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Motion in a Plane NCERT Solutions, Complete PDF Content and Important Questions
NCERT Solutions for Class 11 Physics Chapter 3 Motion in a Plane PDF Download (2026) + Important Questions Welcome, future engineers, doctors, and scientists! If you have just mastered moving in a straight line, it is time to level up. Let's step out of the 1D world and enter the real, two-dimensional world with CBSE Class 11 Physics Chapter 3: Motion in a Plane . Whether it is a cricketer hitting a massive six, a car taking a sharp circular turn, or a satellite orbiting the Earth-everything around us is an example of motion in a plane. This chapter is the absolute backbone of Mechanics. You will be introduced to the magical world of Vectors , which will stick with you through Class 11, Class 12, and competitive exams like JEE and NEET . In this comprehensive guide, we will break down complex concepts like projectile and circular motion into super simple explanations.
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ChapterMotion in a Plane
SubjectPhysics
Class11
BoardCBSE
DifficultyModerate to High (Requires good visualization)
Exam Weightage~7-9 Marks (Combined with Motion in a Straight Line)
Learning Objectives
- Understand the fundamental difference between scalars and vectors. Perform mathematical operations on vectors (Addition, Subtraction, and Resolution into components).
- Apply the kinematic equations to 2D motion.
- Calculate the Time of Flight, Maximum Height, and Horizontal Range for Projectile Motion.
- Understand the dynamics of Uniform Circular Motion and.
- Calculate centripetal acceleration.
Key Concepts, Definitions and Formulas
- Scalars: Physical quantities that have only magnitude and no direction (e.g., Mass, Time, Distance,
- Speed). Vectors: Physical quantities that have both magnitude and direction, and obey the triangle law of addition (e.g., Displacement, Velocity,
- Force). Resolution of Vectors: Splitting a vector $\vec{A}$ into its rectangular components. If $\vec{A}$ makes an angle $\theta$ with the x-axis:
- Horizontal component: $A_x = A \cos\theta$
- Vertical component: $A_y = A \sin\theta$
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Full NCERT Solutions
NCERT Q3.1: 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, angular velocity.
Answer: Scalars (Magnitude only): Volume, Mass, Speed, Density, Number of moles, Angular frequency. Vectors (Magnitude + Direction): Acceleration, Velocity, Displacement, Angular velocity.
NCERT Q3.2: Pick out the two scalar quantities in the following list: force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity.
Answer: The two scalar quantities are Work and Current . Explanation: Work is the dot product of two vectors (Force and Displacement), making it a scalar. Current has direction, but it does not follow the vector laws of addition (it follows ordinary algebra), hence it is a scalar.
NCERT Q3.3: Pick out the only vector quantity in the following list: Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.
Answer: The only vector quantity is Impulse . Explanation: Impulse is the product of Force (vector) and Time (scalar) interval ($J = F \times \Delta t$). Since force is a vector, impulse is also a vector. All other listed quantities do not possess a specific directional property defined by vector laws.
NCERT Q3.4: 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.
Answer: (a) Meaningless: You can only add scalars that represent the same physical quantity (e.g., you cannot add mass to time). (b) Meaningless: A scalar cannot be added to a vector because they have different directional properties. (c) Meaningful: Multiplying a vector by a scalar gives a new vector (e.g., Force = mass $\times$ acceleration). (d) Meaningful: Any two scalars can be multiplied to give a meaningful physical quantity (e.g., Power = Work / Time). (e) Meaningless: Similar to scalars, you can only add vectors representing the same physical quantity (e.g., you cannot add force and velocity). (f) Meaningful: A component of a vector is itself a vector of the same physical dimension. Hence, they can be added.
NCERT Q3.5: 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 is either greater or equal to the magnitude of 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.
Answer: (a) True: The magnitude of a vector simply represents a pure number with a unit, lacking direction, making it a scalar. (b) False: A component of a vector (like $v_x \hat{i}$) is itself a vector quantity. (c) False: Total path length (distance) is greater than displacement unless the particle moves in a straight line without reversing direction. (d) True: Since Distance $\ge$ |Displacement|, dividing both by time gives Average Speed $\ge$ |Average Velocity|. (e) True: To get a null vector, the sum of two vectors must be cancelled by the third. Two vectors always lie in one plane, and their resultant is in the same plane. A third vector not in this plane cannot cancel this resultant.
