Motion of an object in a plane is called two dimensional (2-D) motions. For 2-D motion velocity or acceleration can be described by two components in any two mutually perpendicular directions in cartesian coordinate system i.e. its position, velocity, displacement and acceleration can have two nonzero components.

1. VECTOR & SCALAR:

Those physical quantity in which we know magnitude and direction called vector.

Those physical quantity in which we know magnitude but unknown direction called Scalar.

$\vec{A}+\vec{B}=\vec{R}$

Those vector whose magnitude is zero called zero vector.

Vector Scalar product of vectors

(i) $\vec{A}\cdot \vec{B}=|A||B|\cos \theta$

(ii) $\vec{A}\times \vec{B}=|A||B|\sin \theta$

(iii) $\vec{A}=a_{x}i+a_{y}j+a_{z}k$

(iv) $|A|=\sqrt{{a_x}^2+{a_y}^2+{a_z}^2}$

Illustration 1.

A particle of mass 10^–2kg is moving along the positive X-axis under the influence of a force $F(x)=-\frac{k}{2x^2}$ where k = 10^–2 Nm². At time t = 0, it is at

x = 1.0 m and its velocity is v = 0. find its velocity when it reaches

x = 0.5 m

Solution:

Given

$F(x)=-\frac{k}{2x^2}$

Here k and x² are always positive. Hence F is always negative (whether x is positive or negative)

Now, we know that F = ma

$\therefore a=\frac{F}{m}$

$\therefore$ In this case we have

$a=\frac{-k}{2m{x}^2}=\frac{-{10}^{-2}}{2\times {10}^{-2}\times x^2}=-\frac{1}{2x^2}$ [as F[x] = –k / 2x²]

or $v.\frac{dv}{dx}=-\frac{1}{2x^2}\text{or}\text{vdv}\text{=}\text{-}\frac{\text{1}}{\text{2}}\frac{dx}{x^2}\text{or}\underset{0}{\overset{v}{\int }}vdv=-\frac{1}{2}{\int }_{1.0}^{x=0.5}\frac{dx}{x^2}$

or $\frac{v^2}{2}=\frac{1}{2}{\left(\frac{1}{x}\right)}_{x=1.0}^{x=0.5}$

or $v^2=\left[\frac{1}{0.5}-\frac{1}{1.0}\right]or{v}^2=1.0$

or v = ± 1.0 m/s

So, here v = – 1.0 m/s (because velocity is along negative X-direction)

Illustration 2.

A particle travels according to the equation $a=A-Bv$ where $a$ is the acceleration. A and B are constants, $v$ is the velocity of the particle. Find its velocity as a function of time. Also find terminal velocity.

Key concept: Keep remember that above question is not related with uniform acceleration.

Solution: - $$\begin{gathered} \frac{dv}{dt}=A-Bv \\ or\underset{0}{\overset{V}{\int }}\frac{dv}{A-Bv}=\underset{0}{\overset{t}{\int }}dt\Rightarrow -\frac{1}{B}{\log }_e\left(A-Bv\right)|{}_0^V=t \end{gathered}$$

$$\begin{gathered} or{\log }_e\frac{A-Bv}{A}=-Bt \\ orA-Bv=A{e}^{-Bt}orv=\frac{A}{B}\left(1-e^{-Bt}\right) \end{gathered}$$

At terminal velocity, acceleration = 0, that is, $v_t=A/B$

2. PROJECTILE MOTION:

In two-dimension motion a particle moves in a plane. e.g. a particle going in a circle, a cricket ball thrown in by a fielder (on a windless day). In the first case the particle can go round with a constant speed (uniform circular motion) or it could move with a non-constant speed (Non uniform circular motion). We will study both. In the second case the cricket ball in a projectile i.e. it has been projected (thrown) and it moves under the influence of gravity. This motion is projectile motion. Before we study this specialized case of two-dimension motion, let us first understand the general features of two-dimension motion.

(i) Position and Displacements:

As we had done in one-dimensional motion, the first task for us is to know where the particle is i.e. its position. The position here is a vector intends from a reference point (usually the origin of the coordinate system) to the object.

In the unit vector notation it can be written

$\overrightarrow{r}=x\stackrel{\wedge }{i}+y\stackrel{\wedge }{j}$

Where $x\stackrel{\wedge }{i},y\stackrel{\wedge }{j}$ are the vector components of $\overrightarrow{r}$and the coefficients x and y are its scalar components.

