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GVHSS PAYYOLI- PLUS ONE PHYSICS FOCUS AREA NOTES PART 1, , CHAPTER 1: PHYSICAL WORLD, 1.2 SCOPE AND EXCITEMENT OF PHYSICS, Scope of physics means various range of physical quantities like, mass,length,time etc, Eg: (1)range of length includes 10-14m( electron, protons etc) to, 1026m( astronomical phenomena), (2) range of mass is from 10-30kg(mass of electron) to 1055 kg, (mass of universe), There are two domains in physics (1)macroscopic domain (2), Microscopic domain, (1)macroscopic Domain: It include big quantities or phenomena, at laboratory, terrestrial and astronomical scales, eg:Classical physics which include, mechanics,Electrodyanamics,optics,, thermodynamics etc, (2)Microscopic Domain:It include atomic molecular and nuclear, phenomena (very small quantities ), eg: Quantum theory, , CHAPTER 2: UNITS AND MEASUREMENTS, 2.2 THE INTERNATIONAL SYSTEM OF UNITS :, System of units which commonly used are C.G.S system ,F.P.S, system, M.K.S system, , SI system, System of units which accepted internationally are called SI, system ( internationally accepted System of Units).SI units are, devoloped and recommended by General Conference on Weights, and Measures in 1971, Units of fundamental (base)quantitis are called Fundamental, units or base units and units of derived quantities are called, derived units, There are seven base quantities and two supplimentary, quantities, SREESHYJU K P & JIJESH M T ,HSST PHYSICS GVHSS PAYYOLI

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Base quantities and units, , Supplimentary Quantities and units, , Plane angle, , Solid Angle, , 2.7, 2.7 SIGNIFICA, 2.9 DIMENSIONAL FORMULAE AND DIMENSIONAL, EQUATIONS:, All physical quantities represented by derived units can be, expressed in terms of combination of seven fundamental or base, quantities. These base quantities are called seven dimensions of, physical world. Which are denoted with square brackets [ ], Eg;-, , All physical quantities can be writen in terms of dimensions of, these base quantities, eg 1;- Area

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Then the dimensional formula of area is [ M0L2T0], and the equation obtained by equating physical quantity with, dimensional formula is called dimensional equation ie [ area ] = [, M0L2T0], In this equation the power of mass is zero ,power of length is, two and the power of time is zero., The power of base quantities in an equation are called ‘ the, dimensions’ of the quantity. Therefor area is zero dimension in, mass two dimensions in length and zero dimension in time. Or, area is independent of mass and time, Defenition:-, , eg 2;-, , Volume, , Dimensional equation for volume can be written as, [ volume] = [ M0 L3 T0 ] ie volume is zero dimensions in mass, three dimension in length and zero dimension in time or volume is, independent of mass and time, eg 3 :- Density, 𝒎𝒂𝒔𝒔, Density =, [density]=, , 𝒗𝒐𝒍𝒖𝒎𝒆, [𝑴], , [𝑳𝟑 ], , = [ML-3], , 2.10 PRINCIPLE OF HOMOGENEITY OF DIMENSIONS, , The principle of homogeneity of dimension states that the, magnitudes of physical quantities can be added together or, subtracted from one another only if they have same dimensions, The principle of homogeneity of dimension states that the, dimension of physical quantities on both sides of an equation are, same ie [LHS] = [RHS], PROBLEM 1, , SREESHYJU K P & JIJESH M T ,HSST PHYSICS GVHSS PAYYOLI

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ANSWER:, By principle of homogeneity of dimensions physical, quantities with same dimensions can be added or subtracted., Therefor the dimensions of at2,bt and c must be same and equal to, dimension of velocity as given by equation, ie [v] = [at2], [LT-1]= [a][T2], ∴ [a]=, , [𝑳𝑻−𝟏 ], [𝑻𝟐 ], , = [ LT-1] × [T-2], , [a] = [LT-3], Also [v] = [bt], [LT-1]= [b][T], [b]=, Also, , [𝑳𝑻−𝟏 ], [𝑻], , = [LT-1]× [T-1], , [b]= [LT-2], [v]= [c], [c]= [LT-1], , PROBLEM 2, , and b, , ANSWER:, By principle of homogeneity of dimensions physical, quantities with same dimensions can be added or subtracted, 𝒂, Therefor [P] = [ 𝟐 ], -1, , -2, , [ML T ] = [, , 𝒂, (𝑳𝟑 ), , 𝑽, , 𝟐 ]=[, , [a]= [ML-1T-2][L6], [a] = [ML5T-2], Also [b]= [V]= L3, , 𝒂, 𝑳𝟔, , ]