NCERT Q3.6: Establish the following vector inequalities geometrically or otherwise: (a) $|\vec{a} + \vec{b}| \le |\vec{a}| + |\vec{b}|$ (b) $|\vec{a} + \vec{b}| \ge ||\vec{a}| - |\vec{b}||$ (c) $|\vec{a} - \vec{b}| \le |\vec{a}| + |\vec{b}|$ (d) $|\vec{a} - \vec{b}| \ge ||\vec{a}| - |\vec{b}||$
Answer: Let $\vec{a}$ and $\vec{b}$ form two adjacent sides of a triangle. The third side represents their resultant. (a) In a triangle, the length of one side is always less than or equal to the sum of the other two sides. Thus, the magnitude of resultant $|\vec{a} + \vec{b}|$ is less than $|\vec{a}| + |\vec{b}|$. Equality holds when $\vec{a}$ and $\vec{b}$ are in the same direction. (b) In a triangle, the length of one side is always greater than or equal to the difference of the other two sides. Hence, $|\vec{a} + \vec{b}| \ge ||\vec{a}| - |\vec{b}||$. (c) Replacing $\vec{b}$ with $-\vec{b}$ in equation (a) gives $|\vec{a} + (-\vec{b})| \le |\vec{a}| + |-\vec{b}|$. Since magnitude $|-\vec{b}| = |\vec{b}|$, we get $|\vec{a} - \vec{b}| \le |\vec{a}| + |\vec{b}|$. (d) Replacing $\vec{b}$ with $-\vec{b}$ in equation (b) similarly yields $|\vec{a} - \vec{b}| \ge ||\vec{a}| - |\vec{b}||$.
NCERT Q3.7: Given $\vec{a} + \vec{b} + \vec{c} + \vec{d} = \vec{0}$, which of the following statements are correct: (a) $\vec{a}, \vec{b}, \vec{c}$, and $\vec{d}$ must each be a null vector. (b) The magnitude of $(\vec{a} + \vec{c})$ equals the magnitude of $(\vec{b} + \vec{d})$. (c) The magnitude of $\vec{a}$ can never be greater than the sum of the magnitudes of $\vec{b}, \vec{c}$, and $\vec{d}$. (d) $\vec{b} + \vec{c}$ must lie in the plane of $\vec{a}$ and $\vec{d}$ if $\vec{a}$ and $\vec{d}$ are not collinear, and in the line of $\vec{a}$ and $\vec{d}$, if they are collinear.
Answer: (a) Incorrect: Their vector sum is zero, but individually they do not need to be zero. They could form a closed polygon. (b) Correct: Since $\vec{a} + \vec{b} + \vec{c} + \vec{d} = 0$, we have $(\vec{a} + \vec{c}) = -(\vec{b} + \vec{d})$. Taking magnitudes on both sides gives $|\vec{a} + \vec{c}| = |\vec{b} + \vec{d}|$. (c) Correct: $\vec{a} = -(\vec{b} + \vec{c} + \vec{d})$. The magnitude $|\vec{a}|$ cannot exceed the sum of the magnitudes of the other three vectors (Polygon law). (d) Correct: $(\vec{b} + \vec{c}) = -(\vec{a} + \vec{d})$. This implies the resultant of $\vec{b}$ and $\vec{c}$ is antiparallel to the resultant of $\vec{a}$ and $\vec{d}$. Hence, they must lie in the same plane or along the same line if collinear.
NCERT Q3.8: 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. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of path skate?
Answer: Step 1: Analyze the initial and final positions. All three girls start at point P and end at point Q, which is diametrically opposite. Displacement is the shortest straight-line distance between initial and final points. Step 2: Calculate magnitude of displacement. Displacement = Diameter of the circle = $2 \times \text{Radius} = 2 \times 200 \text{ m} = 400 \text{ m}$. So, the magnitude of the displacement vector for each girl is 400 m. Step 3: Path length comparison. The actual path length is equal to the displacement only when the motion is strictly along a straight line without reversing. The girl who skated along the straight diameter from P to Q has a path length equal to her displacement.
NCERT Q3.9: 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. If the round trip takes 10 min, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist?
Answer: (a) Net Displacement: Since the cyclist returns to the starting point O, the final position is the same as the initial position. Hence, Net Displacement = 0 . (b) Average Velocity: Average velocity = $\frac{\text{Net Displacement}}{\text{Total Time}}$. Since displacement is zero, Average Velocity = 0 . (c) Average Speed: Total Path Length = $OP + \text{Arc } PQ + QO$. $OP = QO = \text{Radius} = 1 \text{ km}$. Arc $PQ$ is one-fourth of the circumference = $\frac{1}{4}(2\pi r) = \frac{\pi \times 1}{2} = 1.57 \text{ km}$. Total path = $1 + 1.57 + 1 = 3.57 \text{ km}$. Time = $10 \text{ min} = \frac{10}{60} \text{ hours} = \frac{1}{6} \text{ hr}$. Average Speed = $\frac{\text{Total Path}}{\text{Time}} = \frac{3.57}{1/6} = 3.57 \times 6 = \textbf{21.42 km/h}$.
