CBSE Class 11 Physics Kinematics Exam Notes

CBSE Class 12 Physics: Kinematics Exam Notes

KINEMATICS

INTRODUCTION

Mechanics deals with the motion of various types of bodies, the relations of forces, the mechanical properties of bodies and matter, etc. Kinematics deals with the motion of particles and rigid bodies without accounting for the forces responsible for their motion. If the body is, however, of a small size such that its motion can be described by a point mass moving along a straight line, such a motion is called rectilinear motion or motion in one dimension.

MOTION

When a body changes position with respect to an observer, the body is said to be in motion. If the body does not change position, then it is said to be at rest with respect to the observer. For example, if two passengers are travelling in a moving train, they are at rest with respect to each other while they are in motion with respect to an observer standing on the ground.

A PARTICLE

A particle is a physical analogue of a point. A body of finite size may be considered as a particle only if all parts of the body undergo the same displacement and have the same velocity and acceleration. Thus, for a particle-like body, its motion can be described by studying the motion of any point on the body.

DISTANCE AND DISPLACEMENT

As a particle moves in space with respect to time, it follows a curve or a straight line, which is called its path or trajectory. Distance is the actual length covered along the path, while displacement is the difference between the final and initial positions of the particle. The position of a particle is given by a position vector drawn from the origin to the particle.

The figure illustrates the difference between distance and displacement. The curve length AB is called the distance; the vector \(\vec r_{AB}=\vec r_B-\vec r_A\) is called the displacement.

1. Distance is a scalar; displacement is a vector.
2. The magnitude of displacement is always less than or equal to the distance.
3. For a moving body, displacement can be zero but distance cannot be zero.

For a body moving in a straight line from point A to point B along the \(x\)-axis such that its initial and final positions are at distances \(x_i\) and \(x_f\) respectively, the minimum distance between the final and initial positions is called the displacement, \(\Delta x\):

\[AB=\Delta x=x_f-x_i\]

Displacement is a vector from the initial position to the final position. If the path of the body is not along a straight line, the total length of the actual path covered by the body is called the distance covered by the body and is a scalar quantity. The displacement is independent of the actual path covered by the body.

AVERAGE SPEED AND AVERAGE VELOCITY

The average speed in a time interval is defined as the total distance travelled by the particle divided by the time interval.

\[\text{Average speed}=\frac{\text{Total distance travelled}}{\text{Total time taken}}\]

The average velocity is defined as

\[\vec v_{av}=\frac{\text{displacement}}{\text{time elapsed}}\quad\text{or}\quad\vec v_{av}=\frac{\vec r_B-\vec r_A}{t_B-t_A}=\frac{\Delta\vec r_{AB}}{\Delta t}\]

1. Average speed is a scalar quantity; average velocity is a vector quantity. Both are expressed in m s⁻¹ or km h⁻¹.
2. Average velocity can be zero but average speed cannot be zero.
3. The magnitude of average velocity is always less than or equal to the average speed. \[ |\vec v_{av}|\le v\quad\text{or}\quad\frac{|\text{displacement}|}{\text{time elapsed}}\le\frac{\text{distance}}{\text{time elapsed}},\ \text{thus}\ |\text{displacement}|\le\text{distance} \]
4. Average speed does not mean the magnitude of the average velocity vector.

The displacement of a body per unit time in a specific direction is called the velocity. It is a vector quantity measured in m/s or similar units. The distance covered by a body per unit time, without any reference to its direction of motion, is the speed.

Consider a body moving on a straight line having the initial position A at time \(t_i\) at a distance \(x_i\) from the origin. It travels a distance \(x_1\) to reach B and thereafter travels a distance \(x_2\) from B to C, reaching at time \(t_f\). C is the final position, at a distance \(x_f\) from the origin. The average speed will be equal to the ratio of total distance covered to the total time.

\[\therefore\ \text{Average speed}=\frac{x_1+x_2}{t_f-t_i},\qquad\text{Average velocity}=\frac{x_f-x_i}{t_f-t_i}\]

