NCERT Solutions for Class 11 Physics Chapter 4 Motion In A Plane

NCERT Solutions for Class 11 Physics Chapter 4 Motion in a Plane are very important for your school exams. The majority of questions asked in tests are from the NCERT textbook only. That's why, to score well, you must learn these class 11 Physics Motion in a Plane exercise solutions. The "Motion in a Plane" chapter teaches the basics of two-dimensional motion, which is an important part of Physics. It helps you understand how objects move when they are not limited to a straight line. 

This chapter also explains vectors and scalars, addition of vectors using triangle, parallelogram, and polygon laws, and breaking vectors into rectangular components. In chapter 3, "Motion In A Straight Line," you studied motion in a straight line, but here you will learn about motion in two directions at the same time. NCERT Solutions for Class 11 Physics Chapter 4 also help you identify all of these key topics so that you do not miss learning any important concept from the chapter.

Physics in Class 11 is also important because many topics from this subject are also asked in competitive exams like JEE and NEET. That is why Class 11 Physics Motion in a Plane exercise solutions are so helpful. They provide step-by-step answers for both theory and numerical questions, making it easier for you to understand the concepts clearly, practice different methods, and prepare effectively for your exams. Keep reading to get all NCERT Solutions for Class 11 Physics Chapter 4 Motion in a Plane.

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NCERT Solutions for Class 11 Physics Chapter 4 Motion in a Plane

NCERT Solutions for Class 11 Physics Chapter 4 Motion in a Plane give detailed answers to all the questions from the exercise. These solutions are explained in a simple way so that you can easily understand both the theory and numerical parts. These Motion in a Plane class 11 NCERT solutions are also helpful in learning the methods and for doing homework and revisions before exams.

Class 11 Physics Motion in a Plane Exercise Solutions

Question 1. State, for each of the following physical quantities, if it is a scalar or a vector: volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, and angular velocity.

Solution:

• Scalar: Volume, mass, speed, density, number of moles, angular frequency

• Vector: Acceleration, velocity, displacement, angular velocity

Question 2. Pick out the two scalar quantities in the following list:
force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, and relative velocity.

Solution: Work and current are scalar quantities.

Question 3. Pick out the only vector quantity in the following list:
Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, and charge.

Solution:

Since Impulse = change in momentum = force × time. As momentum and force are vector quantities, impulse is a vector quantity.

Question 4. State with reasons whether the following algebraic operations with scalar and vector physical quantities are meaningful:
a. adding any two scalars,
b. adding a scalar to a vector of the same dimensions,
c. multiplying any vector by any scalar,
d. multiplying any two scalars,
e. adding any two vectors,
f. adding a component of a vector to the same vector.

Solution:

a. Yes, the addition of two scalar quantities is meaningful only if they both represent the same physical quantity.
b. No, the addition of a vector quantity with a scalar quantity is not meaningful.
c. Yes, a scalar can be multiplied by a vector. For example, force is multiplied by time to give impulse.
d. Yes, a scalar, irrespective of the physical quantity it represents, can be multiplied by another scalar having the same or different dimensions.
e. Yes, the addition of two vector quantities is meaningful only if they both represent the same physical quantity.
f. Yes, components of a vector can be added to the same vector, as they both have the same dimensions.

Question 5. Read each statement below carefully and state, with reasons, if it is true or false:
a. The magnitude of a vector is always a scalar.
b. Each component of a vector is always a scalar.
c. The total path length is always equal to the magnitude of the displacement vector of a particle.
d. The average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater than or equal to the magnitude of the average velocity of the particle over the same interval of time.
e. Three vectors not lying in a plane can never add up to give a null vector.

Solution:

a. True, because magnitude is a pure number.
b. False, each component of a vector is also a vector.
c. True, only if the particle moves along a straight line in the same direction; if not, then it is false.
d. True, because the total path length is either greater than or equal to the magnitude of the displacement vector.
e. True, as they cannot be represented by the three sides of a triangle taken in the same order.