NCERT Q3.10: On an open ground, a motorist follows a track that turns to his left by an angle of $60^\circ$ after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.
Answer: The track is a regular hexagon of side $L = 500 \text{ m}$. Let the motorist start from point A. The turns are B, C, D, E, F, and back to A. At the 3rd turn (Point D): The motorist covers AB, BC, CD. Point D is diagonally opposite to A. Magnitude of Displacement $AD = 2L = 2 \times 500 = \textbf{1000 m}$. Total path length = $3L = 3 \times 500 = \textbf{1500 m}$. Ratio (Displacement / Path) = $1000/1500 = \textbf{2/3}$. At the 6th turn (Point A): The motorist completes one full hexagon and is back at the start. Magnitude of Displacement = 0 . Total path length = $6L = 6 \times 500 = \textbf{3000 m}$. Ratio = 0 . At the 8th turn (Point C): After 6 turns, he is at A. 7th is B, 8th is C. Magnitude of Displacement $AC = L\sqrt{3} = 500\sqrt{3} = \textbf{866 m}$. Total path length = $8L = 8 \times 500 = \textbf{4000 m}$. Ratio = $866 / 4000 \approx \textbf{0.216}$.
Extra Important Questions for 2026 Exams
Important Q1: The angle between two vectors $\vec{A}$ and $\vec{B}$ is $\theta$. The resultant of these two vectors is maximum when $\theta$ is: A) $90^\circ$ B) $180^\circ$ C) $0^\circ$ D) $45^\circ$
Answer: C) $0^\circ$. Explanation: Resultant $R = \sqrt{A^2 + B^2 + 2AB\cos\theta}$. It is maximum when $\cos\theta$ is maximum (i.e., $\cos 0^\circ = 1$).
Important Q2: In a projectile motion, the velocity at the maximum height is: A) Zero B) Equal to the initial launch velocity C) Entirely horizontal D) Entirely vertical
Answer: C) Entirely horizontal. Explanation: At the highest point, the vertical component of velocity becomes zero ($v_y = 0$). Only the horizontal component ($v_x = u \cos\theta$) remains.
Important Q3: A body is moving in a circular path with a constant speed. Its acceleration is: A) Zero B) Directed along the tangent C) Directed towards the center D) Directed away from the center
Answer: C) Directed towards the center. Explanation: In uniform circular motion, speed is constant but direction changes continuously. This creates a centripetal acceleration directed towards the center.
Important Q4: Can two vectors of different magnitudes add up to give a zero resultant?
Answer: No. To get a zero resultant, the two vectors must have equal magnitudes and opposite directions. If their magnitudes are different, they cannot cancel each other out.
Important Q5: What is the angle of projection for which the maximum height and horizontal range of a projectile are equal?
Answer: Given $H = R$. $\frac{u^2 \sin^2\theta}{2g} = \frac{u^2 (2\sin\theta\cos\theta)}{g}$ $\frac{\sin\theta}{2} = 2\cos\theta$ $\tan\theta = 4 \implies \theta = \tan^{-1}(4) \approx 76^\circ$.
Important Q6: Distinguish between distance and displacement in 2D motion.
Answer: Distance is a scalar quantity representing the actual path length travelled by an object in the plane. Displacement is a vector quantity representing the shortest straight-line distance between the initial and final positions in the plane.
Important Q7: Why does a person jumping out of a moving train fall in the forward direction?
Answer: Due to inertia of motion. When the person jumps, their feet touch the ground and come to rest instantly due to friction, but the upper part of their body continues moving forward with the velocity of the train, causing them to fall forward.
Important Q8: State the Parallelogram Law of Vector Addition. Derive the analytical expression for the magnitude and direction of the resultant vector.
Answer: Statement: If two vectors acting at a point are represented in magnitude and direction by the adjacent sides of a parallelogram drawn from a point, their resultant is represented by the diagonal passing through that point. Derivation: Let $\vec{P}$ and $\vec{Q}$ be two vectors at angle $\theta$. By dropping a perpendicular from the tip of $\vec{Q}$ to the extended line of $\vec{P}$, we form a right-angled triangle. Using Pythagoras theorem: $R^2 = (P + Q\cos\theta)^2 + (Q\sin\theta)^2$ $R^2 = P^2 + Q^2\cos^2\theta + 2PQ\cos\theta + Q^2\sin^2\theta$ $R = \sqrt{P^2 + Q^2 + 2PQ\cos\theta}$ Direction ($\alpha$) with respect to $\vec{P}$: $\tan\alpha = \frac{Q\sin\theta}{P + Q\cos\theta}$.
Important Q9: Derive the equation of trajectory for a projectile. Show that the path of a projectile is a parabola.