INSTANTANEOUS SPEED AND INSTANTANEOUS VELOCITY

The instantaneous speed and velocity are the respective average speed and velocity for an infinitesimally small time interval (i.e., \(\Delta t\to0\)). Thus

\[\text{Instantaneous speed }v=\lim_{\Delta t\to0}\frac{|\Delta\vec r|}{\Delta t}=\left|\frac{d\vec r}{dt}\right|\]

\[\text{Instantaneous velocity }\vec v=\lim_{\Delta t\to0}\frac{\Delta\vec r}{\Delta t}=\frac{d\vec r}{dt}\]

1. As the time interval tends to zero (\(\Delta t\to0\)), the displacement vector \(\Delta\vec r\) approaches a limiting direction which is tangential to the path of the particle at that instant. Thus, the instantaneous velocity direction is always tangential to the path followed by the particle.
2. The instantaneous speed is equal to the magnitude of the instantaneous velocity.
3. A particle may have constant speed but variable velocity.
4. For a particle moving with constant velocity, its average velocity and instantaneous velocity are always equal.

The instantaneous velocity of a particle moving on a straight line, say along the \(x\)-axis, is given as

\[v(t)=\lim_{\Delta t\to0}\frac{\Delta x}{\Delta t}=\lim_{\Delta t\to0}v_{av}\]

Instantaneous velocity \(v(t)\) is the derivative of displacement with respect to time:

\[\vec v(t)=\frac{d}{dt}[\vec r(t)]\]

AVERAGE ACCELERATION AND INSTANTANEOUS ACCELERATION

The acceleration is defined as the time rate of change of velocity. The average acceleration is defined as

\[\vec a_{av}=\frac{\vec v_2-\vec v_1}{t_2-t_1}=\frac{\Delta\vec v}{\Delta t}\]

The instantaneous acceleration is defined as

\[\vec a=\lim_{\Delta t\to0}\frac{\Delta\vec v}{\Delta t}=\frac{d\vec v}{dt}\]

1. Acceleration is a vector quantity with S.I. unit m s⁻².
2. The direction of the average acceleration vector is the direction of the change in the velocity vector.
3. There is no definite relation between the direction of the velocity vector and the direction of the acceleration vector.
4. If the magnitude of velocity is constant and only its direction changes with time, then the acceleration is perpendicular to the velocity vector.

If a body is moving on a straight line with uniform acceleration, then for such a body the average acceleration and the instantaneous acceleration will be the same.

MOTION OF BODIES IN THREE DIMENSIONAL SPACE

If a body has coordinates \((x,y,z)\) in space, its position vector \(\vec r\) at any time is given by

\[\vec r=x\hat i+y\hat j+z\hat k\]

The velocity vector \(\vec v\) is given by

\[\vec v=v_x\hat i+v_y\hat j+v_z\hat k\]

where \(v_x,v_y,v_z\) are the magnitudes of the components of the velocity along the respective \(x\), \(y\) and \(z\) axes at time \(t\):

\[v_x=\frac{dx}{dt},\quad v_y=\frac{dy}{dt},\quad v_z=\frac{dz}{dt}\]

If \(a_x,a_y,a_z\) are the components of acceleration along the three axes at time \(t\),

\[\vec a=a_x\hat i+a_y\hat j+a_z\hat k\]

GRAPHS

(a) Displacement–time graph: The slope of the \(x\)–\(t\) curve gives the instantaneous velocity, while the area under the curve represents no physical quantity.

(b) Velocity–time graph: The area under the \(v\)–\(t\) curve represents displacement. The area above the time axis is taken as positive while the area below the \(t\)-axis is taken as negative.

The total distance travelled is given by the sum of the magnitudes of the areas calculated in different time intervals.

(c) Acceleration–time graph: The slope at any point gives no physical quantity (in this chapter). The area under the curve is equal to the change in velocity in the time interval under consideration.

EQUATIONS OF MOTION

When a body does not move with constant acceleration with respect to the reference frame under consideration, then we apply the following equations.

\[\frac{dv}{dt}=a\qquad\ldots(i)\]

\[\frac{dx}{dt}=v\qquad\ldots(ii)\]

Put the values of \(a\) and \(v\) in equations (i) and (ii) and solve to find the desired results.