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From (1) and (2) the [LHS]=[RHS], Therefor the equation is dimensionally correct, (4) H=, , 𝒗𝟐 𝒔𝒊𝒏𝟐 𝜽, 𝟐𝒈, , By principle of homogeneity of dimensions, [H]= [, , 𝒗𝟐, , 𝒔𝒊𝒏𝟐 𝜽, 𝟐𝒈, , [LHS]=[RHS], , ], , [H]= [L] …..(1), [, , 𝒗𝟐 𝒔𝒊𝒏𝟐 𝜽, 𝟐𝒈, , 𝟐, , ] =[, , [(𝑳𝑻−𝟏) ], [𝑳𝑻−𝟐 ], , =, , [𝑳𝟐 𝑻−𝟐 ], [𝑳𝑻−𝟐 ], , = [L]…(2), , ( 𝜽 is a dimensionless quantity ), From (1) and (2) the [LHS]=[RHS], Therefor the equation is dimensionally correct, 2.10.2 DEDUCING RELATION AMONG PHYSICAL QUANTITIES, PROBLEM3, Derive an expression for time period of a simple pendulum., Assume that time period depends on length of the pendulum,, mass of the bob and acceleration due to gravity.( or show that, time period of oscillation of a simple pendulum does not depends, on mass of the bob), ANSWER:, Let the time period ( T )of pendulum depends on, (i) mass of the bob(m), (i) length of the pendulum( l ), (iii) acceleration due to gravity ( g ), Let the equation of Time period T is,, T= k mx l y g z ………(1), Taking dimensions, [T]= [T], [m]= [ M ], [ g ]= [LT-2 ], Substituting in (1), [T]= [M]x[L]y[LT-2]z, [M0L0T1] = [ M]X[L]Y+Z[T]-2Z, By principle of homogeneity [LHS]= [RHS], Equating the powers on both sides, x= 0, y+z = 0, -2z = 1, y-, , 𝟏, 𝟐, , y=, , =0, 𝟏, 𝟐, , substituting in (1), , z=-, , 𝟏, 𝟐

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T= k m0 l 1/2 g – ½, Since the dimension(power) of mass is zero the time period does, not depends on mass, T= k √, , 𝒍, 𝒈, , Let k = 2𝝅 (by experiment), T= 2𝝅√, , 𝒍, 𝒈, , PROBLEM 2, Derive an expression for kinetic energy of an object?, ANSWER, Let the kinetic energy E depends on, (i) mass ( m ) of the object, (ii ) Velocity (v) of the object, Let the equation for kinetic energy E is,, E= k mx v y ……….(1), Taking dimensions, [E ]= [ ML2T-2] , [m] = [M], [v]= [ LT-1], Substituting in (1), [ ML2T-2]= [M]x [ LT-1]y, [ ML2T-2]= [MxLyT-y], By principle of homogeneity [LHS]= [RHS], Equating the powers on both sides, x= 1, y= 2, substituting in (1), E= k m1 v 2 = k mv2, Let k = ½ (by experiment), 𝟏, , E= mv2, 𝟐, , PROBLEM 3, Derive an expression for gravitational potential energy of an, object, ANSWER, Let the gravitational potential energy U depends on, (i) mass ( m ) of the object, (ii) acceleration due to gravity ( g ), (ii)height of the object from the surface of earth( h), Let the equation for gravitational potential energy U is,, U = k mxgyhz ……..(1), Taking dimensions, [U ]= [ ML2T-2] , [m] = [M], [g]= [ LT-2], [h] = [L], Substituting in (1)

