The average speed of an object is the total distance travel divided by the total time requires to cover that distance.

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1 AS Physics 9702 unit 2: Motion, Force and Energy 1 UNIT 2 MOTION, FORCE AND ENERGY This unit includes topics 3, 4, 5 and 6 from the CIE syllabus for AS course. Kinematics: Kinematics is a branch of mechanics which describes the motion of objects without the consideration of its masses or forces that bring that motion. Dynamics is concerned with the forces that produce any motion to an object of mass m. Together, kinematics and dynamics form the branch of physics known as mechanics. Linear and non-linear motion: Linear motion is the change in position of an object moving in a straight line and non-linear motion is change in position of an object without following the straight path for example in circular motion. Displacement and distance: 1 The displacement is a vector quantity that points from an object s initial position to its final position in straight line. It has a magnitude that equals the shortest distance between the two positions. The SI unit of displacement is meters (m). The symbols use to represent displacement are s or x to represent displacement along x-axis. In case of free falling or vertical motion the displacement can be represented by the vertical distance covered and symbol use is h or y. The scalar form of displacement is distance. It refers to how much ground an object has covered during its motion. The SI unit of distance is meters (m) and the symbol use to represent distance is d. Average speed and average velocity ( ): The average speed of an object is the total distance travel divided by the total time requires to cover that distance. The SI unit of average speed is meters per seconds or ms -1. It is a scalar quantity. The average velocity is the total displacement of an object from point A to point B divided by the total time taken. The SI unit of average velocity is also meters per seconds or ms -1. It is a vector quantity. The instantaneous velocity v of an object indicates how fast the object is moving and what is its direction at any particular instant of time when it is moving with non-uniform velocity. This can be found by finding the gradient of the tangent to the curved part of distance-time graph. 1 For further explanation see

2 Prepared by Faisal Jaffer, revised on Nov 2011 Acceleration: Acceleration is a vector quantity which is defined as the rate at which an object changes its velocity. Where v is the change in velocity and t is time elapsed. Also the u and v are the initial and final velocities of the moving object. The SI unit of acceleration is meters per second square or ms -2. Note: Whenever the acceleration and velocity are in same direction then the acceleration is positive and when acceleration and velocity are in opposite direction, it means the body is slowing down and the acceleration is negative. It is also called retardation or deceleration. Graphical representation of velocity, acceleration and displacement: Velocity time graphs: In the following graphs the motion of an object is considered as positive if it is moving along the x-axis towards right or y-axis in upwards direction. (A) moving with constant acceleration in positive direction (B) moving with constant deceleration in positive direction (C) moving with constant deceleration along negative direction (D) object is stationary (E) moving with constant velocity in negative direction (F) moving with constant velocity in positive direction (G) object is changing direction (H) object is momentarily stop at two occasions during its whole motion (I) object is momentarily stop only at one occasion during its whole motion

3 AS Physics 9702 unit 2: Motion, Force and Energy 3 Displacement time graph: (A) moving away from the mean position with constant velocity (B) moving towards the mean position with constant velocity (C) moving towards the mean position from negative displacement with constant velocity (D) object is stationary (E) object is stationary (F) object is stationary Important!!!! The area under the velocity time graph measures total displacement. The slope or gradient of a velocity time graph represents the acceleration of the body. The slope or gradient of a displacement time graphs represents the velocity of the body. If the object is moving with non-uniform acceleration then the acceleration at any particular time t can be found by finding the gradient of the tangent to the curve at time say t=4s as shown in the figure. Derivation of equations of uniformly accelerated motion: From the definition of acceleration, it is the rate of change of velocity. Consider a body moving with initial velocity u and after time t the velocity becomes v. the equation can be expressed as v u a t by arranging the above equation we can get from the definition of average velocity that is sum of initial velocity and final velocity divided by two and by equating two equations we get

