Chapter 3: Motion in a Plane
Unit II – Kinematics
3.1 Scalars and Vectors
| Property | Scalar | Vector |
|---|---|---|
| Definition | Magnitude only | Magnitude + Direction |
| Examples | Mass, speed, time, energy | Velocity, force, displacement, acceleration |
| Addition | Algebraic | Vector addition rules |
3.2 Vector Notation and Operations
A vector \(\vec{A}\) has magnitude \(|\vec{A}|\) and a direction.
3.2.1 Vector Addition – Triangle Law
\[ \vec{R} = \vec{A} + \vec{B} \]
3.2.2 Parallelogram Law
If \(\theta\) is the angle between \(\vec{A}\) and \(\vec{B}\):
\[ |\vec{R}| = \sqrt{A^2 + B^2 + 2AB\cos\theta} \]
\[ \tan\alpha = \frac{B\sin\theta}{A + B\cos\theta} \]
where \(\alpha\) is the angle \(\vec{R}\) makes with \(\vec{A}\).
3.2.3 Dot Product (Scalar Product)
\[ \vec{A} \cdot \vec{B} = AB\cos\theta \]
Properties:
- \(\hat{i}\cdot\hat{i} = \hat{j}\cdot\hat{j} = \hat{k}\cdot\hat{k} = 1\)
- \(\hat{i}\cdot\hat{j} = \hat{j}\cdot\hat{k} = \hat{k}\cdot\hat{i} = 0\)
3.2.4 Cross Product (Vector Product)
\[ |\vec{A} \times \vec{B}| = AB\sin\theta \]
Direction: Right-hand rule (perpendicular to both \(\vec{A}\) and \(\vec{B}\))
Properties:
- \(\hat{i}\times\hat{j} = \hat{k},\quad \hat{j}\times\hat{k} = \hat{i},\quad \hat{k}\times\hat{i} = \hat{j}\)
- \(\vec{A}\times\vec{B} = -\vec{B}\times\vec{A}\)
3.3 Resolution of Vectors
Any vector \(\vec{A}\) can be resolved into components:
\[ \vec{A} = A_x\hat{i} + A_y\hat{j} \]
\[ A_x = A\cos\theta, \quad A_y = A\sin\theta \]
\[ A = \sqrt{A_x^2 + A_y^2}, \quad \theta = \tan^{-1}!\left(\frac{A_y}{A_x}\right) \]
3.4 Projectile Motion
A projectile is an object given an initial velocity and then left to move under gravity alone.
3.4.1 Equations of Projectile Motion
Initial velocity \(u\) at angle \(\theta\) with horizontal:
\[ u_x = u\cos\theta, \quad u_y = u\sin\theta \]
Horizontal motion (no acceleration):
\[ x = u\cos\theta \cdot t \]
Vertical motion (acceleration \(= -g\)):
\[ y = u\sin\theta \cdot t - \frac{1}{2}gt^2 \]
3.4.2 Key Results
\[ \text{Time of flight:}\quad T = \frac{2u\sin\theta}{g} \]
\[ \text{Maximum height:}\quad H = \frac{u^2\sin^2\theta}{2g} \]
\[ \text{Horizontal range:}\quad R = \frac{u^2\sin 2\theta}{g} \]
\[ \text{Range is maximum at } \theta = 45°,\quad R_{max} = \frac{u^2}{g} \]
Trajectory equation (parabola):
\[ y = x\tan\theta - \frac{g,x^2}{2u^2\cos^2\theta} \]
3.4.3 Projectile Motion Diagram
3.5 Uniform Circular Motion
When a particle moves on a circular path with constant speed, it undergoes uniform circular motion (UCM).
- Speed is constant, but velocity changes direction → acceleration exists.
