Table of Contents
How to Use This Page
Read each concept carefully, then check the formula, common mistake, and exam tip before moving to the next. This page completely covers Motion in a Plane (2D Kinematics & Vectors) for CBSE Class 11 Physics.
Key Concepts
Motion in a Plane
Vectors, Projectiles, and Circular Motion
1
Vector Addition (Analytical Method)
Formula
The magnitude ($R$) and direction ($\alpha$ with respect to vector $\vec{A}$) of the resultant of two vectors $\vec{A}$ and $\vec{B}$ separated by an angle $\theta$.
$$R = \sqrt{A^2 + B^2 + 2AB \cos \theta}$$
$$\tan \alpha = \frac{B \sin \theta}{A + B \cos \theta}$$
2
Resolution of Vectors (2D Components)
Formula
Breaking a vector $\vec{A}$ into perpendicular components along the x and y axes. If $\theta$ is the angle made with the x-axis:
$$A_x = A \cos \theta \quad | \quad A_y = A \sin \theta$$
$$\vec{A} = A_x \hat{i} + A_y \hat{j} \quad | \quad |\vec{A}| = \sqrt{A_x^2 + A_y^2}$$
3
Equations of Motion in Vector Form
Formula
The 1D kinematic equations apply to 2D/3D motion when written in vector form, allowing us to solve $x, y,$ and $z$ components completely independently.
$$\vec{v} = \vec{u} + \vec{a}t$$
$$\vec{r} = \vec{r}_0 + \vec{u}t + \frac{1}{2}\vec{a}t^2$$
4
Projectile Motion: Time & Max Height
Formula
For a projectile launched with initial velocity $u$ at an angle $\theta$ to the horizontal. The Time of Flight ($T$) and Maximum Height ($H$) depend entirely on the vertical component ($u \sin \theta$).
$$T = \frac{2u \sin \theta}{g} \quad | \quad H = \frac{u^2 \sin^2 \theta}{2g}$$
5
Projectile Motion: Horizontal Range
Formula
The total horizontal distance ($R$) covered. It depends on both horizontal and vertical components. Maximum range occurs when launched at $45^\circ$.
$$R = \frac{u^2 \sin 2\theta}{g} \quad | \quad R_{\text{max}} = \frac{u^2}{g} \text{ (at } \theta = 45^\circ)$$
6
Equation of Trajectory
Formula
The mathematical relation between the $y$ and $x$ coordinates of a projectile, independent of time. It proves that the path is a parabola.
$$y = x \tan \theta – \frac{g x^2}{2 u^2 \cos^2 \theta}$$
$$y = x \tan \theta \left(1 – \frac{x}{R}\right) \text{ [Alternative Form]}$$
7
Relative Velocity in 2D (Rain-Man Problem)
Case Formula
If rain is falling vertically with velocity $\vec{v}_r$ and a man walks horizontally with $\vec{v}_m$, the relative velocity of rain with respect to the man ($\vec{v}_{rm}$) determines the angle $\theta$ he must hold his umbrella.
$$\vec{v}_{rm} = \vec{v}_r – \vec{v}_m \quad | \quad \tan \theta = \frac{v_m}{v_r} \text{ (from vertical)}$$
8
River-Boat Problems
Case Formulas
Let $v_b$ be velocity of boat in still water, $v_r$ be velocity of river flow, and $d$ be river width.
$$\text{Shortest Time: Cross perpendicular to flow } \Rightarrow t_{\text{min}} = \frac{d}{v_b}$$
$$\text{Shortest Path (Straight Across): } \sin \theta = \frac{v_r}{v_b} \text{ (upstream angle)}$$
9
Uniform Circular Motion
Formula
When a particle moves in a circle of radius $r$ with constant speed $v$. The angular velocity is $\omega$.
$$v = \omega r \quad | \quad \omega = \frac{2\pi}{T} = 2\pi\nu$$
10
Centripetal Acceleration
Formula
Even if speed is constant in circular motion, velocity changes continuously due to changing direction. This requires a centripetal acceleration ($a_c$) directed exactly towards the center.
$$a_c = \frac{v^2}{r} = \omega^2 r$$
Concept Deep Dive
01
The Independence of Motion
Why 2D is just two 1Ds at the same timeThe most powerful concept in this chapter is that horizontal motion and vertical motion are completely independent of each other.
If you fire a bullet perfectly horizontally from a gun at 1000 m/s, and simultaneously drop a second bullet from your hand at the exact same height, both bullets will hit the ground at the exact same time. Gravity pulls them downwards at $9.8 \text{ m/s}^2$ regardless of what they are doing horizontally. You can always split a 2D problem into a pure X-axis problem and a pure Y-axis problem, linked only by the common variable: Time ($t$).
If you fire a bullet perfectly horizontally from a gun at 1000 m/s, and simultaneously drop a second bullet from your hand at the exact same height, both bullets will hit the ground at the exact same time. Gravity pulls them downwards at $9.8 \text{ m/s}^2$ regardless of what they are doing horizontally. You can always split a 2D problem into a pure X-axis problem and a pure Y-axis problem, linked only by the common variable: Time ($t$).
02
Non-Uniform Circular Motion
When you step on the gas in a curveIn Uniform Circular Motion, speed is constant, so the only acceleration is Centripetal ($a_c$), pointing to the center. But what if a car is accelerating while taking a circular turn?
Now, the car has a Tangential Acceleration ($a_t = dv/dt$) along the path, AND a Centripetal (Radial) Acceleration ($a_c = v^2/r$) towards the center. Because these two vectors are perfectly perpendicular ($90^\circ$) to each other, the total net acceleration is their vector sum.