4 Prepared by Faisal Jaffer, revised on Nov 2011 This equation can be derived by using two equations and. Replace the value of v from the first equation into the second equation. The equation can be derived by using equation get. By squaring both sides of the equation we by opening the bracket by taking common term 2a since, replace it in previous expression If an object is falling freely under the influence of gravity then the only force applied on an object is weight in downward direction and acceleration is g = 9.81ms -2. The above equations of uniformly accelerated motion can be rewritten as: Application of the equations of motion: The equations of motion can be used for any moving object as long as the acceleration of the object is constant. However it helps to use following guidelines when solving problems: a) Make a drawing, if require, to represent the situation being studied. b) First decide at the start which directions are to be called positive (+) and negative (-) relative to a conveniently choose coordinate of origin. It is a common understanding that for the direction of motion we use Cartesian coordinates, that is moving upward or right we consider positive direction and moving downward or left we consider negative direction. c) In an organized way write down the values of s, v, u, a, and t with plus or minus signs and appropriate units. This is called data. d) Before attempting to solve a problem, verify that given information contains values for at least three of the five variables. Once the three known and two unknown variables are identified then select the appropriate equation. e) If there are two segments of the motion then make sure that the final velocity of first segment is the initial velocity of second segment.

5 AS Physics 9702 unit 2: Motion, Force and Energy 5 f) Keep in mind that there may be two possible answers to visualize the different physical situation. Exercise: 2.1 A) Solve the following questions. Q1) A racing car has an initial velocity of 100 m/s and it covers a displacement of 725 m in 10 s. Find its acceleration. (Ans: -5.5 ms -2 ) Q2) An object starts from rest, moves in a straight line with a constant acceleration and covers a distance of 64 m in 4 s. Calculate a) its acceleration (ans: 8ms -2 ) b) its final velocity (ans: 32ms -1 ) c) at what time the object had covered half the total distance (ans: 2.8s) d) what distance the object had covered in half the total time. (ans: 16m) Q3) A ball is thrown vertically upwards with a velocity of 10 m/s from the balcony of a tall building. The balcony is 15 m above the ground and gravitational acceleration is 10 m/s 2. Find a) the time required for the ball to hit the ground (ans: 3s), and b) the velocity with which it hits the ground (ans: -20ms -1 ). B) Solve assignment 5 on the website: C) Solve following questions from past papers May/June 2010, Paper 22, question 2 May/June 2008, Paper 2, question 3 Oct/Nov 2008, Paper 2, question 2 Oct/Nov 2007, Paper 2, question 2 Oct/Nov 2007, Paper 1, question 12

6 Prepared by Faisal Jaffer, revised on Nov 2011 Friction forces: What are friction forces? The friction force is the force exerted by a surface as an object moves across it or makes an effort to move across it. The friction force opposes the motion of the object. For example, if a block of mass m moves across the surface of a desk, the desk exerts a friction force in the direction opposite to the motion of the block. As such, friction depends upon the nature of the two surfaces and upon the degree to which they are pressed together. There are two types of frictional forces: a) Static or limiting frictional force f s : This is the maximum force applied by the surface of the desk on the block until just before the block starts to move. This is the force that needs to be overcome by force F in order for block to move. Mathematically we can write the equation as f s = µ s N b) Kinetic or dynamic frictional force f k : This is the frictional force which exists between the two adjacent surfaces which are in relative motion. This is usually slightly less than the static friction. Mathematically f k = µ k N N N f s m F f k m F F g = w = mg F g = w = mg N is the normal reaction force of surface on mass m. μ s and μ k are the coefficient of static and kinetic friction respectively. Both depend on the nature and the condition of the surfaces which are in contact but are independent of the area of contact. For example steel on steel µ s 0.8 and for Teflon on Teflon µ s If two surfaces are assumed to be perfectly smooth, if there is no frictional force and µ s = µ k = 0. For two surfaces which have relative motion the kinetic frictional force is directly proportional to the motion of an object. Exercise: 2.2 Q) A box that weighs 10.0 N is being dragged with constant velocity along a horizontal surface of the table by a rope that is at an angle 45 o with that surface. The tension in the rope is 5.0 N. What is the coefficient of friction? (ans: Q) Solve assignment 3b on the website:

7 AS Physics 9702 unit 2: Motion, Force and Energy 7 Motion of a body in uniform gravitational field without the resistive force (air resistance or viscous drag): When an object is falling only under the influence of gravitational force then it is said to be free falling. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. This force of gravity causes acceleration called acceleration due to gravity and is expressed by letter `g`. The average numerical value for the acceleration due to force of gravity is 9.81ms -2. There are slight variations in this numerical value (to the second decimal place) that are dependent on the altitude of the point under consideration. Motion of body in uniform gravitational field with air resistance or viscous drag: What is air resistance? 2 Air resistance is a special type of frictional force which acts upon objects as they travel through the air in uniform gravitational field. Like all frictional forces, the force of air resistance always opposes the motion of the object. It is most noticeable for objects which travel at high speeds (e.g., a skydiver or a downhill skier) or for objects with large surface areas. What is viscosity of fluid and viscous drag? Any fluid is said to be viscous if it offers a resistance to the motion of any solid body passes through it. Viscous effects are due to the frictional force which exists between two adjacent layers of fluid which are in relative motion. For viscous drag consider a sphere of radius r moving with velocity v through a fluid, it experiences a drag force F which acts in opposite direction to that in which the sphere is moving. The expression for viscous force can be stated as (no derivation of this expression requires in this course) This expression is called Stokes Law. The coefficient of viscosity of a fluid (liquid or gas) is the measure of the degree to which the fluid exhibits viscous effects. Viscosity is the property of the fluid that measures its resistance to flowing. For example honey is more viscous then water. The higher the coefficient of viscosity, the more viscous the fluid is. In liquids, viscosity decreases with the increase of temperature, whereas in gases it increases with the increase of temperature. Its unit is N.m -2 s. Terminal velocity: Consider a sphere falling from rest through a viscous fluid (air or liquid). The forces acting on the sphere are its weight w acting downward, the upthrust U due to displaced fluid 3, and the viscous force R or air resistance, both acting upwards. Initially the downward force w is greater than the upward forces U+R, and the sphere accelerates downwards. As the velocity of the sphere increases so does the viscous force (air resistance) R, and eventually U+R becomes equal to w the weight. Since there is now no net force acting on it, its velocity has a constant maximum value known as terminal velocity. In the beginning w >> U+R and as the body moves down w >U+R and eventually at terminal velocity w=u+r. 2 For further explanation 3 It is the force that liquid exerts on the sphere. It is equal to the weight of the liquid displaced by the sphere when immersed. w = mg

8 Prepared by Faisal Jaffer, revised on Nov 2011 Exercise: 2.3 Solve the following questions from past papers. Oct/Nov 2009, Paper 21, question 2 May/June 2002, Paper 1, question 6 Solve assignment no 6 on the website: Projectile motion: A projectile is any object which once projected, dropped or thrown continues its constant motion along the x-direction (horizontal) and accelerated motion along y-direction (vertical) due to the force of gravity. Projectiles travel with a parabolic trajectory due to the influence of gravity. There are no horizontal forces acting upon projectiles and thus no horizontal acceleration. The horizontal velocity of a projectile is constant. There is a vertical acceleration caused by gravity; its value is 9.81 ms -2, downwards. The vertical velocity of a projectile changes by 9.81 m/s each second. The horizontal motion of a projectile is independent of its vertical motion. If there was any other force acting upon an object, then that object would not be a projectile. Examples of projectile, in the absence of air resistance are: 1. An object dropped from rest is a projectile. 2. An object which is thrown vertically upward is also a projectile. 3. An object which is thrown upward at an angle to the horizontal is also a projectile. F v x v x v y A body projected with a velocity v at an angle above the horizontal has horizontal (v x ) and vertical (v y ) component of velocity y x v y v x v y = v sin and v x = v cos and horizontal (x) and vertical (y) displacements are v y and because vertical initial velocity is zero

9 AS Physics 9702 unit 2: Motion, Force and Energy 9 In similar way to find the resultant velocity v from its components we use Pythagoras theorem v 2 v x v y 2 and direction is If a projectile is projected at an angle θ with the horizontal then the maximum range of projectile can be derived from expressions and Vertical height, m v Replacing y=0 in the first equation and replacing the value of t in the second equation we get the expression. θ y R, Horizontal Range, m From the above expression we can see that the value of R will be maximum when sin 2 =1 or =45 o (because sin 90 o =1). The projectile can have a maximum horizontal range if an object is thrown at angle of 45 o. Exercise: 2.4 Q1) A soccer ball is kicked horizontally off a 22.0-metre high hill and lands a distance of 35.0 metres from the edge of the hill. Determine the initial horizontal velocity of the soccer ball. ans: 16.5ms -1 Q2) A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28 o above the horizontal. Determine a) the time of flight, b) the horizontal distance, and c) the peak height of the long-jumper. Ans (a) 1.1s, (b) 12.2 m, (c) 1.6 m Solve assignment no 7 on the website: Solve following questions from the past papers Oct/Nov 2009, Paper 22, question 3(a) (b) May/June 2011, Paper 12, question 14 May/June 2011, Paper 22, question 1(c) May/June 2009, Paper 1, question 8 Oct/Nov 2010, Paper 21, question 2 Oct/Nov 2010, Paper 22, question 3(b)

