Introduction Mechanics is that branch of science which deals with the state of rest or motion of bodies under the action of forces. The subject of Mechanics is logically divided into two parts.Statics which concerns the equilibrium of bodies under the action of forces,and Dynamics concerns the motion of Bodies. Dynamics is again subdivided into a. Kinematics b. Kinetics k1 | Website for Students | VTU NOTES | QUESTION PAPERS

Study of Kinematics concerns with the motion of a body, without referring to the forces causing the motion of that body. Study of Kinetics concerns with the motion of the body considering the forces causing the motion. Terms and definitions space: is the geometric region occupied by bodies whose positions are described by linear and angular measurements relative to a coordinate system. k2 | Website for Students | VTU NOTES | QUESTION PAPERS

Time is the measure of succession of events and is basic quantity in dynamics. Mass is a measure of the inertia of a body, which is its resistance to a change of velocity Particle. A body of negligible dimensions is called a particle. In the mathematical sense a particle is a body whose dimensions approach to zero so that it may be analyzed as point mass. Rigid body. A body is considered rigid when the relative movements between its parts are negligible for the purpose at hand. k3 | Website for Students | VTU NOTES | QUESTION PAPERS

Newton’s Laws of Motion First Law. A particle remains at rest or continues to move in a straight line with a uniform velocity if there is no unbalanced force acting on it. Second Law. The acceleration of a particle is proportional to the resultant force acting on it and is in the direction of this force. Third Law. The forces of action and reaction between interacting bodies are equal in magnitude, opposite in direction, and collinear. k4 | Website for Students | VTU NOTES | QUESTION PAPERS

Newton’s second law forms the basis for most of the analysis in dynamics. As applied to a particle of mass m, it may be stated as F=ma Where F is the resultant force acting on the particle and a is the resulting acceleration. This equation is a vector equation. Rectilinear Motion: It is the motion of a particle along a straight line. k5 | Website for Students | VTU NOTES | QUESTION PAPERS

A particle moves along a straight line OX as shown in the figure. The moving particle is in position M at any time ‘t’ and it covers a distance ‘s’ from ‘O’. The particle moves to N, through a distance ∆s in a small interval of time ∆t. The velocity v at the instant when the particle is at certain point M, at time ‘t’ is the rate of change of displacement ∆s as the increment of time ∆t approaches to zero as limit is known INSTANTANEOUS VELOCITY and is given by Lt v = ∆t  0 ∆s/∆t = dS/dt Rectilinear motion with uniform acceleration: S t ∆s ∆t X M N O k6 | Website for Students | VTU NOTES | QUESTION PAPERS

Average Velocity : It is the uniform velocity with which the particle may be considered to be moving in order to cover the total distance s in a total time t. k7 (i) When the particle moves with uniform velocity s = u.t (ii) When particle moves with variable velocity v av = Total distance covered (s) Total time (t)

(iii) When particle moves with initial velocity u and constant acceleration a, its velocity changes to v, then v av = (u+v)/2 s = (u+v)/2 x t s = distance covered in time ‘t’ (iv) If the distances moved by the particle from start are s 1 in t 1, s 2 in t 2, then average velocity may also be found by v avg = (s 2 - s 1 ) / (t 2 -t 1 ) k8 | Website for Students | VTU NOTES | QUESTION PAPERS

If the velocity of a particle is v at M and it changes by ∆v, in a small interval of time ∆t then the acceleration of moving particle, a at the instant at which particle is at M i. e. the instantaneous acceleration is given by A particle may move in a straight line with constant acceleration or with variable acceleration. Acceleration : k9 | Website for Students | VTU NOTES | QUESTION PAPERS

If the velocity of a body changes by equal amounts in equal intervals of time, the body is said to move with uniform acceleration. Variable Acceleration: If the velocity of a body changes by unequal amounts in equal intervals of time, the body is said to move with variable acceleration. Note: When the velocity is increasing the acceleration is reckoned as positive, when decreasing as negative (retardation or deceleration). Uniform Acceleration : k10 | Website for Students | VTU NOTES | QUESTION PAPERS

Displacement – Time Variations: Time Displacement Fig. 1 k11 In fig (1)The graph is parallel to the time axis indicating that the displacement is not changing with time. The slope of the graph is zero. The body has no velocity and is at rest. | Website for Students | VTU NOTES | QUESTION PAPERS