The coefficients x and y give the objects location along the axes and relative to the origin. Figure shows particle P whose position vector at an instant is

$\overrightarrow{r}=3\stackrel{\wedge }{i}+6\stackrel{\wedge }{j}$

As the particle moves its position vector (P.V.) changes such that the vector always extends to the object from origin.

Assuming the particle has a P.V. ${\overrightarrow{r}}_1$at time t₁ and P.V. ${\overrightarrow{r}}_2$at time t₂, its displacement is $\Delta \overrightarrow{r}={\overrightarrow{r}}_2-{\overrightarrow{r}}_1$

During time interval $\Delta \overrightarrow{r}$.

Illustration 3.

The P.V. for a particle is initially$\overrightarrow{r_1}=-2\stackrel{\wedge }{i}-2\stackrel{\wedge }{j}$ and then later is $\overrightarrow{r_1}=8\stackrel{\wedge }{i}+10\stackrel{\wedge }{j}$. What is the displacement?

Solution:

$\Delta \overrightarrow{r}=\left(8\stackrel{\wedge }{i}+10\stackrel{\wedge }{j}\right)-\left(-2\stackrel{\wedge }{i}-2\stackrel{\wedge }{j}\right)=10\stackrel{\wedge }{i}+12\stackrel{\wedge }{j}$

(ii) Velocity and Average velocity:

If a particle moves through a displacement $\Delta \overrightarrow{r}$ in a time interval$\Delta t$, then the average velocity is

$⟨\overrightarrow{V}⟩=\frac{\Delta \overrightarrow{r}}{\Delta t}$ ….(1)

$=\frac{\Delta x\stackrel{\wedge }{i}+\Delta y\stackrel{\wedge }{j}}{\Delta t}$

$=\frac{\Delta x}{\Delta t}\stackrel{\wedge }{i}+\frac{\Delta y}{\Delta t}\stackrel{\wedge }{j}$

Þ $⟨\overrightarrow{V}⟩=\overrightarrow{V_x}\stackrel{\wedge }{i}+\overrightarrow{V_y}\stackrel{\wedge }{i}$

Here $\overrightarrow{V_x}\stackrel{\wedge }{i}\text{and}\overrightarrow{V_y}\stackrel{\wedge }{i}$ are x and y components of average velocity.

The instantaneous velocity

$\overrightarrow{V}=\underset{\Delta \text{t}\to \text{0}}{\lim }\frac{\Delta \overrightarrow{\text{r}}}{\Delta \text{t}}=\frac{d\overrightarrow{\text{r}}}{d\text{t}}$

But $\begin{aligned} & \,\,\,\,\,\,\,\,\overrightarrow{r}=\,x\overset{\wedge }{\mathop{i}}\,+y\overset{\wedge }{\mathop{j}}\, \\ & \therefore \,\,\,\,\,\overrightarrow{v}=\,\dfrac{dx}{dt}\overset{\wedge }{\mathop{i}}\,+\dfrac{dy}{dt}\overset{\wedge }{\mathop{j}}\, \\ & or\,\,\,\,\overrightarrow{v}=\,{{V}_{x}}\overset{\wedge }{\mathop{i}}\,+{{V}_{y}}\overset{\wedge }{\mathop{j}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....(6) \\ \end{aligned}$

$V_x=\frac{dx}{d\text{t}},V_y=\frac{dy}{d\text{t}}$, then components of $\overrightarrow{V}$ along x and y respectively.

Let us understand this using a particle, which is moving in 2D.

In following figure particle is at ${\overrightarrow{r}}_1$ at t₁ and at ${\overrightarrow{r}}_2$ at t₂ (= t₁+ t₂). The vector $\Delta \overrightarrow{r}$ is the particle displacement in $\Delta t$time. Then as is clear from (1) average velocity $⟨\overrightarrow{V}⟩$ points in same direction as $\Delta \overrightarrow{r}$

Now if we reduce $\Delta t$ we see that

(1) vector ${\overrightarrow{r}}_2$moves towards ${\overrightarrow{r}}_1$

(2) $\Delta \overrightarrow{r}$ becomes smaller and its direction approaches the tangent

(3) In the limit $\Delta t$$\to$ 0, $\Delta \overrightarrow{r}$, the instantaneous velocity at t₁ i.e. average velocity becomes instantaneous velocity $\overrightarrow{V}$, which has the direction along tangent line.