Question 6. Establish the following vector inequalities geometrically or otherwise:
a. a + b| ≤ a| + b
b. a + b| ≥ |a| − b|
c. a − b| ≤ a| + b
d. a − b| ≥ |a| − b|
When does the equality sign above apply?

Solution:

a. Let two vectorsandbe represented by the adjacent sides of a parallelogram OMNP, as shown in the given figure.

Here, we can write:

In a triangle, each side is smaller than the sum of the other two sides.
Therefore, in ΔOMN, we have:

ON < (OM + MN)

If the two vectorsandact along a straight line in the same direction, then we can write:

Combining equations (iv) and (v), we get:

b. Let two vectorsandbe represented by the adjacent sides of a parallelogram OMNP, as shown in the given figure.

Here, we have:

In a triangle, each side is smaller than the sum of the other two sides.

Therefore, in ΔOMN, we have:

… (iv)

If the two vectorsandact along a straight line in the opposite direction, then we can write:

… (v)

Combining equations (iv) and (v), we get:

c. Let two vectorsandbe represented by the adjacent sides of a parallelogram PORS, as shown in the given figure.

Here we have:

In a triangle, each side is smaller than the sum of the other two sides. Therefore, in ΔOPS, we have:
PR < PS + SR

If the two vectors act in a straight line but in opposite directions, then we can write:
… (iv)
Combining equations (iii) and (iv), we get:

d. Let two vectorsandbe represented by the adjacent sides of a parallelogram PORS, as shown in the given figure.

The following relations can be written for the given parallelogram.

The quantity on the LHS is always positive, and that on the RHS can be positive or negative. To make both quantities positive, we take the modulus on both sides as:

If the two vectors act in a straight line but in the same direction, then we can write:

Combining equations (iv) and (v), we get:

Question 7. Given vectors a + b + c + d = 0, which of the following statements are correct:
a. Vectors a, b, c, and d must each be a null vector,
b. The magnitude of vectors (a + c) equals the magnitude of vectors(b+ d),
c. The magnitude of a can never be greater than the sum of the magnitudes of b, c, and d,
d. Vectors b + c must lie in the plane of a and d if a and d are not collinear, and in the line of a and d, if they are collinear?

Solution:

a. In order to make vectors a + b + c + d = 0, it is not necessary to have all four given vectors to be null vectors. There are many other combinations that can give the sum zero.
b. Correct

a + b + c + d = 0
a + c = – (b + d)
Taking modulus on both sides, we get:
| a + c | = | –(b + d)| = | b + d
Hence, the magnitude of (a + c) is the same as the magnitude of (b + d).
c. Correct
a + b + c + d = 0
a = - (b + c + d)
Taking the modulus of both sides, we get:
| a | = | b + c + d |
| a |  ≤  | a | + | b | + | c | .... (i)
Equation (i) shows that the magnitude of a is equal to or less than the sum of the magnitudes of b, c, and d.
Hence, the magnitude of vector a can never be greater than the sum of the magnitudes of b, c, and d.
d. Correct
For a + b + c + d = 0
a + (b + c) + d = 0
The resultant sum of the three vectors a, (b + c), and d can be zero only if (b + c) lies in a plane containing a and d, assuming that these three vectors are represented by the three sides of a triangle.
If a and d are collinear, then it implies that the vector (b + c) is in the line of a and d. This implication holds only when the vector sum of all the vectors is zero.

Question 8. Three girls skating on a circular ice ground of radius 200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in Fig. 4.20. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of the path skated?

Solution:

Displacement is given by the minimum distance between the initial and final positions of a particle. In the given case, all the girls start from point P and reach point Q. The magnitudes of their displacements will be equal to the diameter of the ground.
Radius of the ground = 200 m
Diameter of the ground = 2 × 200 = 400 m
Hence, the magnitude of the displacement for each girl is 400 m. This is equal to the actual length of the path skated by girl B.

Question 9. A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO as shown in Fig. 4.21. If the round trip takes 10 minutes, what is the

a. net displacement,
b. average velocity, and
c. average speed of the cyclist?