Answer: Assume an object is launched with velocity $u$ at angle $\theta$. Horizontal displacement $x$ at time $t$: $x = (u\cos\theta)t \implies t = \frac{x}{u\cos\theta}$. Vertical displacement $y$ at time $t$: $y = (u\sin\theta)t - \frac{1}{2}gt^2$. Substitute $t$ into the $y$ equation: $y = u\sin\theta \left( \frac{x}{u\cos\theta} \right) - \frac{1}{2}g \left( \frac{x}{u\cos\theta} \right)^2$ $y = x \tan\theta - \left( \frac{g}{2u^2\cos^2\theta} \right) x^2$ This equation is of the form $y = ax - bx^2$, which represents a downward-opening parabola. Hence proved.
Important Q10: What is centripetal acceleration? Derive an expression for it for a particle executing uniform circular motion.
Answer: Centripetal acceleration is the acceleration acting on a body moving in a circle, directed towards the center. Let a particle move from point $P$ to $P'$ on a circle of radius $r$ in time $\Delta t$. Velocity vectors $\vec{v}$ and $\vec{v}'$ at $P$ and $P'$ have the same magnitude $v$. By geometry, the triangle formed by position vectors and the triangle formed by velocity vectors are similar. Therefore, $\frac{|\Delta \vec{v}|}{v} = \frac{|\Delta \vec{r}|}{r}$. $\Delta v = \frac{v}{r} \Delta r$. Dividing by $\Delta t$ and taking limits as $\Delta t \to 0$: $a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{v}{r} \left(\lim_{\Delta t \to 0} \frac{\Delta r}{\Delta t}\right)$ Since $v = \frac{dr}{dt}$, we get $a_c = \frac{v^2}{r}$. A javelin thrower is preparing for the Olympics. He wants to maximize the distance his javelin travels. He can throw the javelin with a maximum initial speed of 30 m/s.
Important Q11: At what angle with the horizontal should he throw the javelin to achieve the maximum horizontal range?
Answer: For maximum range, the angle of projection should be $45^\circ$.
Important Q12: Calculate the maximum horizontal range he can achieve. (Take $g = 10 \text{ m/s}^2$)
Answer: $R_{max} = \frac{u^2}{g} = \frac{30^2}{10} = \frac{900}{10} = \textbf{90 m}$.
Important Q13: What will be the maximum height reached by the javelin when thrown at this angle of maximum range?
Answer: $H = \frac{u^2 \sin^2(45^\circ)}{2g} = \frac{30^2 \times (1/\sqrt{2})^2}{2 \times 10} = \frac{900 \times 0.5}{20} = \frac{450}{20} = \textbf{22.5 m}$.
Important Q14: Assertion (A): In projectile motion, the horizontal velocity remains constant throughout the flight. Reason (R): There is no force acting in the horizontal direction, so acceleration in the x-direction is zero.
Answer: A (Both A and R are true, and R is the correct explanation of A).
Important Q15: Assertion (A): When a body moves in a circular path with uniform speed, its acceleration is zero. Reason (R): Uniform speed means the magnitude of velocity does not change.
Answer: D (A is false, but R is true). The velocity vector's direction is constantly changing, so acceleration is not zero (it is centripetal).
FAQ Section
Is Motion in a Plane Chapter 3 or Chapter 4?
Under the latest rationalized CBSE syllabus (implemented post-2023), "Motion in a Plane" is Chapter 3 . (Previously, it was Chapter 4).
Which topics in Class 11 Physics Chapter 3 are most important for JEE/NEET?
Projectile motion on an inclined plane, Relative velocity in 2D (like the River-Boat and Rain-Man problems), and Vector resolution are extremely crucial for competitive exams.
What is a null vector?
A null vector is a vector with zero magnitude and an arbitrary direction. It usually results from adding a vector to its exact opposite.
How can I download the NCERT PDF for Class 11 Physics?
You can easily download the official and latest NCERT textbook PDF for Class 11 Physics directly from the official NCERT website (ncert.nic.in).
Are the projectile formulas valid if air resistance is considered?
No. All standard projectile formulas ($T, H, R$) assume ideal conditions where air resistance is neglected. If air drag is included, the trajectory becomes non-parabolic and the range decreases
More Class 11 Physics Chapters
Ch 1: Units and MeasurementsCh 2: Motion in a Straight LineCh 3: Motion in a PlaneCh 4: Laws of MotionCh 5: Work, Energy and PowerCh 6: System of Particles and Rotational MotionCh 7: GravitationCh 8: Mechanical Properties of SolidsCh 9: Mechanical Properties of FluidsCh 10: Thermal Properties of MatterCh 11: ThermodynamicsCh 12: Kinetic TheoryCh 13: OscillationsCh 14: Waves