Description Of Projectile Motion:

If an object is given an initial velocity in any direction (except $\theta \ne$ 90°) and then allowed to travel freely under gravity, the motion is called projectile motion. In case of projectile motion it is assumed that acceleration due to gravity is constant and the effect of air resistance is negligible. Projectile motion is a two dimensional motion and can be regarded as simultaneous superposition of two motions one horizontal with velocity ucos$\theta$ and acceleration=0 and other vertical with initial velocity u sin$\theta$ and acceleration = -g. The acceleration due to gravity, g, is uniform and the surface of the earth is considered to be flat.

Following are for main features of a projectile:

(i) Such a particle will move horizontally and as well as vertically i.e. along a curve.

(ii) For convenience, we will take origin at the point from where the particle is thrown and horizontal and vertical components of velocity along x-axis and y-axis respectively.

(iii) The velocity of particle at any instant is directed along the tangent to the path and can have horizontal and vertical components.

(iv) The only force acting on the particle is its weight (mg) directed downwards. Hence acceleration is g directed vertically downwards. The air friction is neglected.

(v) As acceleration does not change with time, the projectile motion is a uniformly accelerated motion at all time instants, a₁ (horizontal)=0 and a₂(vertical) = -g.

Time Of Flight, Horizontal Range And Trajectory Of The Projectile:

Let P(x, y) represent the position of a projectile after a time t from the time of projection.

At the point O, horizontal component of velocity = ucos$\theta$ and

vertical component of velocity = u sin$\theta$

Acceleration due to gravity acts vertically downwards. Hence vertical component of velocity changes. While horizontal component of velocity remains constant (ucosq) during the complete motion.

Time of flight: It is the time taken by the projectile from the instant it is realised till it strikes a position the same horizontal plane as the point of projection.

To calculate time of flight, we can find the time when vertical displacement is zero i.e.

0 = usin$\theta$t – $\frac{1}{2}g{t}^2$ $\Rightarrow$ (usin$\theta$ - $\frac{1}{2}gt$)t = 0

$\Rightarrow$ t = 0 or t = $\frac{2usin\theta }{g}$

Thus

Time of flight T =$\frac{2usin\theta }{g}$

Range: During this time horizontal component of velocity has taken the particle through a distance x horizontally

Where, $x$ = ucos$\theta$ x T = u cos$\theta$ x $\frac{2usin\theta }{g}$ = $\frac{u^{2}sin2\theta }{g}$

This distance, we call as range on horizontal plane

R = $\frac{u^{2}sin2\theta }{g}$

R_max = $\frac{u^2}{g}$ for $\theta$ = 45°

For a given velocity same range can be obtained for an angle $\theta$ and angle (90° – $\theta$) i.e.

R = $\frac{u^{2}sin2\left(90{}^{\circ }-\theta \right)}{g}$ = $\frac{u^{2}sin\left(180{}^{\circ }-2\theta \right)}{g}$

R = $\frac{u^{2}sin2\theta }{g}$

To calculate maximum height, the vertical component of velocity becomes zero, when the particle is at the highest point from the ground. At that time particle has only horizontal component of velocity i.e. = ucos$\theta$

0 = (usin$\theta$)² – 2gH_max

H_max = $\frac{u^{2}si{n}^2\theta }{2g}$

The motion of projectile can be analysed through vector notation also for e.g. velocity of projectile (v) at any time t can be written as

$\vec{v}=ucos\theta \hat{i}+\left(usin\theta -gt\right)\hat{j}$

Hence, the magnitude of velocity (speed) v at any time is given as

v = $\sqrt{{\left(ucos\theta \right)}^2+{\left(usin\theta -gt\right)}^2}$

= $\sqrt{u^2+g^2{t}^2-2ugt\sin \theta }$ (The path of a projectile is called its trajectory)

Trajectory of projectile:

y = usin$\theta$ t – $\frac{1}{2}g{t}^2$ . . . (1) (vertical displacement at any time t)

x = ucos$\theta$t . . . (2) (Horizontal displacement at any time t)

Using t = $\frac{x}{u\cos \theta }$

We get,

y = x tan$\theta$ –$\frac{g{x}^2}{2u^2{\cos }^2\theta }$

Which suggest that trajectory of projectile is parabola

Illustration 4.

The figure shows two position A and B at the same height h above the ground. If the maximum height of the projectile is H, Then determine the time t elapsed between the positions A and B in terms of H.