Solution:

a. Displacement is given by the minimum distance between the initial and final positions of a body. In the given case, the cyclist comes to the starting point after cycling for 10 minutes. Hence, his net displacement is zero.
b. Average velocity is given by the relation:
Average velocity = Net Displacement / Total time
Since the net displacement of the cyclist is zero, his average velocity will also be zero.
c. Average speed of the cyclist is given by the relation:
Average speed = Total path length / Total time
Total path length = OP + PQ + QO
= 1 + (1 (2π × 1) / 4 )  + 1
= 2 + (π / 2)
= 3.570 km
Time taken = 10 min = 10 / 60 = 1 / 6 h
∴ Average speed = 3.570 / (1/6) = 21.42 km/h

Question 10. On an open ground, a motorist follows a track that turns to his left by an angle of 60° after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth, and eighth turns. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.

Solution:

The path followed by the motorist is a regular hexagon with a side 500 m, as shown in the given figure

Let the motorist start from point A.
The motorist takes the third turn at D.
Magnitude of displacement =AD=BC+EF=500+500=1000m
Total path length =AB+BC+CD=500+500+500=1500m
The motorist takes the sixth turn at point A, which is the starting point.
Magnitude of displacement =0
Total path length =6AB
=6×500=3000m
The motorist takes the eighth turn at point C
The magnitude of displacement =AC
From the triangle law, the magnitude of displacement is 866.03m at an angle of 30 ° with AC.
It means AC makes an angle of 30 degrees with the initial direction. Total path length =8×500=4000m\

Read More: NCERT Solutions For Class 11 Chemistry Chapter 1

Question 11. A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min. What is (a) the average speed of the taxi, (b) the magnitude of average velocity? Are the two equal?

Solution:

(a) Total distance travelled = 23 km
Total time taken = 28 min = 28 / 60 h
∴  Average speed of the taxi = Total distance travelled / Total time taken
= 23 / (28/60) = 49.29 km/h
(b) Distance between the hotel and the station = 10 km = Displacement of the car
∴ Average velocity = 10 / (28/60) = 21.43 km/h
Therefore, the two physical quantities (average speed and average velocity) are not equal.

Question 12. Rain is falling vertically with a speed of 30 m s-1. A woman rides a bicycle with a speed of 10 m s-1 in the north to south direction. What is the direction in which she should hold her umbrella ?

Solution:

The described situation is shown in the given figure.

Here,

v c = Velocity of the cyclist
v r = Velocity of falling rain
In order to protect herself from the rain, the woman must hold her umbrella in the direction of the relative velocity (v) of the rain with respect to the woman.
v = v r + (-v c )
= 30 + (-10) = 20 m/s
tan θ = v c / v r = 10 / 30
θ = tan -1 (1 / 3)
= tan-1 (0.333) ≈ 180
Hence, the woman must hold the umbrella toward the south, at an angle of nearly 18° with the vertical.

Question 13. A man can swim with a speed of 4.0 km/h in still water. How long does he take to cross a river 1.0 km wide if the river flows steadily at 3.0 km/h and he makes his strokes normal to the river current? How far down the river does he go when he reaches the other bank?

Solution:

Speed of the man, v m = 4 km/h
Width of the river = 1 km
Time taken to cross the river = Width of the river / Speed of the river
= 1/4 h = 1 × 60 / 4 = 15 min
Speed of the river, v r = 3 km/h
Distance covered with flow of the river = v r × t
= 3 × 1 / 4 = 3 / 4 km
= 3 × 1000 / 4
= 750 m

Question 14. In a harbor, wind is blowing at the speed of 72 km/h and the flag on the mast of a boat anchored in the harbor flutters along the N-E direction. If the boat starts moving at a speed of 51 km/h to the north, what is the direction of the flag on the mast of the boat ?

Solution:

The flag is fluttering in the north-east direction. It shows that the wind is blowing toward the north-east direction. When the ship begins sailing toward the north, the flag will move along the direction of the relative velocity (vwb) of the wind with respect to the boat.