Displacement Time ∆t ∆x In fig (2)The displacement increases linearly with time. The displacement increases by equal amount in equal intervals of time. The slope of the graph is constant. k 12 Fig.2 | Website for Students | VTU NOTES | QUESTION PAPERS

k13 Displacement Time Fig. 3 ∆x 2 ∆t 2 ∆x 1 ∆t 1 In fig(3) The displacement is not changing by equal amounts in equal intervals of time. The slope of the graph is different at different times.The velocity of the body is changing with time. The motion of the body is accelerated. | Website for Students | VTU NOTES | QUESTION PAPERS

Velocity – Time Variations: Time Fig. (a) velocity O ∆v ∆t In fig. (a): The velocity of the body increases linearly with time. The slope of the graph is constant i.e. the velocity changes by equal amounts in equal intervals of time. (acceleration of the body is constant and at t = 0, the velocity is finite. Thus the body moving with a finite initial velocity, and has constant acceleration). Constant velocity k14 | Website for Students | VTU NOTES | QUESTION PAPERS

Velocity – Time Variations: Time Fig. (b) velocity O ∆v ∆t In fig. (b): The body has a finite initial velocity. As time passes, the velocity decreases linearly with time until its final velocity becomes zero i.e it comes to rest. Thus the body at a constant deceleration, since the slope of the graph is negative. k15 Uniform acceleration | Website for Students | VTU NOTES | QUESTION PAPERS

In fig. (c): The velocity –Time graph is a curve. The slope is, therefore, different at different times. In other words, the velocity is not changing at constant rate. The body does not have a uniform acceleration since acceleration is changing with time. Time Fig. (c) velocity O ∆v 2 ∆t 2 ∆v 1 ∆t 1 k16 Variable acceleration | Website for Students | VTU NOTES | QUESTION PAPERS

1. Equation of motion (Relation between v,u, a & t) If we assume a body starts with an initial velocity u and uniform acceleration a. After time t, it attains a velocity v. Therefore the change in velocity in t seconds is (v – u) Change in velocity / sec. = v – u / t = a v = u + at -----(1) Equations of Motion Under Uniform Acceleration k17 | Website for Students | VTU NOTES | QUESTION PAPERS

2. Equation of Motion: (Relation between s, u, a and t) Let a body moving with an initial uniform velocity u be accelerated with a uniform acceleration a for time t. If v is the final velocity, the distance s which the body travels in time t is determined as follows. Now since acceleration is uniform it is obvious that the average velocity = (u + v) /2  Distance traveled = v av x t = (u + v)/2 x t = (u + (u + at))/2 x t (Substituted from 1) s = ut + ½ at (2) k18 | Website for Students | VTU NOTES | QUESTION PAPERS

3. Equation of motion: (Relation between u, v, a and s) s = average velocity x time = (u + v)/2 x t = (u + v)/2 x (v - u)/a for t = (v – u)/a therefore s = (v 2 - u 2 )/2a v 2 = u 2 +2as (3) k19 | Website for Students | VTU NOTES | QUESTION PAPERS

Motion under Gravity It has been observed that bodies falling to the earth (through distances which are small as compared to the radius of the earth) experience entirely unrestricted increase in their velocity by 9.81 m/s for every second during their fall. This acceleration is called the acceleration due to gravity and as conventionally denoted by g. For downward motion: a = +g v = u + gt h = ut + ½ gt 2 v 2 = u 2 + 2gh For upward motion: a = – g v = u – gt h = ut – ½ gt 2 v 2 = u 2 – 2gh h = Distance moved in vertical direction k20 | Website for Students | VTU NOTES | QUESTION PAPERS

Equations Acceleration 123 Uniform v = u + at S = ut + ½ at 2 v 2 = u 2 + 2as t t s Variable v = u+∫ a(t)dt S = ut + 1/2∫ a (t 2 )dt v 2 – u 2 = 2 ∫ a (s)ds Comparison between equations of motion under uniform acceleration and variable acceleration: The slope ds/dt at any point gives the velocity v at that point. Graphical Representation: t1t1 t2t2 t dt ds v = ds/dt s-t curve S k21 | Website for Students | VTU NOTES | QUESTION PAPERS

The slope dv/dt at any point gives the acceleration a at that point. The shaded area under v – t curve shown above gives the incremental displacement ds during the small interval of time dt. v v dt dv dt t2t2 v – t curve k22 t t1t1 | Website for Students | VTU NOTES | QUESTION PAPERS