Solution:

Let T be the time of flight. We can now write

$T^2-t^2=\frac{8h}{g}$

since $\begin{array}{r}T=\frac{2v_0\sin \theta }{g}\\ T^2=\frac{4{v^2}_0{\sin }^2\theta }{g^2}=\frac{8H}{g}\end{array}$

or $t=\sqrt{\frac{8}{g}\left(H-h\right)}$

Time Of Flight & Horizontal Range Of Projectile Motion On Inclined Plane:

Projectile Motion on Inclined Plane: Figure shows an inclined plane at an angle $\alpha$ and a particle at an angle $\theta$ with the direction of plane with initial velocity u. In such cases we take our reference x- and y-axes in the direction along and perpendicular to the inclined as shown.

Unlike to the simple projectile motion, here the x-component of the velocity of the projectile will also be retarded by a gsin$\alpha$. Now y-component of the velocity is retarded by g cos$\alpha$ instead of g. As shown here g is resolved in two directions.

As here y-direction component is retarded by gcos$\alpha$, to find the time of flight and maximum height, we can use equations T= $\frac{2usin\theta }{g}$ and R = $\frac{u^{2}sin2\theta }{g}$, replacing g by gcos$\alpha$,

Time of flight on inclined plane projectile is

$T_{}=\frac{2u\sin \theta }{g\cos \alpha }$

Maximum height of the projectile with respect to inclined plane is

$H=\frac{u^2{\sin }^2\theta }{2g\cos \alpha }$

For evaluation of range on inclined plane we cannot use equation R=$\frac{u^{2}sin2\theta }{g}$, just by replacing g by gcos$\alpha$, as here we also have

Acceleration in x-axis a_x = -g sin$\alpha$

Now we again find the distance traveled by the particle along x-direction in the duration time of flight is

R =$u\sin \theta .T_{}-\frac{1}{2}g\sin \alpha .T_{}^2$

On substituting the value of time of flight T, we get

$R=\frac{u^2\sin 2\theta }{g\cos \alpha }-\frac{2u^2\sin \alpha {\sin }^2\theta }{g{\cos }^2\alpha }$

Students are advised not to apply the above expression of range on inclined plane, as a standard result, it should be processed and evaluated according to the numerical problem. Above results we've derived for the projectile thrown up an inclined plane. If projectile is thrown down an inclined plane, the acceleration along the plane gsin$\alpha$ will increase the velocity of the particle along the plane, thus in the expression for range we should use +ve sign as

$R=\frac{u^2\sin 2\theta }{g\cos \alpha }+\frac{2u^2\sin \alpha {\sin }^2\theta }{g{\cos }^2\alpha }$

To find the maximum range on incline plane, One can use maxima-minima as $\frac{dR}{d\theta }$. The range on inclined plane has a maximum value given as

$R=\frac{u^2}{g\left(1\pm \sin \alpha \right)}$

In above equation +ve sign is used for projectile up the plane and –ve sign is used for projectile down the plane.

Illustration 5.

A ball is dropped from a height h above a point on an inclined plane, with angle of inclination $\theta$. The ball makes an elastic collision with the surface and rebounds off the plane. Determine the distance from the point of first impact to the point where ball hit the plane second time.

Key concept:

Take the point of first impact as the origin. Direction along the plane will be the x-axis and the direction perpendicular to the plane will be the y-axis. It is given that the ball rebounds elastically and implies that no change in kinetic energy of the ball before and after the collision. The ball rebounds with the same speed with which it will strike the plane after falling a distance h, which is $u=\sqrt{2gh}$. After rebound, the horizontal component of velocity u sin$\theta$ will be accelerated by g sin$\theta$ and the vertical component of the velocity ucos$\theta$ will be retarded by gcos$\theta$.

Solution:

Here time of flight from first impact to the second impact is given as $T=\dfrac{2{{u}_{y}}}{{{a}_{y}}}=\dfrac{2u\cos \theta }{g\cos \theta }=\dfrac{2u}{g}$

In this duration the distance travelled by the horizontal component is

R = $R=u\sin \theta .\frac{2u}{g}+\frac{1}{2}g\sin \theta {(\frac{2u}{g})}^2=\frac{4u^2\sin \theta }{g}$

$8h\sin \theta$ ( since u=$\sqrt{2gh}$)

3. UNIFORM AND NON - UNIFORM CIRCULAR MOTION

Uniform circular motion: In uniform circular motion, a particle moves in a circular path of constant radius with constant speed. The velocity of a particle varies continuously as the direction changes. Thus, uniform circular motion is uniformly accelerated motion. Since the acceleration produces a change only in the direction of velocity vector, therefore, it must always be at right angles to the direction of motion. Otherwise, a component in the direction of motion would produce change in speed of the particle.

