MOTION IN A PLANE-Notes
Physics - Notes
SCALARS AND VECTORS
SCALARS QUANTITIES
A scalar
quantity is a quantity with magnitude only. It is specified
completely by a single number, along with the proper unit.
Examples are : the distance between two points,
mass of an object, the temperature of a body and the
time at which a certain event happened. The rules for
combining scalars are the rules of ordinary algebra.
Scalars can be added, subtracted, multiplied and divided just as the ordinary numbers
VECTOR QUANTITIES
A vector quantity is a quantity that has both a magnitude and a direction and obeys the triangle law of addition or equivalently the parallelogram law of addition.
Represntation of vector
To represent a vector, it is written bold faced. Thus, a velocity vector can be
represented by a symbol \(\textbf{v}\). Since bold face is difficult to produce, when written by hand, a
vector is often represented by an arrow placed over a letter, say \(\vec{\mathrm{v}}\).
Thus, both \(\textbf{v}\) and \(\vec{\mathrm{v}}\) represent the velocity vector.
The magnitude of a vector is often called its absolute value,
indicated by \(\boldsymbol{\mid \mathrm{v}\mid} = \mathrm{v}\).
Position and Displacement Vectors
Position Vector
To describe the location of a particle in a plane, we first choose a reference point, usually the origin (O) of a coordinate system.- The position vector of a particle is a vector drawn from the origin to the point where the particle is located.
- It tells us the exact position of the object relative to the origin.
Definition
The position vector \(\vec{r}\) is the vector joining the origin to the position of the particle
Mathematical Form
If a particle is at point P(x, y), then its position vector is: \[\boxed{\;\boldsymbol{\vec{r}=x\,\hat{i} + y\,\hat{j}}\;}\] Where,\(x\) = x-coordinate of the particle
\(y\) = y-coordinate of the particle
\(\hat{i}\) and \(\hat{j}\) are unit vectors along x- and y-axes respectively
Displacement Vector
When a particle moves from one position to another, the change in its position is described by the displacement vector.
Definition
The displacement vector is the vector drawn from the initial position of the particle to its final position.
Mathematical form
If:
Initial position vector = \(\vec{r}\)
Final position vector = \(\vec{r'}\)
then displacement vector \(\vec{s}\) is:
\[\boxed{\;\boldsymbol{\vec{s}=\vec{r'}-\vec{r}}\;}\]
Position Vector vs Displacement Vector
| Position Vector | Displacement Vector |
|---|---|
| Describes location of a particle | Describes change in position |
| Drawn from origin to particle | Drawn from initial to final point |
| Depends on choice of origin | Does not depend on origin |
| Changes if origin changes | Remains the same for same motion |
Equality of Vectors
Two vectors are said to be equal if both their magnitude and direction are exactly the same, even if
they
are located at different positions in space.
The actual starting point of a vector does not matter for equality — only its length and direction
matter.
Conditions for Equality of Vectors
Two vectors \(\vec{A}\) and \(\vec{B}\) are equal if:
- Their magnitudes are equal
\[\mid \vec{A} \mid = \mid\vec{B}\mid\] - They have the same direction
(That is, they are parallel and point in the same sense.)
If both conditions are satisfied, we write: \[\vec{A}=\vec{B}\]
MULTIPLICATION OF VECTORS BY REAL NUMBERS
Multiplying a vector A with a positive number λ gives a vector whose magnitude is changed by the factor λ but the direction is the same as that of A :
\[\begin{aligned}\mid\lambda A\mid&=\lambda\mid A \mid\\ \text{if }&\lambda\gt0\end{aligned}\]The factor λ by which a vector \(\vec{A}\) is multiplied could be a scalar having its own physical dimension. Then, the dimension of \(\lambda \vec{A}\) is the product of the dimensions of λ and \(\vec{A}\).
ADDITION AND SUBTRACTION OF VECTORS — GRAPHICAL METHOD
Vector Addition (Graphical Method)
Vector addition means finding a single vector that has the same effect as two or more vectors acting together. This single vector is called the resultant vector.
Triangle Law of Vector Addition
If two vectors are represented (in magnitude and direction) by two sides of a triangle taken in
order, then the third side of the triangle taken in the opposite order represents their
resultant.
