Class 11 Physics: Motion in a Plane | Physics Q&A

CBSE Class 11 › Unit II: Kinematics

Motion in a Plane

NCERT Chapter 3 • Vectors, Projectile Motion & Circular Motion

NCERT 2025–26 Unit II • ~10 Marks JEE Main • 2-3 Questions

Javelin flight trajectory showing parabolic path and velocity components. — Motion in a Plane: Analyzing 2D trajectories and vectors.

1. Scalars and Vectors

Physical quantities are classified into two categories based on whether they require direction for their complete description.

Scalar quantities have only magnitude and no direction. Examples: mass, time, distance, speed, temperature.
Vector quantities have both magnitude and direction, and obey vector algebra. Examples: displacement, velocity, acceleration, force.

Key Difference: Scalars follow ordinary arithmetic rules. Vectors must be added using geometric or component methods (e.g., triangle law).

1.1 Representation of a Vector

$\vec{A} = |\vec{A}|\hat{A}$

A vector is represented by a directed line segment:

Length ∝ magnitude, $|\vec{A}|$
Arrowhead indicates direction, $\hat{A}$ , a unit vector

In printed text: boldface (A). In handwriting: $\vec{A}$ .

1.2 Magnitude of a Vector

For a vector $\vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$ , its magnitude is:

$|\vec{A}| = A = \sqrt{A_x^2 + A_y^2 + A_z^2}$

1.3 Unit Vector

A unit vector has magnitude 1 and specifies direction only. It is dimensionless.

$\hat{A} = \dfrac{\vec{A}}{|\vec{A}|}$

Standard unit vectors along coordinate axes:

$\hat{i}$ → x-direction
$\hat{j}$ → y-direction
$\hat{k}$ → z-direction

1.4 Types of Vectors

Position Vector ( $\vec{r}$ ): From origin to particle’s location: $\vec{r} = x\hat{i} + y\hat{j}$ .

Displacement Vector ( $\Delta \vec{r}$ ): $\Delta \vec{r} = \vec{r_2} - \vec{r_1}$ ; straight line from initial to final position.

Equal Vectors:

Same magnitude and direction. If $\vec{a}=\vec{b}\implies |\vec{a}|=|\vec{b}|$ and $\hat{a}=\hat{b}$

Parallel Vectors: Same direction, possibly different magnitudes. If $\vec{a} || \vec{b}\implies |\vec{a}|\neq|\vec{b}|$ but $\hat{a}=\hat{b}$

Antiparallel Vectors: Opposite directions. $\hat{a}= -\hat{b}$

Null Vector ( $\vec{0}$ ): Zero magnitude; direction undefined. Occurs when initial = final position.

1.5 Vector Addition and Subtraction

Addition: $\vec{R} = \vec{A} + \vec{B}$ — use head-to-tail or parallelogram method.

Subtraction: $\vec{A} - \vec{B} = \vec{A} + (-\vec{B})$ , where $-\vec{B}$ has same magnitude but opposite direction.

Vector addition is commutative and associative.

1.6 Parallelogram Law of Vector Addition

If two vectors $\vec{A}$ and $\vec{B}$ act at a point with angle $\theta$ between them, their resultant $\vec{R}$ is the diagonal of the parallelogram formed.

Parallelogram law of addition of vectors

Derivation of Resultant Magnitude and Direction Full Proof

Step 1: Geometry Setup
Extend $\vec{A}$ and drop perpendicular from tip of $\vec{B}$ to form right triangle.
Horizontal component: $A + B\cos\theta$
Vertical component: $B\sin\theta$

Step 2: Magnitude of Resultant
$R^2 = (A + B\cos\theta)^2 + (B\sin\theta)^2$
$R^2 = A^2 + 2AB\cos\theta + B^2(\cos^2\theta + \sin^2\theta)$
$\boxed{R = \sqrt{A^2 + B^2 + 2AB\cos\theta}}$

Step 3: Direction of Resultant
Let $\alpha$ = angle between $\vec{R}$ and $\vec{A}$ .
$\tan\alpha = \dfrac{\text{vertical}}{\text{horizontal}} = \dfrac{B\sin\theta}{A + B\cos\theta}$
$\boxed{\tan\alpha = \dfrac{B \sin\theta}{A + B \cos\theta}}$

Problem 1: Two forces, 5 N and 5 N, act at 60°. Find resultant magnitude and direction.

Show Answer

$R = \sqrt{5^2 + 5^2 + 2(5)(5)\cos60^\circ} = \sqrt{75} = 5\sqrt{3} \approx 8.66$ N.
$\tan\alpha = \dfrac{5 \sin60^\circ}{5 + 5 \cos60^\circ} = \dfrac{4.33}{7.5} \Rightarrow \alpha \approx 30^\circ$ .