10 Prepared by Faisal Jaffer, revised on Nov 2011 Dynamics: Dynamics is the study of motion of an object with respect to forces that produce the motion. Force is defined as the one that changes body s state of rest or of uniform motion in straight line. Another definition of force is that it causes a body to accelerate. The unit of force is newton (N). 1 newton is the force required to give a mass of 1 kg an acceleration of 1 ms -2. Issac Newton, one of the greatest scientists ( ) presented the most powerful three laws of motion. Newton s three laws of motion: First Law: A body continues its state of rest or of uniform motion in a straight line unless some external force is applied to change that state. Newton s first law is another way of saying that all matter has a built-in opposition to being moved if it is at rest or, if it is moving, to having its motion changed. This property of matter is called inertia (from Latin word laziness ). Therefore first law of motion is also called law of inertia. Second law: Whenever a net force acts on a body, it produces acceleration in the direction of the net force that is the acceleration is directly proportional to the net force and inversely proportional to the mass of the body. a F and 1 a m Combining both we get F a m or F a k m where k is the constant of proportionality. By the definition of unit of force, newton, k is defined as: 1 N = k x 1 kg x 1 ms -2 thus k = 1 The newton (N) is defined as the force that produces an acceleration of 1 m/s 2 when applied on a body of mass 1 kg. Third law: To every action there is always an equal and opposite reaction force of two bodies acting upon each other. They are always equal and opposite in direction. Consider a book of weight w is placed on the table. The book exerts a force equal to w on the table and table exert equal and opposite reaction force N on the book which is always perpendicular to the surface of the table. N Book w Other examples are: 1. Earth exerts a gravitational force of attraction on Moon; the Moon exerts a force of the same size on the Earth. 2. A rocket moves forward as a result of the push exerted on it by the exhaust gases which the rocket has pushed out. 3. When a man jumps off the ground it is because he a pushed down on the Earth and the Earth has pushed up on him.

11 AS Physics 9702 unit 2: Motion, Force and Energy 11 Exercise: 2.5 Solve the following questions from past papers: May/June 2011, Paper 12, question 10 May/June 2011, Paper 22, question 2 May/June 2010, Paper 1, question 14 Oct/Nov 2008, Paper 1, question 11 Oct/Nov 2010, Paper 12, question 11 Mass and weight: The following table clarifies the difference between mass and weight: Mass The mass is the intrinsic property of matter and does not change as an object moved from one location to another. The mass of a body is measure of its resistance of acceleration that is it is measure of the inertia of the body. Weight The weight of the body is the force acting on its mass due to the gravitational attraction of earth in earth s gravitational field. Weight can change if the body is moved relative to the centre of earth. Weight of the body is where w is the weight of the body, m is the mass and is the acceleration due to gravity or acceleration of free fall and its value is 9.81 ms -2. SI unit of mass is kilogram (kg) which is the base unit. SI unit of weight is newton which is derived unit by multiplication of mass and acceleration. Centre of gravity (or mass): A body behaves as if its whole mass is concentrated at one point, called its centre of mass or centre of gravity, even though the Earth attracts every part of it. 1. The centre of mass of a body is that point at which the mass of the body may thought to be concentrated. 2. If suspended, a body will come to rest with its centre of mass directly below the point of suspension. 3. The centre of mass of a symmetrical body is along the axis of symmetry. 4. If a body is not turning, the total clockwise moment must be exactly balanced by the total anticlockwise moment about its centre of gravity. Stability of an object: Stability is the extent to which an object resists toppling over. Stable objects do not topple over easily. If the line of action of the weight of a body lies outside the base of the body there will be a resultant moment and the body will tend to topple. When designing vehicles, engineers try to design so that the centre of mass is as low as possible. This makes vehicles less likely to turn over when going round corners. For an object to start rotating it needs to have an unbalanced moment acting on it as illustrated in the figure.