The angle between vw and (–v b ) = 90° + 45°
tan β = 51 Sin (90 + 45) / (72 + 51 Cos (90 + 45)
Substituting and solving, we get,
tan β = 51 / 50.80 = 1.0038
∴ β = tan -1 (1.0038) = 45.110
Angle with respect to the east direction = 45.11° – 45° = 0.11°
Hence, the flag will flutter almost due east.

Question 15. 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 m s-1 can go without hitting the ceiling of the hall?

Solution:

Maximum height, h = 25 m
In projectile motion, the maximum height reached by a body projected at an angle θ, is given by the relation:
h = u 2 Sin 2 θ / 2g
25 = (40) 2 Sin 2 θ / 2 × 9.8
sin 2 θ = 0.30625
sin θ = 0.5534
∴ θ = sin–1(0.5534) = 33.60°
Horizontal range, R = u 2 Sin 2θ / g
= (40) 2 × Sin (2 × 33.600) / 9.8
= 1600 × Sin (67.2) / 9.8
= 1600 × 0.922 / 9.8  =  150.53 m

Question 16. A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball?

Solution:

Maximum horizontal distance, R = 100 m
The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection is 45°, i.e., θ = 45°.
The horizontal range for a projection velocity v, is given by the relation:
R = u 2 Sin 2θ / g
100 = u 2 Sin 900 / g
u 2 / g = 100   ....(i)
The ball will achieve the maximum height when it is thrown vertically upward. For such motion, the final velocity v is zero at the maximum height H.
Acceleration, a = –g
Using the third equation of motion:
v 2 - u 2 = -2gH
H = u 2 / 2g  =  100 / 2  =  50 m

Question 17. 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 s, what is the magnitude and direction of acceleration of the stone?

Solution:

Length of the string, l = 80 cm = 0.8 m
Number of revolutions = 14
Time taken = 25 s
Frequency, v = Number of revolutions / Time taken  =  14 / 25 Hz
Angular frequency, ω = 2πν
= 2  × (22/7)  ×  (14/25)  =  88 / 25 rad s -1
Centripetal acceleration, ac = ω 2 r
= (88 / 25) 2 ×  0.8
= 9.91 ms-2
The direction of centripetal acceleration is always directed along the string, toward the centre, at all points.

Question 18. An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.

Solution:

Radius of the loop, r = 1 km = 1000 m
Speed of the aircraft, v = 900 km/h = 900  ×  5 / 18  =  250 m/s
Centripetal acceleration, ac = v 2 / r
= (250) 2 / 1000 = 62.5 ms -2
Acceleration due to gravity, g = 9.8 m/s 2

a c / g  =  62.5 / 9.8  =  6.38
a c =  6.38 g

Question 19. Read each statement below carefully and state, with reasons, if it is true or false:
(a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre
(b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point
(c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector

Solution:

(a) False, the net acceleration of a particle is towards the centre only in case of a uniform circular motion.
(b) True, because while leaving the circular path, the particle moves tangentially to the circular path.
(c) True, The direction of acceleration vector in a uniform circular motion is directed towards the centre of circular path. It is constantly changing with time. The resultant of all these vectors will be zero vector.

Question 20. The position of a particle is given by

where t is in seconds and the coefficients have the proper units for r to be in metres.
a. Find the v and a of the particle?

b. What is the magnitude and direction of velocity of the particle at t = 2.0 s?

Solution:

a. The position of the particle is given by:

Velocity v, of the particle is given as:

Acceleration a of the particle is given as:

b. 8.54 m/s, 69.45° below the x-axis

The magnitude of velocity is given by:

The negative sign indicates that the direction of velocity is below the x-axis.

Read More: NCERT Solutions for Class 11 Physics Chapter-2

Question 21. A particle starts from the origin at t = 0 s with a velocity of 10.0 ĵ m/s and moves in the x-y plane with a constant acceleration of (8.0 î+ 2.0 ĵ) m s -2 .

a. At what time is the x- coordinate of the particle 16 m? What is the y-coordinate of the particle at that time?
b. What is the speed of the particle at the time?