This graphical method is called the head-to-tail method.
Method
- Draw the first vector \(\vec{A}\)
- From the head of \(\vec{A}\) , draw the second vector \(\vec{B}\).
- Join the tail of \(\vec{A}\) to the head of \(\vec{B}\).
- The joining vector represents the resultant \(\vec{R}\).
Parallelogram Law of Vector Addition
If two vectors acting at a point are represented by the two adjacent sides of a parallelogram, then the diagonal of the parallelogram passing through that point represents their resultant.
Method
- Draw vectors \(\vec{A}\) and \(\vec{B}\) from the same point.
- Complete the parallelogram using \(\vec{A}\) and \(\vec{B}\).
- Draw the diagonal from the common point.
- This diagonal gives the resultant vector.
Vector Subtraction (Graphical Method)
Vector subtraction means finding the vector which, when added to one vector, gives another vector.
Basic Idea
\[\vec{A}-\vec{B}=\vec{A}+(-\vec{B})\]So, subtraction is done by adding the negative of a vector.
RESOLUTION OF VECTORS
Resolution of a vector is the process of splitting one vector into two or more simpler vectors, called components, such that their combined effect is exactly the same as the original vector. These components are usually chosen along convenient directions—most commonly along the horizontal (x-axis) and vertical (y-axis).
Think of it like this: instead of dealing with one slanted push or motion, we break it into one part acting sideways and another acting upward or downward. Together, these parts fully describe the original effect.
Why do we resolve vectors?
Resolving vectors makes calculations and understanding much easier because:
- Motion in two dimensions can be treated as two independent one-dimensional motions.
- Forces acting at angles can be analyzed using simple algebra.
- It helps in applying equations of motion separately along each direction.
In short, resolution turns a complicated problem into smaller, manageable pieces.
Resolution of a vector in a plane
Consider a vector \(\vec{A}\) making an angle \(\theta\) with the positive x-axis.
- The component of \(\vec{A}\) along the x-axis is called the horizontal component
- The component along the y-axis is called the vertical component
These two components are perpendicular to each other and together recreate the original vector.
Mathematical expression of components
-
Horizontal component:
\[A_x=A\cos \theta\] - Vertical component:
\[A_y=A\sin \theta\] - Here:
\(\theta\) is the angle the vector makes with the x-axis
\(\cos \theta\) and \(\sin \theta\) decide how much of the vector lies along each direction
If the magnitude of the vector is A, then:
Unit Vectors
A unit vector is a vector that has magnitude 1 and points along a specific direction.
In the rectangular coordinate system:
- \(\hat i\) points along the positive x-axis
- \(\hat j\) points along the positive y-axis
- \(\hat k\) points along the positive z-axis
These three unit vectors are mutually perpendicular and form the basic directional framework of space.
Resolving a vector using unit vectors
Consider a vector \(\vec{A}\)acting in space. When we resolve it, we break it into components along the \(x,\ y,\ \text{and }z\) directions. If:
\({A}_{x}\) is the component along x-axis
\({A}_{y}\) is the component along y-axis
\({A}_{z}\) is the component along z-axis
then the vector can be written as:
This expression tells us how much of the vector acts in each direction.
Resolution in two dimensions (x–y plane)
For motion confined to a plane, the z-component is zero.
If a vector \(\vec{A}\) makes an angle \(\theta\) with the x-axis and has magnitude \({A}\), then:
So the vector becomes:
\[\vec{A}=A\cos \theta\ \hat{i}+A\sin \theta\ \hat{j}\] This form is extremely useful in projectile motion and planar kinematics.The magnitude of the vector is then given by:
\[\mid\vec{A}\mid=\sqrt{A_x^2+A_y^2}\]Resolution in three dimensions
If a vector acts in three-dimensional space, all three components may be present.
