Motion in a Plane: Overview, Questions, Preparation

Motion in a plane involves analyzing a particle's trajectory using vector components along two perpendicular axes, typically $x$  and $y$ .

General Definition: Vectors in a plane, such as position, velocity, and acceleration, have magnitude and direction, resolved into components along perpendicular axes. NCERT Perspective: A particle's position in the $x-y$  plane is given by the position vector $\vec{r}=x\hat{i}+y\hat{j}$ . Velocity is $\vec{v}=\frac{d\vec{r}}{dt}=\frac{dx}{dt}\hat{i}+\frac{dy}{dt}\hat{j}$ , and acceleration is $\vec{a}=\frac{d\vec{v}}{dt}$ . Components are treated independently along each axis (Page 2, Fig. 3.1). Key Features:

• Vector addition follows the parallelogram law or triangle rule.
• Magnitude of a vector $\vec{v}=v_x\hat{i}+v_y\hat{j}$ is $\left|\vec{v}\right|=\sqrt[]{v_x^2+v_y^2}$ .
• Direction is given by $\mathrm{t}\mathrm{a}\mathrm{n}\theta =\frac{v_y}{v_x}$ .

JEE questions often involve resolving vectors into components for analysis.

General Definition: Projectile motion occurs when an object is launched with an initial velocity and moves under gravity alone, following a parabolic trajectory. NCERT Perspective: For a projectile launched at speed $u$  and angle $\theta$  from the horizontal (Page 3, Fig. 3.2), the motion is split into: - Horizontal Motion: $u_x=u\mathrm{c}\mathrm{o}\mathrm{s}\theta ,a_x=0$ , so $v_x=u\mathrm{c}\mathrm{o}\mathrm{s}\theta ,x=u\mathrm{c}\mathrm{o}\mathrm{s}\theta \cdot t$ . - Vertical Motion: $u_y=u\mathrm{s}\mathrm{i}\mathrm{n}\theta ,a_y=-g$ , so $v_y=u\mathrm{s}\mathrm{i}\mathrm{n}\theta -gt,y=u\mathrm{s}\mathrm{i}\mathrm{n}\theta \cdot t-\frac{1}{2}g{t}^2$ . Key parameters
(Page 4): - Time of flight: $T=\frac{2u\mathrm{s}\mathrm{i}\mathrm{n}\theta }{g}$ . - Range: $R=\frac{u^2\mathrm{s}\mathrm{i}\mathrm{n}2\theta }{g}$ . - Maximum height: $H=\frac{u^2{\mathrm{s}\mathrm{i}\mathrm{n}}^2\theta }{2g}$ .

• Trajectory: $y=x\mathrm{t}\mathrm{a}\mathrm{n}\theta -\frac{g{x}^2}{2u^2{\mathrm{c}\mathrm{o}\mathrm{s}}^2\theta }$ . Key Features:
• Trajectory is parabolic due to constant horizontal velocity and uniformly accelerated vertical motion.
• Maximum range occurs at $\theta ={45}^{\circ }$ .
• Range is equal for complementary angles $\theta$ and ${90}^{\circ }-\theta$ .

JEE problems test derivations and applications, like finding range or height.

General Definition: Projectile motion on an inclined plane involves launching a projectile on a surface sloped at an angle, altering time of flight and range. NCERT Perspective: For an incline at angle $\alpha$  (Page 7, Fig. 3.7): - Up the Incline: Initial velocity components along the incline $\left(x\right)$  and perpendicular $\left(y\right)$  are $u_x=u\mathrm{c}\mathrm{o}\mathrm{s}(\theta -\alpha ),u_y=u\mathrm{s}\mathrm{i}\mathrm{n}(\theta -\alpha )$ . Accelerations are $a_x=-g\mathrm{s}\mathrm{i}\mathrm{n}\alpha$ , $a_y=-g\mathrm{c}\mathrm{o}\mathrm{s}\alpha$ . Time of flight: $T=\frac{2u\mathrm{s}\mathrm{i}\mathrm{n}(\theta -\alpha )}{g\mathrm{c}\mathrm{o}\mathrm{s}\alpha }$ . Range: $R=\frac{u^2}{g{\mathrm{c}\mathrm{o}\mathrm{s}}^2\alpha }[\mathrm{s}\mathrm{i}\mathrm{n}(2\theta -\alpha )-\mathrm{s}\mathrm{i}\mathrm{n}\alpha ]$  (Page 8, Fig. 3.8). - Down the Incline: Components are $u_x=u\mathrm{c}\mathrm{o}\mathrm{s}(\theta +\alpha ),u_y=u\mathrm{s}\mathrm{i}\mathrm{n}(\theta +\alpha )$ , with $a_x=$   $g\mathrm{s}\mathrm{i}\mathrm{n}\alpha ,a_y=-g\mathrm{c}\mathrm{o}\mathrm{s}\alpha$ . Time of flight: $T=\frac{2u\mathrm{s}\mathrm{i}\mathrm{n}(\theta +\alpha )}{g\mathrm{c}\mathrm{o}\mathrm{s}\alpha }$ . Range: $R=\frac{u^2}{g{\mathrm{c}\mathrm{o}\mathrm{s}}^2\alpha }[\mathrm{s}\mathrm{i}\mathrm{n}(2\theta +\alpha )+\mathrm{s}\mathrm{i}\mathrm{n}\alpha ]$ (Page 9). Key Features:

• Coordinate system is often aligned with the incline for simplicity.
• Maximum range occurs at $\theta =\frac{\pi }{4}+\frac{\alpha }{2}$ (Page 10).
• Range formulas reduce to flat-ground case when $\alpha =0$ .

JEE questions may involve calculating range or time for specific angles.

General Definition: Relative motion describes the motion of one object as observed from another, using vector differences. NCERT Perspective: The relative velocity of object A with respect to B is ${\vec{v}}_{AB}={\vec{v}}_A-{\vec{v}}_B$ . Similarly, relative acceleration is ${\vec{a}}_{AB}={\vec{a}}_A-{\vec{a}}_B$  (Page 11). Examples include: - Boat crossing a river: Resultant velocity combines boat's velocity and stream's velocity (Page 13, Fig. 3.16). - Rain's apparent direction: Relative velocity of rain with respect to a moving observer (Page 12, Fig. 3.14). Key Features:

• Shortest time to cross a river occurs when the boat's velocity is perpendicular to the stream.
• Drift is the displacement along the stream's direction.

JEE problems focus on vector subtraction and angle calculations.

General Definition: Circular motion involves a particle moving along a circular path, with velocity tangential and acceleration radial (centripetal) in uniform cases. NCERT Perspective: For uniform circular motion (UCM) (Page 15, Fig. 3.21): - Angular velocity: $\omega =\frac{d\theta }{dt}=\frac{v}{r}$ . Centripetal acceleration: $a_r=\frac{v^2}{r}={\omega }^{2}r$ . For non-uniform circular motion (Page 16), tangential acceleration $a_t=\frac{dv}{dt}=r\alpha$ , where $\alpha =\frac{d\omega }{dt}$ . Total acceleration is $a=\sqrt[]{a_r^2+a_t^2}$ . Key Features:

• UCM requires a centripetal force directed toward the center.
• Non-uniform motion includes tangential acceleration due to changing speed.

JEE questions test acceleration calculations and angular-linear relations.

Example 1: A projectile is launched at $12\mathrm{m}/\mathrm{s}$  at ${45}^{\circ }$  from the horizontal on level ground. Find the range ( $g=10\mathrm{m}/{\mathrm{s}}^2$  ). (JEE Main, Page 4, Illustration 1) Using $R=\frac{u^2\mathrm{s}\mathrm{i}\mathrm{n}2\theta }{g}$  : $R=\frac{{12}^2\mathrm{s}\mathrm{i}\mathrm{n}\left(2\cdot {45}^{\circ }\right)}{10}=\frac{144\cdot 1}{10}=14.4\mathrm{m}$

Example 2: A projectile is fired horizontally at $98\mathrm{m}/\mathrm{s}$  from a 490 m high hill. Find the time to reach the ground ( $g=9.8\mathrm{m}/{\mathrm{s}}^2$  ). (JEE Main, Page 6, Illustration 5) Vertical motion: $u_y=0,a_y=-g,s_y=-490$ . Using $s_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$ : $-490=0-\frac{1}{2}\cdot 9.8\cdot t^2⟹t^2=\frac{980}{9.8}=100⟹t=10\mathrm{s}$

Example 3: A boat sails at $2\mathrm{m}/\mathrm{s}$  upstream in a river flowing at $1\mathrm{m}/\mathrm{s}$ . A person walks from front to rear at $1\mathrm{m}/\mathrm{s}$  relative to the boat. Find the person's speed relative to the ground. (JEE Main, Page 14, Illustration 13) Boat's velocity relative to ground: $v_A=v_{AB}+v_B=-2+1=$ $-1\mathrm{m}/\mathrm{s}$ . Person's velocity: $v_C=v_{CA}+v_A=1+(-1)=0\mathrm{m}/\mathrm{s}$ .