RELATIVE VELOCITY

If a body A is travelling with velocity \(\vec v_{AO}\) with respect to origin O and another body B is travelling with velocity \(\vec v_{BO}\) with respect to the origin, then the relative velocity of B with respect to an observer A, \(\vec v_{BA}\), is given by

\[\vec v_{BA}=\vec v_{BO}-\vec v_{AO}\]

Thus the relative velocity of two moving bodies is equal to the vector difference of their velocities with respect to the same origin. Examples are two trains moving in different directions, a boat crossing a river, etc.

The rate of change of \(\vec v_{BA}\) with time gives the relative acceleration of B with respect to A and is given by \(\vec a_{BA}=\vec a_{BO}-\vec a_{AO}\).

RIVER SWIMMER PROBLEMS

Suppose a swimmer jumps into a river of width \(d\) flowing with velocity \(u\), swimming with velocity \(v\) (with respect to water) at an angle \(\theta\) with the shortest distance AB.

The component \(v\cos\theta\) is the velocity with which the swimmer crosses the river, so the time taken to cross the river is

\[t=\frac{d}{v\cos\theta}\qquad\ldots(i)\]

The resultant of \(u\) and \(v\sin\theta\) causes the drift of the swimmer along the flow of water, so the drift or displacement along the \(x\)-axis is

\[x=(u-v\sin\theta)\times\frac{d}{v\cos\theta}\qquad\ldots(ii)\]

Cases:

(a) To cross the river in the shortest time, put \(\cos\theta=1\) in equation (i), i.e., \(\theta=0\):

\[t_{min}=\frac dv\]

i.e., the swimmer has to swim along the shortest path (perpendicular to the bank). The drift in this case is

\[x=(u-v\sin0)\frac{d}{v\cos0}=\frac{ud}{v}\]

(b) To cross the river along the shortest path, i.e., for drift = 0, using equation (ii):

\[0=(u-v\sin\theta)\frac{d}{v\cos\theta}\ \Rightarrow\ \sin\theta=\frac uv\qquad\ldots(iii)\]

If \(u>v\), then \(\sin\theta>1\), which is not possible. Hence the swimmer cannot cross the river along the shortest path if the velocity of the river is greater than the velocity of the swimmer.

EQUATIONS OF MOTION (UNIFORM ACCELERATION)

If a body has an initial velocity \(u\) at time \(t=0\) so that it travels a displacement \(s\) at time \(t\) when it moves with a uniform acceleration \(a\), its equations of motion are given as follows:

\[v=u+at,\qquad s=ut+\tfrac12at^2,\qquad v^2=u^2+2as\]

Note: All the three equations are applicable in all types of reference frames (inertial and non-inertial) provided the acceleration of the body is constant in the reference frame under consideration.

DISPLACEMENT IN THE nth SECOND

If a body travels a displacement \(S_n\) in \(n\) seconds and a displacement \(S_{n-1}\) in \((n-1)\) seconds from the beginning,

\[S_n=un+\tfrac12an^2,\qquad S_{n-1}=u(n-1)+\tfrac12a(n-1)^2\]

Displacement travelled in the \(n\)th second \(=S_n-S_{n-1}\):

\[=un+\tfrac12an^2-u(n-1)-\tfrac12a(n-1)^2\ \Rightarrow\ x_n=u+\frac a2(2n-1)\]

If a body is falling with an acceleration due to gravity \(g\), the corresponding equations of motion are

\[v=u+gt,\qquad s=ut+\tfrac12gt^2,\qquad v^2=u^2+2gs\]

If the upward direction is considered positive, the acceleration \(a\) is taken as \(-g\).

If a body is thrown vertically upwards with an initial velocity \(u\), it reaches a maximum height \(h=\dfrac{u^2}{2g}\) in time \(\dfrac ug\), where it stops, and descends in the same time \(\dfrac ug\), reaching the ground with the same speed \(u\) with which it was projected. If a body is dropped from a point at a height \(h'\) above the ground, it reaches the ground with velocity \(\sqrt{2gh'}\) in a time \(\sqrt{\dfrac{2h'}{g}}\).