\[\vec{A}=A_x\hat{i}+A_y\hat{j}+A_z\hat{k}\]
Here:
Each coefficient represents the projection of the vector along that axis
The unit vectors simply provide direction, not magnitude
The magnitude of the vector is then given by:
VECTOR ADDITION – ANALYTICAL METHOD
Let us consider two vectors \(\vec{A}\) and \(\vec{B}\) in xy-plane with components \(A_x,\ A_y\) and \(B_x,\
B_y\)
We may write vector \(\vec{A}\) and \(\vec{B}\)
Let R be their sum. therfore,
\[\begin{aligned} \vec{R}&=\vec{A}+\vec{B}\\ &=(A_x\hat{i} + A_y\hat{j}) + (B_x\hat{i} + B_y\hat{j}) \end{aligned} \]Since vectors obey the commutative and associative laws, we can arrange and regroup the vectors
\[\vec{R}=(A_x+B_x)\hat{i} + (A_y+B_y)\hat{j}\] Since \[\vec{R}=R_x\hat{i}+R_y\hat{j}\] \[ \Rightarrow\quad\left\{ \begin{aligned} R_x &= A_x + B_x\\ R_y &= A_y + B_y \end{aligned} \right. \]Similarly in three dimension
\[\begin{aligned}\vec{A}&=A_x\hat{i}+A_y\hat{j}+A_z\hat{k}\\ \vec{B}&=B_x\hat{i}+B_y\hat{j}+B_z\hat{k} \end{aligned}\] \[\begin{aligned} \vec{R}&=\vec{A}+\vec{B}\\ &=(A_x\hat{i} + A_y\hat{j} + A_z\hat{k}) + (B_x\hat{i} + B_y\hat{j} + B_z\hat{k}) \end{aligned}\] \[\vec{R}=R_x\hat{i}+R_y\hat{j} + R_z\hat{k}\] \[ \Rightarrow\quad\left\{ \begin{aligned} R_x &= A_x + B_x\\ R_y &= A_y + B_y\\ R_z &= A_z + B_z \end{aligned} \right. \]Resultant of two vectors \(\vec{A}\) and \(\vec{B}\) angle \(\theta\) between them
SN is normal to OP and PM is normal to OS
From Pythagoras Theorem
Substututing these values in Equation (1)
\[\small\begin{aligned} OS^2 &= (A + B \cos \theta)^2 + (B \sin \theta)^2\\\\ &=A^2+B^2\cos^2 \theta+2AB\cos \theta + B^2\sin^2 \theta\\\\ &=A^2+B^2\cos^2 \theta+B^2\sin^2 \theta+2AB\cos \theta\\\\ &=A^2+B^2(\cos^2 \theta+\sin^2 \theta)+2AB\cos \theta\\ &\scriptsize\qquad\qquad\qquad(\cos^2 \theta +\sin^2=1)\\\\ &=A^2+B^2 \theta+2AB\cos \\ \end{aligned}\] but OS=R, therfore, \[\small\begin{aligned} R^2 &= A^2 + B^2 + 2AB \cos \theta\\\\ \Rightarrow R&=\sqrt{A^2 + B^2 + 2AB \cos \theta} \end{aligned}\]In \(\mathrm{\triangle OSN, SN = OS \sin \alpha = R \sin \alpha}\), and
in \(\mathrm{\triangle PSN, SN = PS \sin \theta = B \sin \theta}\)
Similarly,
\(PM = A \sin \alpha = B \sin \beta\)
Combining Eqs. (1) and (2), we get
\[\dfrac{R}{\sin \theta}=\dfrac{A}{\sin \beta}=\dfrac{B}{\sin \alpha}\] Hence, \[\sin \alpha=\dfrac{B}{R}\sin \theta\] or, \[\begin{aligned}\tan \alpha &= \dfrac{\text{SN}}{\text{OP + PN}}\\\\ \tan \alpha&=\dfrac{B \sin \theta}{A + B \cos \theta} \end{aligned}\]MOTION IN A PLANE
The position vector r of a particle P located in a plane with reference to the origin of an x-y reference frame is given by
\[r=x\ \hat{i} + y\ \hat{j}\]where x and y are components of r along x-axis, and y-axis or simply they are the coordinates of the object.