Solution: (a)
(b)

Question 22. For any arbitrary motion in space, which of the following relations are true :
a. v average = (1/2) (v (t 1 ) + v (t 2 ))
b. vaverage = [ r(t 2 ) - r(t 1 ) ] / (t 2 – t1)
c. v (t) = v (0) + a t
d. r (t) = r (0) + v (0) t + (1/2) a t 2
e. aaverage = [ v (t 2 ) - v (t 1 ) ] / ( t 2 – t 1 )
(The ‘average’ stands for the average of the quantity over the time interval t 1 to t 2 )

Solution:

a. It is given that the motion of the particle is arbitrary. Therefore, the average velocity of the particle cannot be given by this equation.
b. The arbitrary motion of the particle can be represented by this equation.
c. The motion of the particle is arbitrary. The acceleration of the particle may also be non-uniform. Hence, this equation cannot represent the motion of the particle in space.
d. The motion of the particle is arbitrary; acceleration of the particle may also be non-uniform. Hence, this equation cannot represent the motion of a particle in space.
e. The arbitrary motion of the particle can be represented by this equation.

Question 23. Read each statement below carefully and state, with reasons and examples, if it is true or false :
A scalar quantity is one that
a. is conserved in a process
b.can never take negative values
c. must be dimensionless
d. does not vary from one point to another in space
e. has the same value for observers with different orientations of axes.

Solution:

a. False. Despite being a scalar quantity, energy is not conserved in inelastic collisions.
b. False. Despite being a scalar quantity, temperature can take negative values.
c. False. Total path length is a scalar quantity. Yet it has the dimension of length.
d. False. A scalar quantity such as gravitational potential can vary from one point to another in space.
e. True. The value of a scalar does not vary for observers with different orientations of axes.

Question 24. 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:

The positions of the observer and the aircraft are shown in the given figure.

Height of the aircraft from ground, OR = 3400 m
Angle subtended between the positions, ∠POQ = 30°
Time = 10 s

In ΔPRO:
tan 15 0 = PR / OR
PR = OR tan 15 0
= 3400 X tan 15 0

ΔPRO is similar to ΔRQO.

∴ PR = RQ
PQ = PR + RQ
= 2PR = 2 × 3400 tan 15°
= 6800 × 0.268 = 1822.4 m
∴  Speed of the aircraft = 1822.4 / 10 = 182.24 m/s

Question 25. A vector has magnitude and direction. Does it have a location in space ? Can it vary with time ? Will two equal vectors a and b at different locations in space necessarily have identical physical effects ? Give examples in support of your answer.

Solution:

(i) A vector in general has no definite location in space because a vector remains unaffected whenever it is displaced anywhere in space provided its magnitude and direction do not change. However, a position vector has a definite location in space.
(ii) A vector can vary with time as example the velocity vector o an unaffected particle varies with time.
(iii) Two equal vectors at different locations in space do not necessarily have same physical effects. For example, two equal forces acting at two different points on a body which can cause the rotation of a body about an axis will not produce equal turning effect.

Question 26. A vector has both magnitude and direction. Does it mean that anything that has magnitude and direction is necessarily a vector ? The rotation of a body can be specified by the direction of the axis of rotation, and the angle of rotation about the axis. Does that make any rotation a vector ?

Solution:

No. A physical quantity having both magnitude and direction need not be considered a vector. For example, despite having magnitude and direction, current is a scalar quantity. The essential requirement for a physical quantity to be considered a vector is that it should follow the law of vector addition.
Generally speaking, the rotation of a body about an axis is not a vector quantity as it does not follow the law of vector addition. However, a rotation by a certain small angle follows the law of vector addition and is therefore considered a vector.

Question 27. Can you associate vectors with (a) the length of a wire bent into a loop, (b) a plane area, (c) a sphere ? Explain.

Solution:

(a) No, one cannot associate a vector with the length of a wire bent into a loop.
(b) Yes, one can associate an area vector with a plane area. The direction of this vector is normal, inward or outward to the plane area.
(c) No, one cannot associate a vector with the volume of a sphere. However, an area vector can be associated with the area of a sphere.