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[ ML2T-2] = [M]x [ LT-2]y [L]z, [ ML2T-2] = [Mx Ly+z T-2y], By principle of homogeneity [LHS]= [RHS], Equating the powers on both sides, x= 1, y+z= 2, -2y = -2, 1+z = 2, y= 1, z= 1, substituting in (1), U = k m1g1h1, Let k = 1(by experiment), U= mgh, PROBLEM 4, Derive an expression for centripetal force, ANSWER, Let the centripetal force(f) depends on, (i) mass of the object (m), ( ii ) velocity of the object (v), (iii) radius of the circular path (r), Let the equation for centripetal force f is, f= kmxvyrz …….. (1), Taking dimensions, [f] = [MLT-2], [m]= [M],[v]= [ LT-1], [r]= [ L], substituting in (1), [MLT-2]= [M]x [ LT-1]y [ L]z, [MLT-2] = [Mx Ly+z T-y], By principle of homogeneity [LHS]= [RHS], Equating the powers on both sides, x=1, y+z = 1, -y= -2, 2+z= 1, y= 2, z= -1, substituting in (1), f= km1v2r -1, Let k =1 ( by experiment), f=, , 𝒎𝒗𝟐, 𝒓, , PROBLEM 5, Derive an expression for orbital velocity of a planet in terms of, gravitational constant G, radius R of the orbit, mass M of the sun, ANSWER, Let the orbital velocity(v) of a planet depends on, (i) gravitational constant (G)

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CHAPTER 3: MOTION IN A STRAIGHT LINE, 3.2 POSITION,PATH LENGTH,AND DISPLACEMENT, Distance or Path length :, Total length of the path travelled by an object is, called path length or distance, (1)If object travels from o to p, , (2)If object travels form o to p and then to Q, , (3) If object travels form o to p and back to O, , (4) If object travels form o to p and then to Q, , Thus path length or distance will be always positive and, continuously increases when object moves. Unit is metre . It is, a scalar quantity and depends on path, DISPLACEMENT, Displacement is change in position, If x1 is the position of an object at an instant t1 and x2 is the, position at an instant t2, Then displacement ∆𝒙 = x2 - x1, ∆𝒙 is positive if x2 > x1, SREESHYJU K P & JIJESH M T ,HSST PHYSICS GVHSS PAYYOLI, ∆𝒙 is negative if x2 < x1, (1) If object travels form o to p

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Here Path length is equal to displacement, (2)If object travels form o to p and then to Q, , Displacement is less than path length, (3) If object travels form o to p and back to O, , Displacement is zero, (4) If object travels form o to p and then to Q, , Displacement is negative, Comparison between distance and displacement, , Qn No:1, , (a), Ans: No, A particle cannot have two different positions at, the same time, (b), (C), , Ans: No,Total path length never, can be decreased or can be zero, , Ans: yes, displacement, decreases with time

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GRAPHICAL REPRESENTATION OF MOTION, (1) Position time graph of an object at rest, , (2) uniform motion along a straight line, , The position time graph of an object in uniform motion is a, straight line with a slope, Qn No:2, , State whether the following statements are true or false, (a), B, , (b), , OP- distance of A from, school to home, OQ- distance of B from, school to home, OP < OQ

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than A, , From fig A starts earlier than B, (c), , A reaches home at t1 and B, reaches home at t2, (d), , (e) B overtakes A, True, , Qn No 3:, , His position as a function of time is shown below

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(a)How far was he from home when started raining, (b)identify the portion of the graph,where he was waiting, for the rain to stop, (c) What is his displacement in 10m, (d) he returned home with a greater speed . is this satemeent, true, Answer:, (a) 0.5 km, (b)BC, (c) zero, (d) true, because he returned home from market in, 3minutes but reched market form home in 5 minutes, 3.3 AVERAGE VELOCITY AND AVERAGE SPEED, AVERAGE VELOCITY, Displacement per unit time is called average velocity., , Let x1 is the position of the particle at an instant t1 and x2 is the, position of the particle at an instant t2, , Average velocity,, , or, , Unit is ms-1, It is a vector quantity and can be positive or negative, Or zero, Position –time graph, , = average velocity, , Slope of position time graph is the average velocity

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Case 1, Here x2 > x1 , slope of graph is positive, If the slope of position time graph is positive ,the object, moves away from origin, Case 2, , If the slope of position time graph is negative ,the object, moves towards the origin ( average velocity is negative), Case 3: object at rest, , If Slope is zero ,object is at rest, AVERAGE SPEED, , Unit is ms-1 and it is a scalar quantity, PROBLEM1