Now, the car has a Tangential Acceleration ($a_t = dv/dt$) along the path, AND a Centripetal (Radial) Acceleration ($a_c = v^2/r$) towards the center. Because these two vectors are perfectly perpendicular ($90^\circ$) to each other, the total net acceleration is their vector sum.
$$a_{\text{net}} = \sqrt{a_c^2 + a_t^2}$$
Compare & Contrast
✗ Scalar Product (Dot Product)
- Yields a Scalar (number).
- $\vec{A} \cdot \vec{B} = |A||B| \cos \theta$.
- Commutative: $\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}$.
- Maximum when parallel ($0^\circ$), Zero when perpendicular ($90^\circ$).
- Example: Work ($W = \vec{F} \cdot \vec{s}$).
✓ Vector Product (Cross Product)
- Yields a Vector (with direction).
- $|\vec{A} \times \vec{B}| = |A||B| \sin \theta$.
- Anti-commutative: $\vec{A} \times \vec{B} = -(\vec{B} \times \vec{A})$.
- Zero when parallel ($0^\circ$), Maximum when perpendicular ($90^\circ$).
- Example: Torque ($\vec{\tau} = \vec{r} \times \vec{F}$).
Remember
If a question asks you to “Prove that two given vectors are perpendicular”, immediately calculate their Dot Product. If it equals $0$, you have proved it!
Common Mistakes to Avoid
Mistake 1
Velocity at the Highest Point is NOT Zero: In 1D motion (throwing straight up), $v=0$ at the top. But in a 2D Projectile, only the vertical velocity ($v_y$) becomes zero at the highest point. The object is still moving horizontally with $v_x = u \cos \theta$.
Mistake 2
Misidentifying the Vector Angle: When resolving vectors, students blindly use $F_x = F \cos \theta$. This is ONLY true if $\theta$ is the angle made with the X-axis. If the question gives the angle made with the Y-axis, then $F_y = F \cos \theta$ and $F_x = F \sin \theta$. Always draw a triangle to be sure!
Mistake 3
River-Boat “Shortest Path” Failure: To cross a river along the shortest possible path (straight across), you CANNOT point your boat straight across. If you do, the river will push you downstream. You must aim the boat upstream at an angle to counteract the river flow ($\sin \theta = v_r / v_b$).
Exam Tips
Tip 1
Complementary Angles give the Same Range! A projectile launched at $30^\circ$ and one launched at $60^\circ$ with the same initial speed will land in the exact same spot ($R_{30} = R_{60}$). In general, angles $\theta$ and $(90^\circ – \theta)$ have the same horizontal range.
Tip 2
If a problem asks for the magnitude of change in velocity ($\Delta v$) when a particle completes half a circle ($180^\circ$) at constant speed $v$: The initial velocity is $v$, the final is $-v$. Change is $\Delta v = |-v – v| = 2v$. It is NOT zero!
Expected Exam Questions
SQ
Board Pattern Questions
Class 11 · Motion in a Plane · CBSE Exam1
A stone is thrown horizontally with a speed of $15 \text{ m/s}$ from the top of a cliff $490 \text{ m}$ high. Find the time taken to hit the ground. ($g = 9.8 \text{ m/s}^2$) [1 mark]
Answer
$10 \text{ s}$
📝
Explanation
Because it is thrown horizontally, its initial vertical velocity is $u_y = 0$.
Using the 2nd equation of motion for the Y-axis: $s_y = u_y t + \frac{1}{2}gt^2$.
$490 = 0 + \frac{1}{2}(9.8)t^2 \implies 490 = 4.9t^2 \implies t^2 = 100 \implies t = 10 \text{ s}$.
2
Show that there are two angles of projection for which the horizontal range is the same. [2 marks]
Answer
Angles are $\theta$ and $90^\circ – \theta$.
📝
Explanation
The formula for Range is $R_1 = \frac{u^2 \sin 2\theta}{g}$.
Let the second angle be $(90^\circ – \theta)$. Calculate its range $R_2$:
$R_2 = \frac{u^2 \sin 2(90^\circ – \theta)}{g} = \frac{u^2 \sin (180^\circ – 2\theta)}{g}$.
Since $\sin(180^\circ – x) = \sin x$, we get $R_2 = \frac{u^2 \sin 2\theta}{g} = R_1$.
Therefore, ranges are equal for complementary angles.
3
Rain is falling vertically with a speed of $30 \text{ m/s}$. A woman rides a bicycle with a speed of $10 \text{ m/s}$ in the North to South direction. What is the direction in which she should hold her umbrella? [3 marks]
Answer
$\tan^{-1}(1/3)$ or $18^\circ 26’$ South of vertical
📝
Explanation
Let vertical downward be the $-y$ direction ($\hat{j}$) and South be the $+x$ direction ($\hat{i}$).
Velocity of rain $\vec{v}_r = -30\hat{j} \text{ m/s}$.
Velocity of woman $\vec{v}_m = 10\hat{i} \text{ m/s}$.
Relative velocity of rain w.r.t woman $\vec{v}_{rm} = \vec{v}_r – \vec{v}_m = -30\hat{j} – 10\hat{i}$.
The umbrella must be held in the direction opposite to $\vec{v}_{rm}$. The angle $\theta$ with the vertical is given by:
$\tan \theta = \frac{|v_m|}{|v_r|} = \frac{10}{30} = \frac{1}{3}$.
Therefore, $\theta = \tan^{-1}(1/3)$ towards the South from the vertical.
Concept Map
Motion in a Plane connects to →
Mechanics
1D Kinematics
Vector Algebra
Newton’s Laws
Work & Energy (Dot Product)
Rotational Motion (Cross Product)
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