Suppose a particle moves along the curve shown by the thick line and is at P at time t and P′ at time t′ . Then, the displacement is :
\[\Delta \vec{r} = \vec{r′} – \vec{r}\] Writing this equation in component form \[\begin{aligned} \Delta \vec{r} &= \vec{r′} – \vec{r}\\ \Delta \vec{r} &=(x'\hat{i} + y'\hat{j}) - (x\hat{i} + y\hat{j})\\ \Delta \vec{r} &=(x'-x)\hat{i} + (y'-y)\hat{j}\\ \Delta \vec{r} &=\hat{i}\Delta x + \hat{j}\Delta y\\ \end{aligned}\]Velocity
The average velocity \(\vec{v}\) of an object is the ratio of the displacement and the corresponding time interval :
\[\begin{aligned}\vec{v}&=\frac{\Delta \vec{r}}{\Delta t}\\\\ &=\dfrac{\Delta x\ \hat{i} + \Delta y\ \hat{j}}{\Delta t}\\\\ &=\hat{i}\dfrac{\Delta x}{\Delta t} + \hat{j} \dfrac{\Delta y}{\Delta t} \end{aligned}\] or, \[\vec{v}=v_x\ \hat{i} + v_y\ \hat{j}\]Since \(\vec{v}=\frac{\Delta \vec{r}}{\Delta t}\) the direction of the average velocity is the same as that of \(\Delta r\)>br> The velocity (instantaneous velocity) is given by the limiting value of the average velocity as the time interval approaches zero
\[\begin{aligned}\vec{v}&=\lim_{\Delta t \to 0}\dfrac{\Delta \vec{r}}{\Delta t}\\ &=\dfrac{d\vec{r}}{dt}\end{aligned}\] \[\boxed{\;\vec{v}=\dfrac{d\vec{r}}{dt}\;}\]We can express \(\vec{v}\) in a component form
\[\begin{aligned} v&=\dfrac{d\vec{r}}{dt}\\\\ &=\lim_{\Delta t \to 0}\left(\dfrac{\Delta x}{\Delta t}\ \hat{i} + \dfrac{\Delta y}{\Delta t}\ \hat{j}\right)\\\\ &=\dfrac{dx}{dt}\ \hat{i} + \dfrac{dy}{dt}\ \hat{j}\\\\ \end{aligned}\]Where,
\[\begin{aligned} v_x=\dfrac{dx}{dt}\\\\ v_y=\dfrac{dy}{dt}\end{aligned}\]The magnitude of \(\vec{v}\) is then
\[v\sqrt{v_x^2+v_y^2}\]and the direction of v is given by the angle \(\theta\)
\[\begin{aligned} \tan \theta&=\dfrac{v_y}{v_x}\\\\ \theta &= \tan^{-1}\left(\frac{v_y}{v_x}\right)\end{aligned}\]Acceleration
The average acceleration \(\overline{a}\) of an object for a time interval ∆t moving in x-y plane is the change in velocity divided by the time interval :
\[\begin{aligned} \overline{a}&=\dfrac{\Delta v}{\Delta t}\\\\ &=\dfrac{v_x\ \hat{i} + v_y\ \hat{j}}{\Delta t}\\\\ &=\dfrac{v_x\ \hat{i}}{\Delta t} + \dfrac{v_y\ \hat{j}}{\Delta t}\\\\ &=a_x\ \hat{i} + a_y\ \hat{j} \end{aligned}\] \[\overline{a}=a_x\ \hat{i} + a_y\ \hat{j}\]The acceleration (instantaneous acceleration) is the limiting value of the average acceleration as the time interval approaches zero :
\[\vec{a}=\lim_{\Delta t \to 0}\dfrac{\Delta v}{\Delta t}\]Since \(\vec{v}=\Delta v_x\ \hat{i} + \Delta v_y\ \hat{j}\) we have
\[\begin{aligned}\vec{a}&=\lim_{\Delta t \to 0}\dfrac{\Delta v_x\ \hat{i} + \Delta v_y\ \hat{j}}{\Delta t}\\\\ &=\lim_{\Delta t \to 0}\left(\dfrac{\Delta v_x}{\Delta t}\right)\ \hat{i} + \lim_{\Delta t \to 0}\left(\dfrac{\Delta v_y}{\Delta t}\right)\ \hat{j}\\\\ &=\left(\dfrac{dv_x}{dt}\right)\ \hat{i} + \left(\dfrac{dv_y}{dt}\right)\ \hat{j}\\\\ \vec{a}&=a_x\ \hat{i} + a_y\ \hat{j} \end{aligned}\]MOTION IN A PLANE WITH CONSTANT ACCELERATION
Suppose that an object is moving in x-y plane and its acceleration \(\vec{a}\) is constant. Over an interval of time, the average acceleration will equal this constant value. Now, let the velocity of the object be \(\vec{v_o}\) at time \(t = 0\) and \(\vec{v}\) at time \(t\). Then, by definition