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ANSWER:, (a) From O to P, , Average velocity, , Average speed, , (b) From O to P and then back to Q, , Average velocity

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ANSWER:, , PROBLEM 1:

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3.5 ACCELERATION, Time rate of change of velocity is called acceleration, Average acceleration:, , Unit of average acceleration is ms-2 and it is a vector quantity, Instantaneous acceleration:, =, , 𝒅𝒗, 𝒅𝒕, , Note:, , Velocity is the first differential of position and acceleration is the, second differential of position, PROBLEM 2, , ANSWER:, V=, a=, , 𝒅𝒙, 𝒅𝒕, 𝒅𝒗, 𝒅𝒕, , = 3t2 – 12, = 6t, , when v = 0, 0= 3t2 – 12, 3t2 =12, SREESHYJU K P & JIJESH M T ,HSST PHYSICS GVHSS PAYYOLI

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t2 = 12/3 = 4, t= 2 s, a= 6x2= 12 ms-2, Uniform acceleration: If acceleration of an object is constant, throughout the motion , then acceleration is called uniform, acceleration or constant acceleration, Slope of velocity time graph:, , Slope of velocity time graph gives acceleration of the object, Area under velocity time graph:, , Area under velocity time graph gives displacement, PROBLEM 3, , SREESHYJU K P & JIJESH M T ,HSST PHYSICS GVHSS PAYYOLI

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ANSWER:, (i ) ox, , (ii) XY, Since slope is zero, acceleration aXY = 0, (iii) YZ, , = area under velocity time graph, , The direction of acceleration and velocity are towards right., , Then the velocity increase in positive direction, Velocity- Time graph, , The direction of acceleration and velocity are towards left., , Then the velocity increase in negative direction

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Velocity- Time graph, , Velocity is positive and acceleration is negative. So velocity of the, object decreases since acceleration is negative, , When a body moves up its acceleration is downwards and, therefor velocity decreases and becomes zero. Since acceleration, is continuing the direction of velocity reversed and body moves, downwards and velocity increasers., , PROBLEM 1:, Velocity- time graph of an object is shown below.draw the, corresponding acceleration time graph

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Answer:, , …..(1), , a= ?, , Aceeleration can be calculated using the equation, , Sustituting the value of a in equation (1) we get, , PROBLEM 4, A person is running at his maximum speed of, 10 ms-1 to catch a bus .when he is 48 m from the door of the bus it, starts moving away with a constant acceleration of 1 ms-2.The, minimum time after which he can enter the bus is, , ANSWER:, Distance the person has to travel= 48m+ distance travelled by the, bus in time t, , X=, , 𝒕𝟐, 𝟐, , The minimum time after which he can enter the bus is 8s, FREE FALL: If an object is released near the surface of earth, and if air resistance is neglected, then the object is said to be in, free fall.The constant acceleration of the object is called, acceleration due to gravity( g= 9.8 ms-2), SREESHYJU K P & JIJESH M T ,HSST PHYSICS GVHSS PAYYOLI

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Acceleration – time graph of free fall:, , Velocity – time graph of free fall:, , Position time graph of free fall, , Note: The acceleration of a freely falling body is always in, downward direction and a = -g (9.8) ms-2, PROBLEM 5, 29.4m𝒔−𝟏, , ( Take g= 9.8 ms-2 and neglect air resistance), ANSWER:, , (a) During the upward motion acceleration is vertically, downwards, (b) At the highest point velocity is zero and acceleration is 9.8ms-2, acting vertically downwards., (c)

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(d), , m𝒔−𝟏, , Note.Total time taken by the ball to reach the ground is 3s+ 3s= 6s, PROBLEM 6, , ANSWER:, , Height of the tower is 57.3 m, Galileo’s Law of Odd numbers, , Ratio of distances traversedis 1:3:5:7:……

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EQUATIONS OF MOTION FOR CONSTANT, ACCELERATION USING METHOD OF CALCULUS:, (1) Velocity- Time relation, , (2) Displacement-Time relation, We have, , (3) Velocity- Displacement relation:, , SREESHYJU K P & JIJESH M T ,HSST PHYSICS GVHSS PAYYOLI