2. Resolution of Vectors

Any vector can be split into components along chosen axes. In 2D rectangular coordinates:

$\vec{A} = A_x \hat{i} + A_y \hat{j}$

$A_x = A \cos\theta,\quad A_y = A \sin\theta$

$A = \sqrt{A_x^2 + A_y^2},\quad \tan\theta = \dfrac{A_y}{A_x}$

Vector resolved into horizontal and vertical components — Resolving a vector simplifies 2D motion into independent 1D motions.

Problem 2: Find the magnitude and direction of vector $\vec{A} = 3\hat{i} + 4\hat{j}$ .

Show Answer

Magnitude $= \sqrt{3^2 + 4^2} = 5$ .
Direction $\theta = \tan^{-1}(4/3) \approx 53^\circ$ with x-axis.

Products of Vectors

1. Scalar (Dot) Product

$\vec A \cdot \vec B = |\vec A|\,|\vec B| \cos\theta$

It gives a scalar quantity. Cases

$\theta = 0^\circ \Rightarrow \vec A \cdot \vec B = AB$

$\theta = 90^\circ \Rightarrow \vec A \cdot \vec B = 0$

$\theta = 180^\circ \Rightarrow \vec A \cdot \vec B = -AB$

Component Form

$\vec A = A_x \hat i + A_y \hat j + A_z \hat k$

$\vec B = B_x \hat i + B_y \hat j + B_z \hat k$

$\vec A \cdot \vec B = A_xB_x + A_yB_y + A_zB_z$

Applications

$W = \vec F \cdot \vec s = Fs\cos\theta$

$P = \vec F \cdot \vec v$

2. Vector (Cross) Product

$\vec A \times \vec B = |\vec A|\,|\vec B| \sin\theta \, \hat n$

Cases

$\theta = 0^\circ \Rightarrow \vec A \times \vec B = 0$

$\theta = 90^\circ \Rightarrow |\vec A \times \vec B| = AB$

$\theta = 180^\circ \Rightarrow \vec A \times \vec B = 0$

Component Form

$\vec A \times \vec B = \begin{vmatrix} \hat i & \hat j & \hat k\\ A_x & A_y & A_z\\ B_x & B_y & B_z \end{vmatrix}$

Applications

$\vec\tau = \vec r \times \vec F$

$\vec L = \vec r \times \vec p$

3. Scalar Triple Product

$\vec A \cdot (\vec B \times \vec C) = \text{Volume of parallelepiped}$

Coplanar Condition

$\vec A \cdot (\vec B \times \vec C) = 0$

Determinant Form

$\vec A \cdot (\vec B \times \vec C) = \begin{vmatrix} A_x & A_y & A_z\\ B_x & B_y & B_z\\ C_x & C_y & C_z \end{vmatrix}$

4. Vector Triple Product

$\vec A \times (\vec B \times \vec C) = \vec B(\vec A \cdot \vec C) - \vec C(\vec A \cdot \vec B)$

3. Velocity and Acceleration in a Plane

For a particle with position vector $\vec{r}(t) = x(t)\hat{i} + y(t)\hat{j}$ :

Instantaneous Velocity: $\vec{v} = \dfrac{d\vec{r}}{dt} = v_x\hat{i} + v_y\hat{j}$
Instantaneous Acceleration: $\vec{a} = \dfrac{d\vec{v}}{dt} = a_x\hat{i} + a_y\hat{j}$

Problem 3: If $\vec{r}(t) = 2t^2\hat{i} + 3t\hat{j}$ , find $\vec{v}$ and $\vec{a}$ at $t=2$ s.

Show Answer

$\vec{v} = 4t\hat{i} + 3\hat{j} \Rightarrow$ at $t=2$ , $\vec{v} = 8\hat{i} + 3\hat{j}$ m/s.
$\vec{a} = 4\hat{i}$ m/s².

4. Projectile Motion

An object projected into the air with some initial velocity, moving under the influence of gravity alone, is called a projectile.

Key Concept: Horizontal motion ( $v_x = \text{constant}$ ) and vertical motion ( $a_y = -g$ ) are independent.

Diagram of projectile motion defining Range, Maximum Height, and Time of Flight. — Key parameters of Projectile Motion: Range, Height, and Time.

Equation of Path (Trajectory)

Horizontal: $x = (v_0 \cos \theta)t \Rightarrow t = \dfrac{x}{v_0 \cos \theta}$

Vertical: $y = (v_0 \sin \theta)t - \dfrac{1}{2}gt^2$

Substitute $t$ : $y = (\tan \theta)x - \dfrac{g}{2(v_0 \cos \theta)^2} x^2$

This is the equation of a parabola.