The shaded area under a – t curve shown above gives the incremental velocity dv during the small interval of time dt. a t2t2 t1t1 dt dv=adt t k23 | Website for Students | VTU NOTES | QUESTION PAPERS

PROJECTILES INTRODUCTION Assumptions: 1.Mass of the projectile is not considered. 2.Air resistance is neglected. 3.The trajectory of the particle is in the vertical plane. k24 | Website for Students | VTU NOTES | QUESTION PAPERS

A Projectile is a particle moving in space under the action of gravity. The velocity with which the particle is projected into space has horizontal and vertical components. The combined effect of both components is to move the particle along a parabolic path. The parabolic path traced by the projectile is known as Trajectory of the Projectile. The horizontal component remains constant ( as air resistance is ignored) while the vertical component of motion is always subjected to acceleration due to gravity. k25 | Website for Students | VTU NOTES | QUESTION PAPERS

DEFINITIONS: Velocity of Projection is the velocity with which a body is projected into space. Angle of projection is the angle which the initial velocity vector makes with the horizontal, or the angle at which a projectile is projected with respect to horizontal. Range is the distance along the reference plane between the point of projection and the point at which it strikes the plane. Time of Flight is the total time during which the particle remains in motion. Maximum Height is the maximum vertical distance covered by the projectile from the point of projection. k26 | Website for Students | VTU NOTES | QUESTION PAPERS

MOTION OF A PROJECTILE  u R trajectory x y O P(x,y) M Consider a particle thrown upwards from a point O, with an initial velocity u, at an angle  with the horizontal as shown in the figure above. After attaining maximum height h, it descends and finally hits the reference plane. k27 h Q | Website for Students | VTU NOTES | QUESTION PAPERS

Equation of the Trajectory From the figure above x = u x. t = u (Cos  ) t t = x / (u Cos  ) (1) y = u y. t – ½ gt 2 =u (Sin  )t – ½ gt (2) Sub (1) in (2) we get y = x tan  - gx 2 / (2u 2 Cos 2  ) This is an equation for a parabola. Hence the path of the projectile is a parabola. k28 | Website for Students | VTU NOTES | QUESTION PAPERS

The horizontal distance covered by the projectile is known as Range of Projectile denoted by R. Resolving the initial velocity u into horizontal and vertical components u x = u Cos  u y = uSin  (constant) Time of Flight: We know that v y = u y + at At Q v y = 0 0 = u y - gt t m = u y / g = u Sin  / g  Time of flight T = 2t m = 2 (u Sin  /g) Where t m is the time taken in seconds to reach maximum height. k29 | Website for Students | VTU NOTES | QUESTION PAPERS

Maximum height attained (h): v y 2 - u y 2 =- 2gh (upward motion) v y = 0 h = u y 2 /2g ; h = u 2 (Sin 2  ) / 2g Range: R= u x x T (time of flight) = 2u 2( Sin  Cos  ) / g R = u 2 Sin2  /g (Sin2  = 2Sin  Cos  ) k30 | Website for Students | VTU NOTES | QUESTION PAPERS

From the above equation it is clear that range will be maximum if Sin2  = 1 i.e. 2  = 90 or  = 45 R max = u 2 /g Projectile will cover a maximum range when it is directed at an angle of 45°. Two angles of Projections for a given range: We know that Range R = u 2 Sin2  /g Sin  = Sin (  -  )  u 2 Sin (  - 2  )/g = u 2 Sin (2  1 )/g (say)  where 2  1 = (  - 2  ) k31 | Website for Students | VTU NOTES | QUESTION PAPERS

Thus for same range R = u 2 Sin2  /g = u 2 Sin (  - 2  )/g = u 2 Sin2  1 /g Which shows that the horizontal range remains the same when  is replaced by  1. Or  1 =  - 2  =(  / 2 )-  2 k32 | Website for Students | VTU NOTES | QUESTION PAPERS

Projection on an Inclined Plane  u  R B O R(Sin  ) Q R(cos  ) R= Range along incline; α=angle of projection; β =angle of inclined plane S x =u x + ½ a x t 2 P  P1P1 k33 | Website for Students | VTU NOTES | QUESTION PAPERS