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PROBLEM 7, , ANSWER:, (a), , (b), , (c), , CHAPTER 4: MOTION IN A PLANE, Motion in a plane is a motion in two dimensions. Two coordinates, ( eg: x and y) will change during motion, 4.2 SCALARS AND VECTORS, A Scalar quantity is a quantity which have only magnitude eg:, mass, time ,volume, density, work etc, A vector quantity is a quantity that has both magnitude and, direction. Eg: force ,displacement etc, SREESHYJU K P & JIJESH M T ,HSST PHYSICS GVHSS PAYYOLI

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Representation of a vector:, The length of the line represents magnitude and arrow head, represents direction., ⃗, Vector A can be represented as ⃗𝑨, Magnitude of vector A can be represented as ,, , =, , Two vectors are said to be equal only if they have the same, magnitude and the same direction, POSITION AND DISPLACEMENTVECTORS, Consider a particle at P, ⃗ is the position vector of the point p from, 𝒓, origin o, Let the particle moves from P to p’, ⃗⃗⃗, ⃗ is the, 𝒓′ is the new position vector and ∆𝒓, displacement vector which depends only, on initial and final position and does not, depends on path, PROBLEM 1, , (a), (b), ANSWER:, (a) displacement depends only on initial and final point . There, for displacement for each is 400m, SREESHYJU K P & JIJESH M T ,HSST PHYSICS GVHSS PAYYOLI

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(b) for the girl B, PROBLEM 2, , ANSWER:, (a) total displacement is zero since initial and final positions are, same ie ∆𝒙 = 𝟎, (b), , Since ∆𝒙 = 𝟎, (c) average speed =, , 𝒕𝒐𝒕𝒂𝒍 𝒑𝒂𝒕𝒉 𝒍𝒆𝒏𝒈𝒕𝒉, 𝒕𝒊𝒎𝒆 𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍, , Total path length = OP+PQ+QO, = 1km+ 1.57km+1km= 3.57 km, , 4.10 PROJECTILE MOTION, An object that is in flight after being thrown or projected is called, a projectile. The path or trajectory of a projectile is parabola

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𝒗𝒚, , 𝒗𝒚 =0, , 𝑣0𝑥, , 𝒗𝟎𝒙, 𝒂𝒚 = −𝒈, , 𝒗𝒐𝒙, , 𝒗𝒚, , Consider a projectile projected with a velocity V0 making an, angle 𝜽 with x axis., Since there is no force along x direction, ax = 0, The components of initial velocity V0 are;, V0X = v0 cos 𝜽, V0y = v0 sin 𝜽, Since ax = 0, the horizontal component of velocity remains constant, But the acceleration due to gravity acts on the object which is, vertically downward, ay = -g, Since acceleration is downwards the vertical component of, velocity decreases and horizontal component remains constant, when object moves upwards. At the highest point vertical, component of velocity become zero and the horizontal component, remains same. When the object moves down wards the vertical, component of velocity increases, Therefor a projectile motion can be considered as a uniform, horizontal motion and an accelerated vertical motion, PROBLEM 2, , ANSWER:, , 15 m𝒔−𝟏

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Speed or magnitude, of velocity, , Trajectory of a projectile is Parabolic:-, , Substituting for t, , This equation represents the equation of a parabola

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Therefor the trajectory of a projectile is a parabola, PROBLEM 3, , ANSWER:, , PROBLEM 4, , ANSWER, , (i) y= -300m , v0y = 0, v0x = 60 m/s , a= -9.8 ms-2,t=?, Considering the vertical motion, 𝟐𝒚

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TIME OF FLIGHT (T), Time required to reach the same horizontal level as that of, projection ( vertical displacement ,y =0) is called time of flight (T), or it is the total time during which the projectile is in flight., y, , Vertical displacement y = 0 at t=T, , Time of flight,, Note : Time required to reach maximum height is, , HORIZONTAL RANGE (R), Horizontal distance travelled by the projectile from its initial, position (x=y=0) to position where it passes y = 0 is called, horizontal range. Then the time taken is time of flight, T., , SREESHYJU K P & JIJESH M T ,HSST PHYSICS GVHSS PAYYOLI