Question 28. 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 the fixed, and neglect air resistance.

Solution:

Range, R = 3 km
Angle of projection, θ = 30°
Acceleration due to gravity, g = 9.8 m/s 2
Horizontal range for the projection velocity u 0 , is given by the relation:
R = u 0 2 Sin 2θ / g
3 = u 0 2 Sin 60 0 / g
u 0 2 / g = 2√3      .......(i)
The maximum range (Rmax) is achieved by the bullet when it is fired at an angle of 45° with the horizontal, that is,
Rmax = u 0 2 / g ....(ii)
On comparing equations (i) and (ii), we get:
Rmax = 3√3
= 2 X 1.732 = 3.46 km
Hence, the bullet will not hit a target 5 km away.

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Question 29. A fighter plane flying horizontally at an altitude of 1.5 km with speed 720 km/h passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed 600 m s -1 to hit the plane ? At what minimum altitude should the pilot fly the plane to avoid being hit ? (Take g = 10 m s -2 ).

Solution:

Height of the fighter plane = 1.5 km = 1500 m
Speed of the fighter plane, v = 720 km/h = 200 m/s
Let θ be the angle with the vertical so that the shell hits the plane. The situation is shown in the given figure.

Muzzle velocity of the gun, u = 600 m/s
Time taken by the shell to hit the plane = t
Horizontal distance travelled by the shell = u x t
Distance travelled by the plane = vt
The shell hits the plane. Hence, these two distances must be equal.
uxt = vt
u Sin θ = v
Sin θ = v / u
= 200 / 600 = 1/3 = 0.33
θ = Sin -1 (0.33) = 19.50

In order to avoid being hit by the shell, the pilot must fly the plane at an altitude (H) higher than the maximum height achieved by the shell.
∴ H = u 2 Sin 2 (90 - θ) / 2g
= (600) 2 Cos 2 θ / 2g
= 360000 X Cos 2 19.5 / 2 X 10
= 16006.482 m
≈ 16 km

Question 30. A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?

Solution: 0.86 m/s 2 ; 54.46° with the direction of velocity
Speed of the cyclist, v = 27 km/h = 7.5 m/s
Radius of the circular turn, r = 80 m
Centripetal acceleration is given as:
ac = v 2 / r
= (7.5) 2 / 80 = 0.7 ms -2
The situation is shown in the given figure:

Suppose the cyclist begins cycling from point P and moves toward point Q. At point Q, he applies the breaks and decelerates the speed of the bicycle by 0.5 m/s 2 .

This acceleration is along the tangent at Q and opposite to the direction of motion of the cyclist.
Since the angle between ac and aT is 90 0 , the resultant acceleration a is given by:
a = (ac 2 + aT 2 ) 1/2
= ( (0.7) 2 + (0.5) 2 ) 1/2
= (0.74) 1/2 = 0.86 ms -2
tan θ = ac / aT
where θ is the angle of the resultant with the direction of velocity.
tan θ = 0.7 / 0.5 = 1.4
θ = tan -1 (1.4) = 54.560

Question 31.

a. Show that for a projectile, the angle between the velocity and the x-axis as a function of time is given by
θ(t) = tan -1 ((v 0y -  gt) / v 0x )
b. Shows that the projection angle θ 0 for a projectile launched from the origin is given by
θ 0 = tan -1 (4hm / R)
where the symbols have their usual meaning.

Solution:

a.  Let v 0x and v 0y, respectively, be the initial components of the velocity of the projectile along horizontal (x) and vertical (y) directions.
Let vx and vy, respectively, y be the horizontal and vertical components of velocity at a point P.