PROJECTILES AND CIRCULAR MOTION

MOTION IN A PLANE

The kinematics of one-dimensional motion can be applied to motion in two as well as three dimensions. If a particle is moving in a plane, its motion can be split into two rectilinear motions along two perpendicular directions, which are called the rectangular components of the displacement, velocity and acceleration for the two-dimensional motion. If a particle is moving along the path AB in the XY plane, then at any time \(t\) it reaches point \(P(x,y)\), where its radius vector \(\vec r\) makes an angle \(\phi\) with the \(x\)-axis. Then its components are

\[x=r\cos\phi,\quad y=r\sin\phi,\quad r=\sqrt{x^2+y^2},\quad\tan\phi=\frac yx\]

If the velocity vector \(\vec v\) at \(P\) makes an angle \(\theta\) with the \(x\)-direction, its rectangular components are

\[v_x=v\cos\theta,\quad v_y=v\sin\theta,\quad v=\sqrt{v_x^2+v_y^2},\quad\tan\theta=\frac{v_y}{v_x}\]

Similarly, the acceleration \(\vec a\) at any time can be split up into components \(a_x\) and \(a_y\) along the respective \(x\) and \(y\) directions.

Thus the two-dimensional motion along the \(x\) and \(y\) directions can be independently calculated for a particle moving in a plane using the following relations:

\[v_x=u_x+a_xt,\qquad v_y=u_y+a_yt\]

\[x=u_xt+\tfrac12a_xt^2,\qquad y=u_yt+\tfrac12a_yt^2\]

\[v_x^2=u_x^2+2a_xx,\qquad v_y^2=u_y^2+2a_yy\]

In vectorial notation, the two-dimensional motion can be described by the following relations:

\[\vec r=x\hat i+y\hat j,\qquad\vec v=v_x\hat i+v_y\hat j,\qquad\vec a=a_x\hat i+a_y\hat j\]

The two-dimensional motions of a body in a plane in two important cases, projectile motion and circular motion, are described in the following sections.

PROJECTILE MOTION

When a body is projected at an angle to the direction of a uniform force field (such as gravity), the path followed by the body is known as the projectile path.

(i) Horizontal projection: If a body is thrown horizontally from a height \(h\) with an initial speed \(u\), its horizontal velocity remains unchanged, whereas its vertical component increases due to the acceleration of gravity in the downward direction. If a particle is projected from a height \(h\) with horizontal velocity \(u\), its velocity \(v\) at any point B after time \(t\) has components \(v_x=v\cos\theta=u\) and \(v_y=v\sin\theta=gt\), where \(\theta\) is the angle made by the instantaneous velocity with the horizontal direction, given by

\[\tan\theta=\frac{gt}{u},\qquad v=\sqrt{u^2+g^2t^2}\]

Horizontal distance travelled by the particle: \(x=ut\). Vertical distance travelled by the particle: \(y=\tfrac12gt^2\). Eliminating \(t\),

\[y=\left(\frac{g}{2u^2}\right)x^2\]

The resultant path or trajectory of the particle is a parabola. If \(T\) is the time taken by the particle to reach the ground from height \(h\), then

\[h=\tfrac12gT^2\quad\text{or}\quad T=\sqrt{\frac{2h}{g}}\]

The horizontal distance OC covered to reach the ground is called the horizontal range \(R\):

\[R=uT=u\sqrt{\frac{2h}{g}}\]

(ii) Projection at an angle: If a particle is projected from origin O with velocity \(u\) at angle \(\theta\) to the horizontal, its horizontal component \(u\cos\theta\) remains constant whereas its vertical component \(u\sin\theta\) changes continuously because of the acceleration due to gravity, which acts in the downward direction. The vertical velocity decreases due to the retardation \(g\) and becomes zero at the highest point, at height \(H\). The vertical speed then increases when the body moves downwards until it reaches the \(x\)-axis, covering a distance \(R\) in the horizontal direction, which is called the horizontal range.