Centripetal Acceleration: The acceleration of a particle in uniform circular motion is directed towards the centre of the circular path. This radially inward acceleration is called the centripetal acceleration (means centre seeking). Its magnitude is given by $a_c=\frac{v^2}{r}$

where v is the speed of the particle, and r is the radius of the circular path. In terms of angular speed $\omega =\frac{v}{r}$, we may write a_c = $\omega$² r

If a body makes N revolution per minute, then its angular speed is given by

$\omega =\frac{2\pi N}{60}=\frac{\pi N}{30}$

In uniform circular motion

(i) Velocity remains constant in magnitude but varies in direction

(ii) The acceleration is always normal to the velocity vector.

(iii) The acceleration is always directed towards the centre of the circular path.

Comparison of uniform circular motion with straight line motion and projectile motion:

Non-uniform circular motion:

(i) The velocity changes both in magnitude as well as in direction

(ii) The velocity vector is always tangential to the path

(iii) The acceleration vector is not perpendicular to the velocity vector

(iv) The acceleration vector has two components

(a) Tangential acceleration a_t changes the magnitude of velocity vector, i.e. $a_t=\frac{dv}{dt}$

(b) Normal acceleration or centripetal acceleration a_c changes the direction of the velocity vector, i.e. $a_c=\frac{v^2}{r}$

(v) The total acceleration is the vector sum of the tangential and centripetal acceleration $a=\sqrt{{a_t}^2+{a_c}^2}$

Illustration 6.

A particle is moving in a circular path of radius 10 cm. Its linear speed is given by $v=4t$ cm/s. Find the angle between acceleration and radius at t = 2s.

Solution:

Radial explanation $a_R=\frac{v^2}{R}=6.4cm/s^2$

Tangential acceleration $a_t=\frac{dv}{dt}=4cm/s^2$

$\therefore \tan \theta =\frac{a_t}{a_R}=\frac{4}{6.4}$

$\theta ={\tan }^{-1}\frac{5}{8}$

Equations of motion:

1. $v=u+at$

2. $s=ut+\frac{1}{2}a{t}^2$

3. $v^2=u^2+2as$

4. The distance travelled by the body in n^th second

$s_n=u+\frac{a}{2}(2n-1)$

Projectile motion:

5. Time of flight T =$\frac{2usin\theta }{g}$

6. Range R = $\frac{u^{2}sin2\theta }{g}$

7. R_max = $\frac{u^2}{g}$ for $\theta$ = 45°

8. Hight H_max = $\frac{u^{2}si{n}^2\theta }{2g}$

9. Trajectory of projectile

y = x tan$\theta$ –$\frac{g{x}^2}{2u^2{\cos }^2\theta }$ (i.e. parabolic)

For Inclined plane:

10. Time of flight $T_{}=\frac{2u\sin \theta }{g\cos \alpha }$

11. Hight $H=\frac{u^2{\sin }^2\theta }{2g\cos \alpha }$

12. Range (for down inclined) $R=\frac{u^2\sin 2\theta }{g\cos \alpha }-\frac{2u^2\sin \alpha {\sin }^2\theta }{g{\cos }^2\alpha }$

13. Range (for up inclined) $R=\frac{u^2\sin 2\theta }{g\cos \alpha }+\frac{2u^2\sin \alpha {\sin }^2\theta }{g{\cos }^2\alpha }$

14. Maximum range

$R_{\max }=\frac{u^2}{g\left(1\pm \sin \alpha \right)}$

+ve sign is used for projectile up the plane and –ve sign is used for projectile down the plane.

Circular motion:

15. Centripetal acceleration (normal acceleration) $a_c=\frac{v^2}{r}$

16. Tangential acceleration = $a_t=\frac{dv}{dt}$

Relative motion:

17. Relative velocity of a particle while moving in the same direction.

Relative velocity $=V_A-V_B$

18. Elative velocity of a particle while moving in the opposite direction.

Relative velocity $=V_A+V_B$