\[\begin{aligned}\vec{a}&=\dfrac{\vec{v}-\vec{v_o}}{t-0}\\\\ &=\dfrac{\vec{v}-\vec{v_o}}{t}\end{aligned}\] or, \[\vec{v}=\vec{v_o}+\vec{a}t\]In terms of components :
\[\begin{aligned} v_x=v_{ox} +a_xt\\\\ v_y=v_{oy} +a_yt \end{aligned}\]Let \(\vec{r_o}\) and r be the position vectors of the particle at time 0 and \(t\) and let the velocities at these instants be \(\vec{v_o}\) and \(\vec{v}\). Then, over this time interval \(t\), the average velocity is \[(\vec{v_o}+ \vec{v})/2\] The displacement is the average velocity multiplied by the time interval :
\[\begin{aligned}\vec{r}-\vec{r_o}&=\left(\dfrac{\vec{v_o}+ \vec{v}}{2}\right)\ t\\\\ &=\left(\dfrac{(\vec{v_o}+\vec{a}t) + \vec{v_o}}{2}\right)\ t\\ &\scriptsize\qquad\qquad\quad(\vec{v}=\vec{v_o}+\vec{a}t)\\\\ &=\vec{v_o}t + \dfrac{1}{2}\vec{a}t^2 \end{aligned}\]writing in component form as
\[\begin{aligned} x=x_o+v_{ox}t+\dfrac{1}{2}a_xt^2\\\\ y=y_o+v_{oy}t+\dfrac{1}{2}a_yt^2 \end{aligned}\]PROJECTILE MOTION
A projectile is any object that, after being given an initial push or throw, moves only under the effect of gravity. Air resistance is neglected in ideal projectile motion. Examples include a stone thrown into the air, a football kicked at an angle, or a ball released from a moving vehicle.
Projectile motion can be understood by resolving the motion into two independent parts:
- Horizontal motion: uniform motion with constant velocity
- Vertical motion: uniformly accelerated motion under gravity
These two motions occur simultaneously but do not interfere with each other.
Suppose that the projectile is launched with velocity \(\vec{v}_o\) that makes an angle \(\theta_o\) with the x-axis
After the object has been projected, the acceleration acting on it is that due to gravity which is directed vertically downward:
\[\vec{a}=-g\ \hat{j}\] or, \[a_x=0,\; a_y=-g\]The components of initial velocity \(\vec{v}_o\) are :
\[\begin{aligned} v_{ox}=v_o \cos\ \theta_o\\ v_{oy}=v_o \sin\ \theta_o \end{aligned}\]If we take the initial position to be the origin of the reference frame, we have :
\[x_o=0,\; y_o=0\] substituting these values in Equations \[\begin{align} x&=x_o+v_{ox}t\\\\&=(v_o \cos \theta_o)\ t\tag{1}\\\\ y&=y_o+v_{oy}t-\dfrac{1}{2}gt^2\\\\ y&=(v_o \sin\ \theta_o)\ t-\dfrac{1}{2}gt^2\tag{2} \end{align}\] The components of velocity, \[\begin{aligned} v_x=v_{ox}=v_o \cos \theta_o\\\\ v_y=v_o \sin \theta_o -gt \end{aligned}\]Equation of path of a projectile
Eliminating thetime between the expressions for x and y as given in Equations (1) and (2) We obtain:
\[\begin{aligned}x&=v_o\ \cos \theta_o\ t\\ \Rightarrow t&=\dfrac{x}{v_o\ \cos \theta_o}\end{aligned}\]Substituting value of \(t\) in equation-(2)
\[\begin{aligned}y&=v_o \sin \theta_o \left(\dfrac{x}{\cos \theta_o}\right)-\dfrac{1}{2}g \left(\dfrac{x}{\cos \theta_o}\right)^2\\\\ &=x\ \tan \theta_o - \dfrac{g}{2(v_o \cos_o \theta_o)^2}\ x^2\end{aligned}\] \[\boxed{\;y=x\ \tan \theta_o - \dfrac{g}{2(v_o \cos_o \theta_o)^2}\ x^2 \;}\]Time of maximum height
Let it take time \(t_m\) to reach at the highest point, At this point \(v_y=0\), therfore,