12 Prepared by Faisal Jaffer, revised on Nov 2011 Determining the centre of gravity of irregular shape cardboard (lamina) The center of gravity of an irregular shaped cardboard can be found experimentally by suspending the cardboard from different points, with a plum-line (a mass hanging on a thread) and marking the points along the line of the thread. The point of intersection of two lines is the centre of gravity of the cardboard. A doughnut s CM is in the center of the hole. Exercise: 2.6 Solve the following questions from past papers. May/June 2002, Paper 2, question 3(a) Oct/Nov 2005, Paper 2, question 2

13 AS Physics 9702 unit 2: Motion, Force and Energy 13 Linear momentum: Definition of linear momentum: Momentum is defined as the product of mass and velocity of a body that is moving in straight line. It is a vector quantity and express by unit kg ms -1 or N s. p = m v Where p (small case) is the momentum, m is the mass of the body and v is the velocity. The direction of the momentum is same as the direction of the velocity. The conservation of linear momentum: m A u A u B m B F A m A m B F B v A m A m B v B When two bodies, A and B of mass m A and m B, are involved in a collision, the body A exert force F B on B and body B exerts force F A on body A, then the changes in the momentum, are oppositely directed and the total change in momentum is zero. Hence the principal of conservation of linear momentum can be stated as The total linear momentum of a system of interacting (eg colliding) bodies remains constant if no external forces is applied. From above diagram if the body A is moving towards right with the initial velocity u A and body B is moving towards left with initial velocity -u B then Initial momentum of body A before collision p ia = m A u A Initial momentum of body B before collision p ib = m B (- u B ) and Total initial momentum before collision p i = m A u A - m B u B now after collision body A start moving with final velocity -v A towards left and B towards right with final velocity v B Final momentum of body A after collision p fa = m A (-v A ) Final momentum of body B after collision Total final momentum after collision Principle of conservation of linear momentum is p fb = m B v B p f = -m A v A + m B v B Total initial momentum before collision = Total final momentum after collision p i = p f

14 Prepared by Faisal Jaffer, revised on Nov 2011 Elastic and inelastic collision: When two bodies collide, their total momentum is conserved unless there are external forces acting on them. The total kinetic energy E k, however, usually decreases, since the collision converts some of its kinetic energy to heat and/or sound. The collision in which some kinetic energy lost is known as an inelastic collision. A collision is elastic if there is no loss of kinetic energy. Hence Inelastic collision: Total E k before collision > Total E k after collision For elastic collision Total E k before collision == Total E k after collision Conservation of velocities in elastic momentum: Divide equation 2 by 1 The above equation must satisfied for two bodies to collide elastically provided the object A and B are moving towards each other and after collision they move in opposite direction. The sign convention must be change if the bodies are moving in direction other than the one used in the derivation of the equation. Exercise: 2.7 Solve the following question from past papers May/June 2010, Paper 21, question 3(b) Oct/Nov 2009, Paper 21, question 3 May/June 2009, Paper 21, question 2 May/June 2009, Paper 22, question 2(c) (d) Oct/Nov 2008, Paper 1, question 10 May/June 2008, Paper 1, question 10, 11 May/June 2007, Paper 1, question 10, 11, 12 Solve assignment no 8 on the website:

15 AS Physics 9702 unit 2: Motion, Force and Energy 15 Examples of linear momentum: A car and a truck are approaching each other and about to collide Two trolleys of different masses travelling in the same direction one with 1ms -1 and other with 4ms -1 are about to collide A ball travelling with 20 ms -1 hit by tennis racket and bounce back with the velocity 30ms -1 An object of mass 3m explodes Two trolleys connected to each other by spring are about to move in opposite direction when release. A cannon ball is fired from a cannon. Ball moves in one directon and cannon move in opposite direction. Cases of linear momentum: Case no 1 Case no 2 Case no 3 Case no 4 Case no 5 Initial conditions before collision m A = m B = m and collision is elastic u A = initial velocity of body A u B = initial velocity of body B m A m B and collision is elastic u A = initial velocity of body A u B = 0 m A = m B = m and collision is elastic u A = initial velocity of body A u B = 0 mass of body A is much smaller than B m A << m B and collision is elastic u A = initial velocity of body A u B = 0 m A >> m B and collision is elastic u A = initial velocity of body A u B = 0 Results after collision velocities of the bodies will interchange v A =u B and v B =u A ( ) ( ) body A will be stationary and body B will be start moving with same velocity as initial velocity of body A and body A will bounce back with same velocity in opposite direction and body B will remain stationary and body A will continue with the same velocity and body B will start moving with twice the velocity if body A and