Projection on an Inclined Plane R(cosβ)=u(cosα)(t)+(0) ; a x =0 t= Rcosβ / ucosα…..(1) S y =u y + ½ a y t 2 R sinβ=u sinα (t)- ½ gt 2 Substituting eqn.(1) and simplifying we get The Range along the inclined plane R={2u 2 cos 2 α [sin(α- β)] } / gcos 2 β..(2) k34 | Website for Students | VTU NOTES | QUESTION PAPERS

Projection on an Inclined Plane: To get maximum range on the incline, Differentiating R w.r.t α and equating it to zero we get α= π/4 + β/4 Substituting this value of α in eqn.(2) We get maximum Range R max = u 2 /g(1+sin β) To find the time of flight: using the relation v=u + at k35 | Website for Students | VTU NOTES | QUESTION PAPERS

Projection on an Inclined Plane: 0=u x sin(α- β)-gx(cos β) x t t= {usin(α- β)}/ g(cos β) a=(g cos β) is the acceleration due to gravity along inclined plane t=time taken by the projectile to reach Q where QP is perpendicular distance to the incline plane Time of Flight T= 2xt T={2 u sin(α- β)} / g(cos β) k36 | Website for Students | VTU NOTES | QUESTION PAPERS

Projection on an Inclined Plane: Maximum Height attained h, (PQ) using the relation v 2 -u 2 =2as 0-u 2 sin 2 (α- β)=-2 gcos β x h h={ u 2 sin 2 (α- β)} / 2gcos β Vertical height P 1 Q=h/ cos β ={u 2 sin 2 (α- β)} / 2gcos 2 β k37 | Website for Students | VTU NOTES | QUESTION PAPERS

1.On turning a corner, a motorist rushing at 15m/s, finds a child on the road 40m ahead. He instantly stops the engine and applies brakes,so as to stop the car within 5m of the child, calculate: (a) retardation (b) time required to stop the car Ans: (a) -3.21m/s 2, (b) 4.67s 2.A stone is dropped from the top of a tower 100m high. Another stone is projected upward at the same time from the foot of the tower, and meets the first stone at a height of 40m. Find the velocity,with which the second stone is projected upwards. Ans: u=28.6 m/s k38 Practice Problems: | Website for Students | VTU NOTES | QUESTION PAPERS

k39 Practice Problems: 3. A train starting from rest is accelerated and acceleration at any instant is 3/(v+1) m/s. where v is the velocity of the body in meters per second at any instant. Find the distance in which the train attains velocity of 48kmph. Ans: S=293m 4. A projectile is fired from the edge of a 150m high cliff with an initial velocity of 180m/s at an angle of elevation of 30 o with the horizontal. Neglecting air resistance, find (a) the horizontal distance from the gun to the point where the projectile strikes the ground,and (b) the greatest elevation above the ground reached by the projectile. (a) 3125m (b) 563m | Website for Students | VTU NOTES | QUESTION PAPERS

k40 Practice Problems: 3. A train starting from rest is accelerated and acceleration at any instant is 3/(v+1) m/s. where v is the velocity of the body in meters per second at any instant. Find the distance in which the train attains velocity of 48kmph. Ans: S=293m 4. A projectile is fired from the edge of a 150m high cliff with an initial velocity of 180m/s at an angle of elevation of 30 o with the horizontal. Neglecting air resistance, find (a) the horizontal distance from the gun to the point where the projectile strikes the ground,and (b) the greatest elevation above the ground reached by the projectile. (a) 3125m (b) 563m | Website for Students | VTU NOTES | QUESTION PAPERS

k41 Practice Problems: 5. An aero plane is flying at a height of 300m,with a velocity of 360kmph. A Shell is fired from the ground exactly when the aeroplane is above the gun.what should be the minimum initial velocity of the shell and the angle of inclination in order to hit the aeroplane ? Ans:( 126m/s, o ) 6. A projectile is fired from a point ‘o’ at a velocity of 125m/s has to strike a point located on the top of a tower of 200m high. The horizontal distance betweem the point ‘o’ and the tower is 1000m. Neglect the air resistance and take g= 9.8m/s 2 | Website for Students | VTU NOTES | QUESTION PAPERS

k42 Practice Problems: Contd: Calculate: a)The angle, to the horizontal,at which the projectile must be fired in order to strike the point on tower in minimum time. b) the time taken for flight c)The maximum height above ‘o’ reached by the projectile. Ans: (a) or ,( b) 9.54 s, (c) 233.8m | Website for Students | VTU NOTES | QUESTION PAPERS