\[\begin{aligned}v_y&=v_o \sin \theta_o -gt_m\\ \Rightarrow gt+m&=v_o \sin \theta_o\\ t_m&=\dfrac{v_o \sin \theta_o}{g}\end{aligned}\]Let Total time of flight be \(T_f\), then
\[T_f=2T_m\] Therfore, \[\boxed{\;T_f=\dfrac{2\ (v_o \sin \theta_o)}{g}\;}\]Maximum height of a projectile
The maximum height \(h_m\) reached by theprojectile can be calculated by substituting \(t = t_m\) in Eq. (2)
\[\begin{aligned} y&=(v_o \sin\ \theta_o)\ t-\dfrac{1}{2}gt^2\\ h_m&=(v_o \sin\ \theta_o)\times\dfrac{v_o \sin \theta_o}{g}-\dfrac{1}{2}g\times\left(\dfrac{v_o \sin \theta_o}{g}\right)^2\\\\ &=\dfrac{(v_o \sin\ \theta_o)^2}{2g} \end{aligned}\] \[\boxed{\;h_m=\dfrac{(v_o \sin\ \theta_o)^2}{2g}\;}\]Horizontal range of a projectile
The horizontal distance travelled by a projectile from its initial position \((x = y = 0)\) to the position where it passes \(y = 0\) during its fall is called the horizontal range, R. It is the distance travelled during the time of flight \(T_f\). Therefore, the range R is
\[\begin{aligned} R&=(v_o\ \cos \theta)\times T_m\\\\ &=(v_o\ \cos \theta)\times\left(\dfrac{2\ (v_o \sin \theta_o)}{g}\right)\\\\ &=\dfrac{2v^2\sin 2\theta_o}{g} \end{aligned}\] \[\boxed{\;R=\dfrac{2v^2\sin 2\theta_o}{g}\;}\]\(R\) is maximum when \(\sin\ 2\theta_o\) is maximum, i.e., when \(\theta_o=45^\circ\)
The maximum horizontal range is, therefore,
UNIFORM CIRCULAR MOTION
When an object moves along a circular path with constant speed, its motion is known as uniform circular motion. Although the speed remains unchanged, the motion is far from simple because the direction of motion keeps changing at every point on the circle.
Velocity in uniform circular motion
At any point on the circular path:
- The velocity of the particle acts along the tangent to the circle.
- The direction of velocity changes continuously as the particle moves.
- The magnitude of velocity remains constant.
This continuous change in direction is the key reason behind acceleration in circular motion.
Centripetal acceleration
To keep an object moving in a circular path, an acceleration is required that always acts towards the centre of the circle. This acceleration is called centripetal acceleration.
Its magnitude is given by:
\[a_c=\dfrac{v^2}{r}\]where:
- \(v\) is the uniform speed
- \(r\) is the radius of the circular path
The direction of centripetal acceleration is radially inward, perpendicular to the instantaneous velocity.
Centripetal force
According to Newton’s laws, acceleration must be produced by a force. The force responsible for centripetal acceleration is called the centripetal force.
\[F_c=m\dfrac{v^2}{r}\]This force does not create motion, but only changes the direction of motion. Examples include:
- Tension in a string for a stone tied and rotated
- Gravitational force in planetary motion
- Friction for a vehicle turning on a curved road
Angular variables in circular motion
Uniform circular motion is often described using angular quantities:
- Angular displacement \((\theta)\): angle traced at the centre
- Angular velocity \((\omega)\): \[\omega=\dfrac{v}{r}\]
- Time period (T): \[T=\dfrac{2\pi r}{v}\]
- Rate of change of angular displacement \[\omega=\dfrac{\Delta \theta}{\Delta t}\]
- Centripetal acceleration \(a_c\) in terms of angular speed: \[a_c=\omega^2 R\]