16 Prepared by Faisal Jaffer, revised on Nov 2011 Momentum in terms of Newton s second law: If the steady force F acts on a body of mass m and increases its velocity from u to v in time t, the acceleration a is given by Substituting the value of a ( v u) a and force applied is given by F ma t m( v u) F or t mv and mu are final and initial momentum then mv mu F t mv mu is the change in momentum and mv mu t is the rate of change of momentum. From equation it is clear that the rate of change of momentum is equal to net force F applied which is another form of Newton s second law of motion Impulse: The impulse of a constant force F acting for a time t is defined as the force applied on a body for small fraction of time t. or from equation or or Hence the impulse is the rate of change of momentum of a body. Examples of impulse of force are the batsman hitting the cricket ball or golfer striking the golf ball. In both situations the concept of impulse is used where a large variable force is acting for only a short period of time. Since the impulse of force is the product of force and time of collision, therefore if the force applied for longer time then it increases the impulse and that will also result in gain of momentum. DENSITY: The density of a substance is defined as m V where is the density of substance in kg/m 3, m is the mass in kg and V is the volume in m 3. The relative density of a substance is define as Relativedensity Density of substance Density of water at Relative density has no units. Densities of some common substances are given in the table. Exercise: 2.8 Solve the following questions from the past papers. May/June 2007, Paper 2, question 3 May/June 2007, Paper 1, question 15 May/June 2010, Paper 1, question 17 Oct/Nov 2007, Paper 1, question 17 Oct/Nov 2006, Paper 1, question 19 0 o C Densities in (g cm -3 ) Water at 4 o C 1.0 Mercury 13.6 Aluminium 2.7 Gold 19.3 Ice at 0 o C 0.92 Air CO Hydrogen Helium

17 AS Physics 9702 unit 2: Motion, Force and Energy 17 Pressure: The pressure is defined as the force per unit area acting at right angles to the surface. where p is the pressure in Pa (pascal), F is the force in N and A is the area in m 2 Atmospheric pressure is equal to 1 atm = 101 x 10 5 Nm -2 or Pa = 760 mm Hg = 1.01 bar Pressure in fluids: Consider a cylindrical object of cross-sectional area A and height h is immersed in a fluid of density. The top of the cylinder is at the surface of the fluid and vertical forces acting on it are its weight w and force pa. If the cylinder is in equilibrium then pa = w = m g = V g because m = Vρ =h A g because V = h A or p A = h A g A cancels both sides or p = h g a) The pressure in a fluid increases with depth. All points at the same depth in the fluid are at the same pressure. b) Any surface in a fluid F experiences a force due to the pressure of the fluid. This force is perpendicular to the surface no matter what the orientation of the surface. The magnitude of the force is independent of the orientation of the surface. Pressure acts equally in all direction. c) The above equation is not valid for gas and when h is very high. The density of gas decreases with height and above equation is derived on the assumption that density of fluid (gas or liquid) is constant. d) Mercury barometer is used to measure the atmospheric pressure and manometer is used to measure gas pressure. F F F h A w pa Exercise: 2.9 Solve the following questions from past papers. May/June 2010, Paper 12, question 18 May/June 2009, Paper 1, question 11 May/June 2009, Paper 1, question 18 Oct/Nov 2006, Paper 1, question 21 May/June 2006, Paper 2, question 4 May/June 2008, Paper 2, question 4 (b) (c)

