Motion in a PlaneNEET Physics · Class 11 · NCERT Chapter 3

Introduction

Motion in a straight line covered the simplest case: a particle moving along one axis. In real NEET problems the motion almost always lives in a plane. A cricket ball follows a parabolic path, a cyclist turns a corner, a satellite orbits the earth. To handle all of these we need vectors.

For NEET, this chapter is tested heavily. Projectile motion alone shows up almost every year, often in two forms: a standard launch from the ground and a horizontal projectile from a cliff. Vectors and uniform circular motion are tested almost as often. Master this chapter and you also unlock Laws of Motion, Work-Energy, and Rotational Motion which all build on it.

Scalars and vectors

A scalar is a quantity that has only a magnitude — temperature, mass, energy, time. A vector has both magnitude and direction — displacement, velocity, force, acceleration.

Notation

We write a vector as $\vec{A}$ or in bold A. Its magnitude is $|\vec{A}|$ or just $A$. A unit vector along $\vec{A}$ is $\hat{A}=\vec{A}/|\vec{A}|$.

The Cartesian unit vectors are $\hat{i}$ (x), $\hat{j}$ (y), $\hat{k}$ (z). Any vector in the xy-plane can be written as $\vec{A}=A_x\hat{i}+A_y\hat{j}$.

Equality and types of vectors

• Two vectors are equal if they have the same magnitude AND the same direction.
• The null (zero) vector has zero magnitude and any direction. $\vec{A}-\vec{A}=\vec{0}$.
• A negative vector $-\vec{A}$ has the same magnitude as $\vec{A}$ but opposite direction.

Vector addition

Triangle law

Place the tail of $\vec{B}$ at the head of $\vec{A}$. The resultant $\vec{R}=\vec{A}+\vec{B}$ goes from the tail of $\vec{A}$ to the head of $\vec{B}$.

Parallelogram law

If two vectors of magnitude $A$ and $B$ have an angle $\theta$ between them, the magnitude of the resultant is:

The angle $\alpha$ that the resultant makes with $\vec{A}$ is:

Component method (the NEET shortcut)

Resolve each vector into x and y components. Add the x-components to get $R_x$, and the y-components to get $R_y$. Then:

This works for any number of vectors and is far faster than drawing parallelograms.

Resolving a vector into components

For a vector $\vec{A}$ of magnitude $A$ making angle $\theta$ with the x-axis:

Recovering the magnitude and angle from components:

Dot and cross products

Dot product (scalar product)

• The dot product is a scalar.
• $\vec{A}\cdot \vec{A}=A^2$.
• If two vectors are perpendicular, $\vec{A}\cdot \vec{B}=0$.
• NEET use: work $W=\vec{F}\cdot \vec{d}$, power $P=\vec{F}\cdot \vec{v}$.

Cross product (vector product)

• The cross product is a vector perpendicular to both $\vec{A}$ and $\vec{B}$.
• Direction follows the right-hand rule.
• $\vec{A}\times \vec{A}=\vec{0}$, $\vec{A}\times \vec{B}=-\vec{B}\times \vec{A}$.
• NEET use: torque $\vec{\tau }=\vec{r}\times \vec{F}$, angular momentum $\vec{L}=\vec{r}\times \vec{p}$.

Standard cross products of unit vectors

Position, velocity and acceleration in two dimensions

For a particle in a plane, the position vector is $\vec{r}=x\hat{i}+y\hat{j}$.

Crucially, the x and y motions are independent: solve them separately using the kinematic equations from Motion in a Straight Line. This is the single most important idea in projectile motion.

Projectile motion

A projectile is any object that, after being launched, moves only under gravity (no air resistance). The classic example: a ball kicked at angle $\theta$ with initial speed $u$.

Set up

Take the launch point as the origin, x horizontal, y vertical (upward positive).

• Initial velocity components: $u_x=u\cos \theta$, $u_y=u\sin \theta$.
• Acceleration: $a_x=0$, $a_y=-g$.

The x-motion is uniform (constant velocity). The y-motion is uniformly accelerated (free fall). Apply the kinematic equations to each direction.

Position at time t

Velocity at time t

The horizontal velocity never changes. Only the vertical velocity does.

Trajectory equation (parabola)

Eliminate $t$ from the position equations:

This is a parabola opening downward — the standard result you should be able to derive from memory.

Range, maximum height and time of flight

Time of flight

The projectile lands when $y=0$. Solving the y-equation:

Maximum height

Reached when $v_y=0$:

Range

Horizontal distance covered in time $T$:

Maximum range

$\sin 2\theta$ is largest when $2\theta =90°$, i.e. $\theta =45°$:

Complementary angles trick

Since $\sin 2\theta =\sin (180°-2\theta )=\sin 2(90°-\theta )$, two angles $\theta$ and $(90°-\theta )$ always give the same range. So $30°$ and $60°$ produce the same range, just different times of flight and heights.

Horizontal projectile (off a cliff)

Special case: an object launched horizontally with speed $u$ from a height $h$. Equivalent to projectile motion with $\theta =0$, but the y-motion starts from height $h$ instead of zero.

• Time to hit the ground: $T=\sqrt{2h/g}$ (free fall, since $u_y=0$).
• Horizontal distance: $R=uT=u\sqrt{2h/g}$.
• Speed on impact: $v=\sqrt{u^2+2gh}$.

Uniform circular motion

A particle moving in a circle at constant speed has uniform circular motion. Speed is constant but velocity is not — direction changes continuously, so there must be an acceleration.

Angular variables

• Angular velocity $\omega =d\theta /dt$, in rad/s.
• Time period $T=2\pi /\omega$, frequency $f=1/T$.
• Linear and angular speeds: $v=r\omega$.

Centripetal acceleration

The acceleration is directed toward the centre of the circle and has magnitude:

This acceleration changes only the direction of velocity, not its speed. The accompanying force is the centripetal force $F_c=m{v}^2/r$.

Tangential acceleration

If the speed of the particle also changes, there is a tangential component $a_t=dv/dt$. The total acceleration has magnitude $a=\sqrt{a_c^2+a_t^2}$. In uniform circular motion, $a_t=0$.

Worked NEET problems

Summary cheat sheet

• Vector addition (component method): $R_x=\sum A_x$, $R_y=\sum A_y$, $R=\sqrt{R_x^2+R_y^2}$.
• Resolution: $A_x=A\cos \theta$, $A_y=A\sin \theta$.
• Dot product: $\vec{A}\cdot \vec{B}=AB\cos \theta$ — gives a scalar.
• Cross product: $|\vec{A}\times \vec{B}|=AB\sin \theta$ — gives a vector perpendicular to both.
• Projectile (launched from ground): $T=2u\sin \theta /g$, $H=u^2{\sin }^2\theta /(2g)$, $R=u^2\sin 2\theta /g$.
• Maximum range at $\theta =45°$: $R_{\text{max}}=u^2/g$.
• Complementary angles: $\theta$ and $(90°-\theta )$ give the same range.
• Horizontal projectile: $T=\sqrt{2h/g}$, $R=u\sqrt{2h/g}$.
• Centripetal acceleration: $a_c=v^2/r={\omega }^{2}r$, directed toward the centre.
• Linear-angular link: $v=r\omega$, $T=2\pi /\omega$.

Next: try the interactive widgets for vectors, projectile motion and circular motion, or work through the 30+ NEET PYQs with full solutions. To time yourself, take the free 10-question mock test.