- This acceleration is called centripetal acceleration, directed towards the centre.
\[ a_c = \frac{v^2}{r} = \omega^2 r \]
\[ \text{Centripetal force:}\quad F_c = \frac{mv^2}{r} = m\omega^2 r \]
Angular velocity:
\[ \omega = \frac{2\pi}{T} = 2\pi f \]
Relation between linear and angular quantities:
\[ v = r\omega, \quad a = r\alpha \]
3.6 Uniform Circular Motion Diagram
Key Formulas Summary
| Formula | Quantity |
|---|---|
| \(R = \dfrac{u^2\sin 2\theta}{g}\) | Horizontal range |
| \(H = \dfrac{u^2\sin^2\theta}{2g}\) | Maximum height |
| \(T = \dfrac{2u\sin\theta}{g}\) | Time of flight |
| \(a_c = \dfrac{v^2}{r}\) | Centripetal acceleration |
| \(v = r\omega\) | Linear-angular speed relation |
| \(\vec{R} = \sqrt{A^2 + B^2 + 2AB\cos\theta}\) | Resultant of two vectors |
Practice Questions
Section A – MCQ (1 mark each)
Q1. The trajectory of a projectile is a:
(a) Straight line (b) Circle (c) Parabola (d) Hyperbola
Answer
(c) Parabola — from \(y = x\tan\theta - \dfrac{gx^2}{2u^2\cos^2\theta}\)
Q2. For a projectile, the range is maximum when the angle of projection is:
(a) 30° (b) 45° (c) 60° (d) 90°
Answer
(b) 45° — since \(R = \dfrac{u^2\sin 2\theta}{g}\), maximum when \(\sin 2\theta = 1\), i.e., \(2\theta = 90°\), \(\theta = 45°\).
Q3. A particle moves in a circle of radius \(r\) with uniform speed \(v\). Its centripetal acceleration is:
(a) \(vr\) (b) \(v^2r\) (c) \(\dfrac{v^2}{r}\) (d) \(\dfrac{v}{r^2}\)
Answer
(c) \(\dfrac{v^2}{r}\)
Section B – Short Answer (2–3 marks)
Q4. A ball is thrown horizontally from the top of a cliff 80 m high with a speed of \(20 \text{ m/s}\). Find:
(i) Time to reach the ground.
(ii) Horizontal distance travelled.
(\(g = 10 \text{ m/s}^2\))
Answer
(i) Vertical: \(h = \frac{1}{2}gt^2 \Rightarrow t = \sqrt{\frac{2 \times 80}{10}} = 4 \text{ s}\)
(ii) \(x = u_x \times t = 20 \times 4 = \mathbf{80 \text{ m}}\)
Q5. Two vectors of magnitude 3 N and 4 N make an angle of 90° with each other. Find the magnitude and direction of their resultant.
Answer
\[ R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = \mathbf{5 \text{ N}} \]
\[ \tan\alpha = \frac{4}{3} \Rightarrow \alpha = \tan^{-1}(4/3) \approx 53.1° \]
(with respect to the 3 N vector)
Section D – Competency-Based Questions
Q6. (Case Study) A cricket ball is hit at an angle of 30° to the horizontal with a speed of 40 m/s. (\(g = 10 \text{ m/s}^2\))
(i) Find the time of flight.
(ii) Find the maximum height.
(iii) Find the horizontal range. Would the range increase or decrease if the angle were 60°?
(iv) At what angle should the ball be hit to achieve the same range as at 30°?
Answer
(i) \(T = \dfrac{2 \times 40 \times \sin 30°}{10} = \dfrac{2 \times 40 \times 0.5}{10} = \mathbf{4 \text{ s}}\)
(ii) \(H = \dfrac{(40)^2 \sin^2 30°}{2 \times 10} = \dfrac{1600 \times 0.25}{20} = \mathbf{20 \text{ m}}\)
(iii) \(R = \dfrac{(40)^2 \sin 60°}{10} = \dfrac{1600 \times 0.866}{10} \approx \mathbf{138.6 \text{ m}}\)
At 60°: \(R = \dfrac{1600 \sin 120°}{10} = \dfrac{1600 \times 0.866}{10} \approx 138.6 \text{ m}\) — Same range!
(iv) \(R\) is the same at complementary angles. Same range at \(\theta’ = 90° - 30° = \mathbf{60°}\).