Time taken by the projectile to reach point P = t
Applying the first equation of motion along the vertical and horizontal directions, we get:
v y = v 0y = gt
And v x = v 0x
∴ tan θ = v y / v x = (v 0y - gt) / v 0x
θ = tan -1 (v 0y - gt) / v 0x

b. Maximum vertical height, hm = u 0 2 Sin 2 θ / 2g   ...(i)
Horizontal range, R = u 0 2 Sin 2 2θ / g   ... (ii)
hm / R = Sin 2 θ / 2Sin 2 2θ
= Sin θ X Sin θ / 2 X 2Sinθ Cosθ
= Sin θ / 4 Cos θ = tan θ / 4
tan θ = (4hm / R)
θ = tan-1 (4hm / R)

Check Out: CBSE Class 11 Chapterwise 20 Most Probable Questions 

NCERT Class 11 Chapter 4 Motion in a Plane Brief Summary

The Motion in a Plane chapter of class 11 Physics explains how to study motion in two dimensions with the help of vectors. Unlike motion in a straight line, here both horizontal (x-axis) and vertical (y-axis) directions are involved together.

The concepts of scalars, vectors, vector addition, and resolution are explained step by step and then applied to real-life motions like projectiles and circular paths. Some of the key concepts explained in the chapter that can help you better understand Class 11 Physics chapter 4 NCERT solutions are:

• Scalars and Vectors: Scalars have only magnitude, like distance or speed. Vectors have both magnitude and direction, like displacement and velocity.

• Vector Representation: Motion in a plane is described with vectors along the x and y axes, such as position vector r = xi + yj.

• Vector Addition: Vectors can be combined using the triangle rule or parallelogram law.

• Resolution of Vectors: Any vector can be split into horizontal and vertical components to make problem-solving easier.

• Equations of Motion in a Plane: The same equations of motion are applied separately along the x and y directions.

• Projectile Motion: A body projected into the air under gravity shows curved motion, where topics like time of flight and horizontal range are important.

• Uniform Circular Motion: An object moving in a circle at constant speed has angular velocity and needs centripetal acceleration.

NCERT Motion in a Plane is an important chapter of the Class 11 Physics subject, as it helps you to understand the basics of vector mathematics.

Advantages of NCERT Solutions of Motion in a Plane

NCERT Solutions of Motion in a Plane are a great study resource that explains difficult answers step-by-step for both theory and numerical problems. You can not only use these class 11 Physics chapter 4 NCERT solutions as a guide while doing homework to understand how to answer the question, but also to enhance your exam preparations.

Here are some key advantages of using these NCERT Solutions for Class 11 Physics Chapter 4:

• Complete answers: The solutions cover the majority of important questions from the NCERT textbook with correct and detailed steps.

• Simple language: These Class 11 Physics Chapter 4 NCERT solutions are given in an easy to understand way so that you can learn without any confusion.

• Good for revision: These well-arranged Motion in Plane class 11 NCERT solutions make your last-minute exam preparation more effective and easy.

• Exam-focused: Many exam questions come from NCERT, so solving these increases your chances of scoring better marks.

• Improves answer writing skills: Step-by-step methods used in these class 11 Physics Motion in a Plane exercise solutions teach you how to frame your answers clearly in exams.

• CBSE pattern-based: The solutions are based on the CBSE syllabus, which makes them reliable for school as well as competitive exams.

Also Check, CBSE Class 11 Formula Handbook For 2026 Exams

NCERT Solutions for Class 11 Physics Chapter 4 FAQs

Q.1. What are the important concepts in Class 11 Physics Chapter 4?

Ans. Class 11 Physics Chapter 4 Motion in a Plane explains vectors, scalar and vector quantities, vector addition, resolution of vectors, projectile motion, and uniform circular motion.

Q.2. Are the Motion in a Plane class 11 NCERT solutions sufficient for exams?

Ans. NCERT Solutions for Class 11 Physics Chapter 4 are important for exams because they give complete answers and cover the key concepts. But you should read the chapter and practice past years' exam questions.

Q.3. What is the difference between Scalars and Vectors?

Ans. Scalars have only magnitude, like distance or speed. On the other hand, Vectors have both magnitude and direction, like displacement or velocity.

Q.4. Where can I get detailed Motion in Plane Class 11 NCERT solutions?

Ans. You can find detailed Motion in Plane class 11 NCERT solutions in this article. They include both theory and numerical answers, explained step by step.