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Horizontal range ,, Note: Maximum horizontal range, , We have, Range is maximum when, , Angle of projection for maximum range is, Maximum range ,, QUESTION:, Show that for a given velocity of projection range will be same for, angles 𝜽 and ( 90-𝜽 ), , = 𝑹𝟏, Thus for a given velocity of projection range will be same for, angles 𝜽 and ( 90-𝜽 ), SREESHYJU K P & JIJESH M T ,HSST PHYSICS GVHSS PAYYOLI, , PROBLEM:1

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Equations of motion from Newton’s second law of motion, , SREESHYJU K P & JIJESH M T ,HSST PHYSICS GVHSS PAYYOLI, , Definition of unit of force (Newton), , NOTES:1. From Newtons second law F= ma, If F= 0 , ma = 0 ( since m ≠ 0), a=0 ie body is at rest or moves with constant velocity, which is, Newtons first law, Second Law is cosistant with the first Law

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Eg: For projectile motion, the gravitational force acts along, vertical direction. Therefor the verical component of velocity ( v0, sin𝜽) changes and horizontal component (v0 cos 𝜽 ) remains same, , PROBLEM 1, , ANSWER:

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PROBLEM 2, , by, , The motion, ANSWER:, , V= U, , F= mg, IMPULSE, The product of force and time is called impulse

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Unit of impulse is kgms-1, , PROBLEM 3, , ANSWER:, , 5.7 LAW OF CONSERVATION OF MOMENTUM, , APPLICATIONS:, The backward movement of the gun when a shot is fired from a, gun is called recoil of gun, Velocity with which the bullet moves is called muzzle velocity, SREESHYJU K P & JIJESH M T ,HSST PHYSICS GVHSS PAYYOLI

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COLLISION OF TWO BODIES, collision

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FRICTION, Friction is the resistive force that opposes the relative motion, between two surfaces in contact, Friction are of two types, 1. static friction, Sliding friction, 2. Kinetic friction, Rolling friction, STATIC FRICTION ( fs), , SREESHYJU K P & JIJESH M T ,HSST PHYSICS GVHSS PAYYOLI

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Laws of Kinetic Friction:, , (1), , (2), , (3), Graphical representation of friction with applied force:, , Rolling friction and sliding friction, The frictional force between two surfaces when one body rolls, over the surface of another is called sliding friction eg: when a, wooden block is moved on a level floor , the opposing force is, sliding friction, When one body rolls over the surface of another body ,the, opposing force is rolling friction, Eg: a ball rolling over a surface, the opposing force is rolling, friction, Rolling friction is less than sliding friction, Boll bearings are used to reduce friction, , SREESHYJU K P & JIJESH M T ,HSST PHYSICS GVHSS PAYYOLI

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Angle of repose: Angle of repose is the angle of inclination of an, inclined surface over which a body just begins to slide down, , ie , tangent of angle of repose is coefficient of static friction, , When body moves down an inclined plane with an acceleration a ,, then frictional force is kinetic friction, , SREESHYJU K P & JIJESH M T ,HSST PHYSICS GVHSS PAYYOLI

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Thus for a given value of 𝝁𝒔 and r the maximum speed of, circular motion of car is given by, , CIRCULAR MOTION OF A VEHICLE ON A BANKED ROAD, In banking of roads the outer edge of the road is raised a little, more than the inner edge. Here 𝜽 is the angle of banking., , From fig

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The maximum possible speed without skidding is given by, , If friction is absent, 𝝁𝒔 = 0, (Optimum speed ), Thus maximum speed can be increased by increasing the angle of, banking without considering friction which will cause little wear, and tear of tires, SREESHYJU K P & JIJESH M T ,HSST PHYSICS GVHSS PAYYOLI

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Work can be zero, positive or negative, (1) work can be zero, (i) if displacement is zero, eg: a weight lifter holding a weight on his shoulder for some, seconds does not do any work, (2) if force is zero, eg: for a block moving on a smooth horizontal table, work done is, zero, (3) If force and displacement are mutually perpendicular, eg: for a man carrying a load on his head and moving, horizontally, the work done by gravity is zero, (2) work can be positive if angle 𝜽 between force and, displacement is between 00 and 900, eg: a man pushing a block by applying force and moving it in the, forward direction, (3) work can be negative if angle 𝜽 between force and, displacement is between 900 and 1800, eg: when a body is moved over a rough surface , the work done by, friction is negative,  SI unit of work is joule, ,  Dimension of work/energy is [ ML2T-2], PROBLEM 1:

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ANSWER:, , ENERGY, Energy is the ability to do work., (1) Kinetic energy (2) Potential Energy, 6.4 KINETIC ENERGY, Energy due to motion is known as kinetic energy, For body of mass m moving with velocity v ,, 𝟏, , kinetic energy , K = m𝒗𝟐, 𝟐, ,  Kinetic energy is a scalar quantity, SREESHYJU K P & JIJESH M T ,HSST PHYSICS GVHSS PAYYOLI, , Expression for Kinetic energy

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PROBLEM 2, , ANSWER:, , If vf is the emergent speed of the bullet,, , ANSWER:

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Work done = Force x displacement= change in kinetic, energy, Both bus and the car had same kinetic energy and hence, same amount of work is needed to be done. As retarding, force applied is same for both, both the bus and the car, travel same distance before coming to rest, 6.7 POTENTIAL ENERGY, Potential energy is the stored energy by virtue of the position or, configuration of the body, eg: (1)A body at a height h above the ground possesses potential, energy due to its position, (2) A compressed spring possesses potential energy due to its, state of strain, * dimension of potential energy is [ M L2T-2] and unit is Joule, Expression for potential Energy;, We have, , or – dV= F(x) dx, , 𝒗𝒇, , −, , 𝒗𝒇, , 𝒅𝑽 = −[𝑽]𝒗𝒊, , 𝒗𝒊, 𝒗, , [𝑽]𝒗𝒊 = 𝒗𝒊 − 𝒗𝒇, 𝒇, Potential energy of a body in a Gravitational field:

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CONSERVATIVE AND NON CONSERVATIVE FORCES, , Eg: spring force ,gravitational force etc, Properties:, (1) The work done by a conservative force depends only upon, initial and final positions of the body, (2) The work done by a conservative force around a closed path is, zero, Non conservative force, The work done by a non conservative force depends upon the, path of the displacement of the body, Eg: Frictional force, viscous force, SREESHYJU K P & JIJESH M T ,HSST PHYSICS GVHSS PAYYOLI

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ANSWER:, (a)Force displacement Graph is a straight line, , (b) Force is directly proportional to displacement, ( c) Area under graph gives work done, 𝟏, , Area of ∆𝑶𝑨𝑩 = x OA x AB, 𝟐, , 𝟏, , = x5 x 10 = 25 J, 𝟐, , 6.8 CONSERVATION OF MECHANICAL ENERGY, Mechanical Energy is sum of potential energy and kinetic energy., Let a body undergoes displacement ∆𝒙 under the action of, conservative force F. Then,, By work energy theorem ,kinetic energy∆𝑲 = F(x)∆𝒙, Also potential energy function, -∆𝑽 = F(x)∆𝒙, ∆𝑲 = −∆𝑽, ∆𝑲 + ∆𝑽 =0, ∆(𝑲 + 𝑽) = 0, Which means that sum of potential energy and kinetic energy is, constant or total mechanical energy is constant, The principle of conservation of mechanical energy can be stated, as ‘ The total mechanical energy of a system is conserved if the, forces ,doing work on it ,are conservative’, SREESHYJU K P & JIJESH M T ,HSST PHYSICS GVHSS PAYYOLI

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PROOF OF CONSERVATION OF ENERGY IN THE CASE OF, A FREELY FALLING BODY, Consider a body of mass m falling freely from a height h, , …(1), , ….(2), , …..(3), From equations (1) , (2) and (3), , PROBLEM 1, , SREESHYJU K P & JIJESH M T ,HSST PHYSICS GVHSS PAYYOLI

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ANSWER:, , 6.11 POWER, Power is defined as the time rate at which the work is done, P=, , 𝑾, 𝒕, , Average power:it is the ratio of work done to total time, Pav =, , 𝑾, 𝒕, , Instantaneous power:, , Power is a scalar quantity. Dimension of power is, [ ML2T-3].SI unit of power is watt, Another unit of power is horse power(hp), 1hp= 746 W, Unit of electrical energy is Kilowatt hour(kwh), 1 kwh= 3.6 x 106J, SREESHYJU K P & JIJESH M T ,HSST PHYSICS GVHSS PAYYOLI

