-1 = 250 ms^-1

Centripetal acceleration, \(\mathrm{a}=\frac{\mathrm{v}^2}{\mathrm{r}}=\frac{(250)^2}{1000}\)

Now, \(\frac{\mathrm{a}}{\mathrm{g}}=\frac{(250)^2}{1000} \times \frac{1}{9.8}=6.38\)

Question 14. An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is 30°, what is the speed of the aircraft?
Solution:

Figure O is the observation point at the ground. A and B are the positions of aircraft for which ∠AOB = 30°.

Draw a perpendicular OC on AB. Here OC = 3400 m and ∠AOC = ∠COB = 15°. Time taken by aircraft from A to B is 10 s.

In ΔAOC, AC = OC tan 15° = 3400 x 0.2679

AC = 910.86 m.

AB = AC + CB = AC + AC = 2 AC

= 2 x 910.86 m

Speed of the aircraft, v = \(\frac{\text { distance } A B}{\text { time }}\)

= \(\frac{2 \times 910.86}{10}=182.17 \mathrm{~ms}^{-1}=182.2 \mathrm{~ms}^{-1}\).

KSEEB Class 11 Physics Motion in a Plane Numericals

Question 15. A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away? Assume the muzzle speed to be fixed and neglect air resistance.
Solution:

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

or, 3 = \(\frac{u^2 \sin 60^{\circ}}{g}\) or \(\frac{u^2}{g}=2 \sqrt{3}\)

3 = \(\frac{u^2}{g} \times \frac{\sqrt{3}}{2}\)

2\(\sqrt{3}=\frac{u^2}{g}\)

∴ \(R_{\max }=2 \sqrt{3}\)

∴ \(R_{\max }=3.464 \mathrm{~m} \).