If \(v_x\) and \(v_y\) are the horizontal and vertical components of the velocity after time \(t\), then \(v_x=u\cos\theta\), \(v_y=u\sin\theta-gt\). If \(\alpha\) is the angle subtended by \(\vec v\) with the horizontal,

\[\tan\alpha=\frac{v_y}{v_x}=\frac{u\sin\theta-gt}{u\cos\theta}\]

EQUATION OF TRAJECTORY

If \(P(x,y)\) is the position of the particle after time \(t\),

\[x=u\cos\theta\,t\ \text{ or }\ t=\frac{x}{u\cos\theta},\qquad y=u\sin\theta\,t-\tfrac12gt^2\]

\[\Rightarrow\ y=\frac{x\,u\sin\theta}{u\cos\theta}-\frac{gx^2}{2u^2\cos^2\theta}\]

\[\Rightarrow\ y=x\tan\theta-\frac{gx^2}{2u^2\cos^2\theta}\]

It is the equation of the trajectory of the projectile, which represents a parabola.

MAXIMUM HEIGHT

If \(H\) is the maximum height of the projectile, its velocity in the vertical direction is zero at this height.

\[v^2-u^2=-2gH\ \Rightarrow\ 0-u^2\sin^2\theta=-2gH\ \Rightarrow\ H=\frac{u^2\sin^2\theta}{2g}\]

TIME OF FLIGHT

If \(T\) is the time of flight from O to B, for which the vertical displacement \(y=0\), then

\[y=v_yT-\tfrac12gT^2\ \Rightarrow\ 0=(u\sin\theta)T-\tfrac12gT^2\ \Rightarrow\ T=\frac{2u\sin\theta}{g}\]

Time to reach maximum height \(=\dfrac T2=\dfrac{u\sin\theta}{g}\).

HORIZONTAL RANGE

It is the distance \(x=OB\) covered by the particle with horizontal velocity \(u\cos\theta\) in time \(T\).

\[\therefore\ x=R=u\cos\theta\times T=u\cos\theta\times\frac{2u\sin\theta}{g}=\frac{u^2\sin2\theta}{g}\]

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

The range \(R\) of a particle is maximum when \(\sin2\theta=1\) or \(2\theta=\dfrac\pi2\) or \(\theta=\dfrac\pi4=45^\circ\). The maximum value of the horizontal range is \(\dfrac{u^2}{g}\).

There are two angles of projection, i.e., \(\theta\) and \(\dfrac\pi2-\theta\), for which the horizontal range of a particle is equal, provided the initial velocity \(u\) is the same in each case.

(iii) Projection on an inclined plane: Consider a particle projected with velocity \(u\) at an angle \(\alpha\) to the horizontal on a plane inclined at angle \(\beta\) to the horizontal. The particle strikes the plane at point \(P(x,y)\), having range \(R=OP\) on the plane. The initial velocity of projection will have two components:

1. \(u\cos(\alpha-\beta)\) along the plane
2. \(u\sin(\alpha-\beta)\) perpendicular to the plane

Similarly, the acceleration \(g\sin\beta\) is along the negative \(x\)-axis (down the plane) and \(g\cos\beta\) is perpendicular to the plane in the negative \(y\)-direction. Let \(t\) be the time taken by the particle to go from O to P, so that the distance travelled perpendicular to OP is equal to zero. Applying the equation of motion along the \(y\)-axis,

\[S=ut+\tfrac12at^2;\ \ \text{at }t=t_0,\ S=0:\quad0=u\sin(\alpha-\beta)\,t_0-\tfrac12g\cos\beta\,t_0^2\]

\[t_0=\frac{2u\sin(\alpha-\beta)}{g\cos\beta}\]

where \(t_0\) is the time of flight. The distance OQ travelled along the horizontal direction with velocity \(u\cos\alpha\) in time \(t_0\) is given by

\[OQ=x=\frac{u\cos\alpha\times2u\sin(\alpha-\beta)}{g\cos\beta}=\frac{2u^2\sin(\alpha-\beta)\cos\alpha}{g\cos\beta}\]

The distance OP travelled along the inclined plane gives the range \(R\) along the inclined plane. In \(\Delta OPQ\), \(\cos\beta=\dfrac{OQ}{OP}\), so

\[R=OP=\frac{OQ}{\cos\beta}=\frac{2u^2\sin(\alpha-\beta)\cos\alpha}{g\cos^2\beta}\]