Key Formulae

Time of Flight ( $T_f$ )

At $t = T_f$ , $y = 0$ :

$0 = (v_0 \sin \theta)T_f - \dfrac{1}{2}g T_f^2$

$T_f = \dfrac{2v_0 \sin \theta}{g}$

Maximum Height ( $H_m$ )

At max height, $v_y = 0$ :

$0 = (v_0 \sin \theta)^2 - 2g H_m$

$H_m = \dfrac{v_0^2 \sin^2 \theta}{2g}$

Horizontal Range ( $R$ )

$R = (v_0 \cos \theta) \cdot T_f$

$R= (v_0 \cos \theta)\cdot \dfrac{2v_0 \sin \theta}{g}$

$R= \dfrac{v^2_0 (2\sin \theta \cos \theta)}{g}$

$R = \dfrac{v_0^2 \sin 2\theta}{g}$

Max Range Condition:

$R_{\text{max}}$ when $\theta = 45^\circ$ .

Range = Max Height :

when $\theta = 76^\circ$ .

Problem 4: A cricket ball is thrown at 28 m/s at 30° to the horizontal. Calculate (a) Max Height, (b) Time of flight, (c) Range. ( $g=9.8$ m/s²)

Show Answer

(a) $H = \dfrac{28^2 \sin^2 30^\circ}{2(9.8)} = 10.0$ m.
(b) $T = \dfrac{2(28)(0.5)}{9.8} = 2.9$ s.
(c) $R = \dfrac{28^2 \sin 60^\circ}{9.8} = 69.3$ m.

Could you try to solve this problem?

5. Uniform Circular Motion (UCM)

When an object moves in a circular path at constant speed, it is in UCM. Velocity changes direction continuously → accelerated motion.

Derivation of Centripetal Acceleration

Consider a particle moving with constant speed $v$ in a circle of radius $R$ . Although speed is constant, the direction of velocity changes continuously, so the particle accelerates.

Step-by-Step Derivation 3 Marks

The geometry gives the ratio

$\dfrac{\Delta v}{v} = \dfrac{\Delta s}{r}$

Divide both sides by $\Delta t$ :

$\dfrac{\Delta v}{\Delta t} = \dfrac{v}{r} \cdot \dfrac{\Delta s}{\Delta t}$

Since $\Delta v / \Delta t = a$ and $\Delta s / \Delta t = v$ , we obtain

$a_c = \dfrac{v \cdot v}{r} = \dfrac{v^2}{r}$

Direction is always toward the center.

uniform circular motion, acceleration is always directed towards centre

$a_c = \dfrac{v^2}{R}$

Angular Speed ( $\omega$ )

The rate at which the particle sweeps angle is called angular speed:

$\omega = \dfrac{\Delta\theta}{\Delta t} \quad \text{(in rad/s)}$

Relation Between Linear and Angular Speed

Arc length: $\Delta s = R \Delta\theta$ . Divide by $\Delta t$ :

$v = \dfrac{\Delta s}{\Delta t} = R \dfrac{\Delta\theta}{\Delta t} = R \omega$

$\boxed{v = R \omega}$

Centripetal Acceleration in Terms of $\omega$

Substitute $v = R\omega$ into $a_c = v^2/R$ :

$a_c = \dfrac{(R\omega)^2}{R} = R \omega^2$

$\boxed{a_c = \omega^2 R}$

Connection to Time Period ( $T$ ) and Frequency ( $\nu$ )

In one full revolution: $\Delta\theta = 2\pi$ , time taken = $T$ (time period).
So, $\omega = \dfrac{2\pi}{T} = 2\pi\nu$ , where $\nu = \dfrac{1}{T}$ is frequency (Hz).

Problem 5: An insect sits on a record player rotating at 33 rpm. Radius = 10 cm. Find acceleration.

Show Answer

$\nu = 33/60 = 0.55$ Hz.
$\omega = 2\pi(0.55) = 3.45$ rad/s.
$a_c = (3.45)^2 (0.1) \approx 1.2$ m/s².

Key Concept Notes (Points to Ponder)

Path length ≥ |displacement|; equality only for straight-line motion without reversal.
2D motion with constant acceleration = two independent 1D motions along x and y.
Kinematic equations do NOT apply to UCM (acceleration direction changes).
In UCM, velocity is tangential; acceleration is centripetal (center-seeking).

← Previous Chapter Next Chapter: Laws of Motion →

1. Scalars and Vectors

1.1 Representation of a Vector

1.2 Magnitude of a Vector

1.3 Unit Vector

1.4 Types of Vectors

1.5 Vector Addition and Subtraction

1.6 Parallelogram Law of Vector Addition

2. Resolution of Vectors

3. Velocity and Acceleration in a Plane

4. Projectile Motion

Equation of Path (Trajectory)

Key Formulae

5. Uniform Circular Motion (UCM)

Derivation of Centripetal Acceleration

Angular Speed ()

Relation Between Linear and Angular Speed

Centripetal Acceleration in Terms of

Connection to Time Period () and Frequency ()

Key Concept Notes (Points to Ponder)

Angular Speed ( $\omega$ )

Centripetal Acceleration in Terms of $\omega$

Connection to Time Period ( $T$ ) and Frequency ( $\nu$ )