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Answer;, , CHAPTER 7: SYSTEM OF PARTICLES & ROTATIONAL MOTION, , 7.6 ANGULAR VELOCITY AND ITS RELATION WITH, LINEAR VELOCITY, Consider a rigid body rotating about a fixed axes (z axes). The, velocity of the particle v changes with its distance from the axes., But the angular velocity of all the particles of the rigid body, remains constant at an instant., , The relation between linear velocity and angular velocity is given, by, ⃗ =𝝎, ⃗⃗⃗ × 𝒓, ⃗, 𝒗, The direction is perpendicular to both 𝝎 and r and is directed, along the tangent to the circle described by the particle., , ANGULAR ACCELERATION

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Relation between moment of inertia and angular momentum:, , The direction of angular momentum and angular velocity will be, same, Relation between Torque and angular Momentum, , Thus rate of change of angular momentum of the particle is equal, to torque acting on it., This is rotational analogue of the equation, , TORQUE AND ANGULAR MOMENTUM OF A SYSTEM OF, PARTICLES, If 𝝉𝟏 , 𝝉𝟐 , 𝝉𝟑 ,……., 𝝉𝒏 , are the torque acting on each particle , then, the total toque on the system of particle is, 𝝉= 𝝉𝟏 +𝝉𝟐 + 𝝉𝟑 +…..+ 𝝉𝒏, 𝝉=∑ 𝝉𝒊, Similarly angular momentum, SREESHYJU K P & JIJESH M T ,HSST PHYSICS GVHSS PAYYOLI

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Also total torque, , By Newton’s third Law, , 7.9 MOMENT OF INERTIA OF A RIGID BODY, , SREESHYJU K P & JIJESH M T ,HSST PHYSICS GVHSS PAYYOLI

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⃗⃗⃗ = constant, Thus if external torque is zero , I𝝎, ie if moment of inertia increases angular velocity decreases, MOMENT OF INERTIA OF RIGID BODIES, , 7.10 THEOREM OF MOMENT OF INERTIA, (1) PERPENDICULAR AXES THEOREM, , SREESHYJU K P & JIJESH M T ,HSST PHYSICS GVHSS PAYYOLI

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𝐼𝒕𝒂𝒏𝒈𝒆𝒏𝒕, (2) MOMENT OF INERTIA OF A RING ABOUT THE EDGE, OF THE CIRCLE OF THE RING, , RADIUS OF GYRATION:

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I= Mk2, Eg: in the case of moment of inertia of ring about a diameter, 𝑴𝑹𝟐, 𝑰𝒙 =, 𝟐, k2=, , 𝑹𝟐, 𝟐, , thus k=, , 𝑹, √𝟐, , Note : In the case of a ring and a disc of same mass and diameter ,, moment of inertia will be large for ring because for ring mass is, distributed at maximum distance from the axis of rotation, CHAPTER 8: GRAVITATION, 8.3 NEWTONS(UNIVERSAL) LAW OF GRAVITATION, , SREESHYJU K P & JIJESH M T ,HSST PHYSICS GVHSS PAYYOLI

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Note1:, , Note2 :, , 8.5 ACCELERATION DUE TO GRAVITY OF EARTH (g), Consider a body of mass m on the surface of earth of mass ME, m, and radius RE, 𝐑𝐄, , …….(1), From (1) and (2), , 𝐌𝐄, …..(2), SREESHYJU K P & JIJESH M T ,HSST PHYSICS GVHSS PAYYOLI

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Note: Acceleration due to gravity does not depends on the mass of, the body (m), MASS OF EARTH, , 8.6 ACCELERATION DUE TO GAVITY BELOW & ABOVE, THE SURFACE OF EARTH, , (a), , Comparing with g on the surface of earth, , Thus acceleration due to gravity decreases as we go above the, surface of earth, SREESHYJU K P & JIJESH M T ,HSST PHYSICS GVHSS PAYYOLI

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Thus acceleration due to gravity decreases as we go below the, surface of earth, Note : 1, The value of acceleration due to gravity is maximum on the surface, of earth, Note ;2, At the center of earth d = RE, thus g = 0, , ANSWER:, , SREESHYJU K P & JIJESH M T ,HSST PHYSICS GVHSS PAYYOLI