Class 11 — Motion in a Plane

Chapter Overview & Weightage

Motion in a Plane is the bridge from one-dimensional kinematics to vector physics. It introduces vector addition, projectile motion, and uniform circular motion — concepts that resurface throughout Class 11 and 12 mechanics. CBSE board exams allocate $4$ – $6$ marks here, typically as one numerical and one short-answer question.

Year	Marks Allocated	Question Type
2024	5	Projectile range derivation
2023	3	Vector resolution numerical
2022	5	Circular motion + projectile combo
2021	4	Maximum range angle proof
2020	3	Two-vector resultant

CBSE prefers vector-based and projectile questions. Circular motion appears occasionally but is more developed in Class 11 Laws of Motion. Always show vector diagrams for full marks on derivations.

Key Concepts You Must Know

Scalars vs vectors — magnitude only vs magnitude with direction.
Vector addition — triangle law, parallelogram law, polygon law.
Vector resolution — components along x and y axes.
Position vector and displacement vector — $\vec{r}$ and $\Delta\vec{r}$ .
Velocity and acceleration as vectors — derivatives of $\vec{r}$ .
Projectile motion — horizontal launch, oblique launch, range, time of flight.
Uniform circular motion — centripetal acceleration $v^2/r$ .

Important Formulas

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

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

For two vectors at angle $\alpha$ :

$|\vec{R}| = \sqrt{A^2 + B^2 + 2AB\cos\alpha}$

When to use: any time a problem asks for resultant magnitude or direction.

For initial speed $u$ at angle $\theta$ :

$T = \frac{2u\sin\theta}{g}, \quad H = \frac{u^2\sin^2\theta}{2g}, \quad R = \frac{u^2\sin 2\theta}{g}$

Maximum range at $\theta = 45°$ .

When to use: ground-to-ground projectile problems.

$a_c = \frac{v^2}{r} = \omega^2 r$

Direction: always toward the centre.

When to use: any uniform circular motion problem.

Solved Previous Year Questions

PYQ 1 (CBSE 2024, 5 marks)

A projectile is fired with velocity $u$ at angle $\theta$ . Derive expressions for time of flight, maximum height, and horizontal range. Find the angle for maximum range.

Solution. Decompose $u$ into $u_x = u\cos\theta$ and $u_y = u\sin\theta$ .

Time of flight: vertical motion gives $0 = u\sin\theta\cdot T - \tfrac{1}{2}gT^2$ , so $T = 2u\sin\theta/g$ .

Maximum height: at apex, $v_y = 0$ , so $0 = u^2\sin^2\theta - 2gH$ , giving $H = u^2\sin^2\theta/(2g)$ .

Range: $R = u\cos\theta \cdot T = u^2\sin 2\theta/g$ .

For maximum $R$ , $\sin 2\theta = 1$ , so $\theta = 45°$ , $R_{\max} = u^2/g$ .

PYQ 2 (CBSE 2022, 3 marks)

Two vectors of magnitudes $3\,\text{units}$ and $4\,\text{units}$ are inclined at $60°$ . Find the magnitude and direction of the resultant.

Solution.

$R = \sqrt{9 + 16 + 2(3)(4)(0.5)} = \sqrt{37} \approx 6.08\,\text{units}$

$\tan\alpha = \frac{4\sin 60°}{3 + 4\cos 60°} = \frac{2\sqrt{3}}{5} \approx 0.693$

$\alpha \approx 34.7°$ from the $3$ -unit vector.

PYQ 3 (CBSE 2021, 4 marks)

A stone is thrown horizontally from a tower of height $80\,\text{m}$ with velocity $20\,\text{m/s}$ . Find (a) time to hit ground, (b) horizontal distance covered, (c) speed at impact. Take $g = 10\,\text{m/s}^2$ .

Solution. (a) Vertical motion: $80 = \tfrac{1}{2}(10)t^2 \implies t = 4\,\text{s}$ . (b) Horizontal distance $= 20 \times 4 = 80\,\text{m}$ . (c) At impact: $v_x = 20$ , $v_y = 10 \times 4 = 40\,\text{m/s}$ . Speed $= \sqrt{400 + 1600} = \sqrt{2000} \approx 44.7\,\text{m/s}$ .

Difficulty Distribution

Easy ( $\sim 30\%$ ): vector addition, simple resolution, definitions.
Medium ( $\sim 50\%$ ): projectile motion calculations, derivations.
Hard ( $\sim 20\%$ ): combined motion (projectile + circular), oblique launches with elevation differences.

Expert Strategy

Drill the projectile derivation cold — toppers can write $T$ , $H$ , $R$ in $90$ seconds. CBSE asks the same derivation almost every year.

The fastest way to crack this chapter:

Memorise the three projectile formulas and prove them once.
Practice $20$ vector resolution problems until you can resolve in any direction without thinking.
For circular motion, internalise $a_c$ direction (always to centre, even when speed is constant).

Common Traps

Trap 1: Confusing $\sin\theta$ with $\sin 2\theta$ in range formula. Always write $R = u^2\sin 2\theta/g$ exactly.

Trap 2: Treating projectile motion as one-dimensional. Always split into x and y components first.

Trap 3: Forgetting that horizontal velocity is constant during projectile motion (no horizontal acceleration).

Trap 4: Using degrees and radians inconsistently. CBSE prefers degrees; calculator answers default to radians.

Trap 5: Confusing speed with velocity for uniform circular motion. Speed is constant; velocity (a vector) keeps changing direction.

Quick Revision Notes

Resultant of two perpendicular vectors $A$ and $B$ : $\sqrt{A^2 + B^2}$ .
For maximum range on level ground, launch at $45°$ .
For two angles $\theta$ and $90° - \theta$ , the range is the same.
Projectile path is a parabola; circular path is a circle (obviously, but stated for clarity).
In uniform circular motion, speed is constant but velocity changes — there is non-zero acceleration.

This chapter is a high-yield scoring topic if you drill the formulas. Students who internalise the three projectile equations rarely lose marks here.