Motion in a Plane
Class‑11 Physics (CBSE)
“Motion in a plane” is the study of the displacement of a particle that can move in two dimensions (x‑y plane). It involves kinematics, dynamics, and the analysis of forces acting on the particle.”
1. Simple Explanation
- Plane motion means the particle’s trajectory is confined to a flat surface.
- Two independent directions (usually taken as x and y) are used to describe the motion.
- Equations of motion are derived by applying Newton’s laws separately along each axis.
- Many real‑world problems (projectile, circular motion, motion of trains, etc.) can be analysed in this framework.
2. Key Points & Equations
| Concept | Vector Representation | Key Equations |
|---|---|---|
| Displacement | (\vec r = x,\hat i + y,\hat j) | (\Delta \vec r = \vec r_f - \vec r_i) |
| Velocity | (\vec v = \frac{d\vec r}{dt} = v_x,\hat i + v_y,\hat j) | (v_x = \dfrac{dx}{dt}), (v_y = \dfrac{dy}{dt}) |
| Acceleration | (\vec a = \frac{d\vec v}{dt} = a_x,\hat i + a_y,\hat j) | (a_x = \dfrac{dv_x}{dt}), (a_y = \dfrac{dv_y}{dt}) |
| Uniform Rectilinear Motion (URM) | (\vec a = 0) | (x = x_0 + v_0 t), (y = y_0 + u_0 t) |
| Uniformly Accelerated Rectilinear Motion (UAR) | (\vec a =) constant | (v = u + at) (s = ut + \frac12 at^2) (v^2 = u^2 + 2as) |
| Projectile Motion | (g = 9.8,\text{m/s}^2) downward | Horizontal: (x = u \cos\theta, t) Vertical: (y = u \sin\theta, t - \frac12 g t^2) |
| Range | (R = \frac{u^2 \sin 2\theta}{g}) | Max Range when (\theta = 45^\circ) |
| Time of Flight | (T = \frac{2u\sin\theta}{g}) | |
| Maximum Height | (H = \frac{u^2 \sin^2\theta}{2g}) | |
| Circular Motion | (r) constant, (\vec a = - \frac{v^2}{r},\hat r) | (v = r\omega) (a_c = r\omega^2 = \frac{v^2}{r}) |
| Relative Motion | (\vec v_{AB} = \vec v_A - \vec v_B) | |
| Vector Addition | (\vec A + \vec B = (A_x+B_x)\hat i + (A_y+B_y)\hat j) |
Important: Always resolve forces and accelerations along the chosen axes before applying Newton’s second law.
3. Worked Example
Problem:
A stone is thrown from the top of a cliff of height (h = 45,\text{m}) with an initial speed of (u = 25,\text{m/s}) at an angle of (30^\circ) above the horizontal.
(a) Find the time of flight until the stone reaches the ground.
(b) Find the horizontal range.
(c) Find the maximum height above the ground the stone reaches.
Solution
- Decompose the initial velocity
[ u_x = u\cos 30^\circ = 25 \times \frac{\sqrt{3}}{2} = 21.65,\text{m/s} ] [ u_y = u\sin 30^\circ = 25 \times \frac{1}{2} = 12.5,\text{m/s} ]
- Vertical motion (take upward as positive, (g = 9.8,\text{m/s}^2))
Equation of motion in y‑direction: [ y = u_y t - \frac12 g t^2 ] The stone starts from the top of the cliff, so the vertical displacement to the ground is (-h = -45,\text{m}).
Thus: [ -45 = 12.5,t - \frac12 (9.8), t^2 ] [ 0 = 4.9 t^2 - 12.5 t - 45 ] Solve the quadratic: [ t = \frac{12.5 \pm \sqrt{12.5^2 + 4 \times 4.9 \times 45}}{2 \times 4.9} ] [ t = \frac{12.5 + \sqrt{156.25 + 882}}{9.8} ] [ t = \frac{12.5 + \sqrt{1038.25}}{9.8} ] [ t \approx \frac{12.5 + 32.2}{9.8} \approx \frac{44.7}{9.8} \approx 4.57,\text{s} ] (negative root discarded)
Answer (a): (T \approx 4.57 , \text{s})
- Horizontal range
[ x = u_x , T = 21.65 \times 4.57 \approx 99.0,\text{m} ]
Answer (b): (R \approx 99,\text{m})
- Maximum height
Maximum height occurs when vertical velocity becomes zero: [ v_y = u_y - g t_{\text{up}} = 0 \implies t_{\