CIRCULAR MOTION

ANGULAR VELOCITY

Consider a particle A moving along a circular path of radius \(r\) and centre O with a constant linear speed \(v\) which is tangential to every point on the circular path; then the particle moves with uniform circular motion. Let the particle travel a distance AB, traversing an angle \(\theta\) in time \(t\). Length of the arc \(AB=r\theta=vt\). If \(\Delta\theta\) is the angular displacement in a very small time interval \(\Delta t\),

\[r\Delta\theta=v\Delta t\ \Rightarrow\ \frac{\Delta\theta}{\Delta t}=\frac vr\]

The angular velocity \(\omega\) of the particle is given by

\[\omega=\lim_{\Delta t\to0}\frac{\Delta\theta}{\Delta t}=\frac{d\theta}{dt}\]

The angular velocity \(\omega\) of the particle is defined as the rate of change of angle with respect to time. Its unit is radian/second.

\[v=r\omega\]

Linear velocity = radius \(\times\) angular velocity.

If \(T\) is the time taken to complete one revolution by traversing an angle of \(2\pi\) radians,

\[T=\frac{2\pi}{\omega}=\frac{2\pi r}{v}\]

If the particle completes \(n\) revolutions per second,

\[n=\frac1T=\frac{\omega}{2\pi}\quad\text{or}\quad\omega=2\pi n\]

where \(n\) is the frequency of the particle.

ANGULAR ACCELERATION

The rate of change of angular velocity is called angular acceleration, \(\alpha\).

\[\alpha=\lim_{\Delta t\to0}\frac{\Delta\omega}{\Delta t}=\frac{d\omega}{dt}=\frac{d}{dt}\left(\frac vr\right)=\frac1r\left(\frac{dv}{dt}\right)=\frac ar\]

\(\therefore\) Linear acceleration \(a\) = radius \(\times\) angular acceleration.

If a particle traverses a small distance \(ds\) along the length of the arc when it moves through an angle \(d\theta\), \(ds\simeq r\,d\theta\). Differentiating with respect to \(t\),

\[\frac{dv}{dt}=\frac{d}{dt}(r\omega)=r\frac{d\omega}{dt}=r\frac{d^2\theta}{dt^2}\]

\(\therefore\) Linear (tangential) acceleration \(a_T=r\alpha\), where \(a_T\) is the tangential linear acceleration and \(\alpha=\dfrac{d^2\theta}{dt^2}\). Its unit is radian/sec².

For a body moving with initial angular velocity \(\omega_0\) and constant angular acceleration \(\alpha\), its angular velocity becomes \(\omega\) after time \(t\) and its angular displacement is \(\theta\); then

\[\omega=\omega_0+\alpha t,\qquad\omega^2=\omega_0^2+2\alpha\theta,\qquad\theta=\omega_0t+\tfrac12\alpha t^2\]

These relations are similar to the relations for linear motion.

When a particle moves in a circular path with constant speed \(v\), its speed is constant but the direction of its velocity vector is continuously changing. It implies that the motion is such that the particle is accelerated. Such an acceleration is directed towards the centre of the circle and is called the centripetal acceleration. The magnitude of the centripetal acceleration is

\[a_c=\frac{v^2}{r}=\omega v=r\omega^2\]

but its direction at every point along the circular path is towards the centre O. The net force due to centripetal motion, \(F\), is equal to the product of the mass of the particle and the centripetal acceleration.

\[F=m\left(\frac{v^2}{r}\right)=mr\omega^2=mr\left(\frac{2\pi}{T}\right)^2=\frac{4\pi^2mr}{T^2}=4\pi^2mrn^2\]

where the frequency \(n=\dfrac1T\).

If a stone is tied with a string and rotated along a circular path, the direction of the velocity of the stone changes, producing a centripetal acceleration \(a_c\) directed towards the centre due to the centripetal force. If the string is released, the stone will fly off in a tangential direction. The accelerations \(a_c\) and \(a_t\) are mutually perpendicular to each other, and the resultant acceleration \(a\) of the particle is given by

\[a=\sqrt{a_c^2+a_t^2}\]