18 Prepared by Faisal Jaffer, revised on Nov 2011 Archimedes principle (upthrust): A body immersed in a fluid (totally or partially) experiences an upthrust (upward force) which is equal to the weight of the fluid displaced by the body. Consider a cylinder of height h and cross-sectional area A, a distance h o below the surface of the fluid of density. According to the Archimedes Principle: h o p x A volume of fluid displaced = Volume of cylinder pressure p x V = Ah mass of fluid = Ah weight of fluid = Ah g [ = m/v] [W = m/g] h A The fluid exerts forces p x A and p y A on the top and bottom faces of the cylinder. The upthrust that is the resultant force due to the fluid is therefore given by: upthrust = p y A - p x A = (h + h o ) ga - h o ga = h ga Upthrust = weight of fluid displaced p y A pressure p y It is clear that upthrust on a body does not depend on the shape of the body. If the body is more dense than the fluid in which it is immersed, then its weight is greater than the weight of the fluid displaces and body will sink. Similarly, if the body is less dense than the fluid in which it is immersed then it will float. Measuring the density of liquid using Archimedes principle: The density of liquid can be found by determining the upthrust on some suitable object when it is immersed in the liquid and then when it is immersed in water. Density of Density of liquid water Upthrust in liquid Upthrust in water From the above equation the density of unknown liquid can be found. Exercise: 2.10 Solve the following questions from the past papers May/June 2002, Paper 1, question 12, 14 May/June 2008, Paper 1, question 15 Oct/Nov 2003, Paper 2, question 2

19 AS Physics 9702 unit 2: Motion, Force and Energy 19 TORQUE OR MOMENT OF FORCE: Consider a force F acting on a rod of length d so as to cause it to turn about a fix point called pivot or fulcrum. This effect of force is called turning effect or moment of force. It is depend on the size and direction of force as well as the perpendicular distance from the fulcrum. The turning effect or moment of force is also called torque. Mathematically it is defined as: τ is the torque in (N.m) this is a greek letter called tau. d the perpendicular distance of the line of action of the force from the axis (m). If F and d are not perpendicular to each other then the expression is, where θ is the angle between F and d. Torque due to a couple Two forces which are equal in magnitude and which are anti-parallel are called couple. i. There is no direction, in which couple can give rise to a resultant force, and therefore a couple can produce only a turning effect, it cannot produce translational motion. ii. Since a single force is bound to produce translation, it follows that a couple cannot be represented by single force. From the figure, Total torque about O = F x OA +F x OB= F(OA+OB) = Fd Thus the torque about O does not depend on the position of O and therefore it follows that: The torque due to a couple is same about any axis and is given by If F and d are not perpendicular than F= = A O d= = F= = B In the given example the couple is Equilibrium system and moment of force: A body is in equilibrium if it follows the following conditions: a) The net resultant force (and acceleration) on the centre of mass of the body is zero. The above equation means that vector sum of all the force along the x-axis is equal to zero and vector sum of all the force along the y-axis is equal to zero. b) Total torque about all axes is zero i.e. Exercise: 2.11 Solve the following questions from past papers. Oct/Nov 2009, Paper 11, question 13 Oct/Nov 2008, Paper 2, question 3 Oct/Nov 2007, Paper 1, question 14 Oct/Nov 2001, Paper 2, question 3(a), 3(b) Oct/Nov 2007, Paper 1, question 13 May/June 2002, Paper 2, question 3(b) May/June 2001, Paper 1, question 15, 17

20 Prepared by Faisal Jaffer, revised on Nov 2011 WORK: When a force F is applied to an object than it covers a displacement s in the direction of the force applied. It is said that the work has been done on an object. The unit of work done joules or J, F is the constant force applied in newton, s is the displacement covered by the body in the direction of the force in metre. When the force and the displacement covered are not in the same direction then we use the component of force along the direction of force. N where is the angle between the force applied and direction of motion. External work done by expanding gas: Consider a gas enclosed in a cylinder by a frictionless piston of cross-sectional area A. Suppose that the piston is in equilibrium under the action of the force pa exerted by the gas and an external force F. s Suppose the gas now expands and moves the piston outwards through a distance of x, where x is so small that p, the pressure of the gas can be considered constant. The external work done w is given by the equation: but is the change in the volume of gas in the cylinder Work done in stretching a spring: The tension F in a spring whose extension is x and which obeys Hooke s law is given by F kx where k is the spring constant. If the extension is increased by x where x is very small that F can be considered constant then the work done can be expressed as or w F x w kx x The total work done in increasing the extension from 0 to x i.e. the elastic potential energy stored in

21 AS Physics 9702 unit 2: Motion, Force and Energy 21 the spring when its extension is x is given by area under the F-x graph. Since this work done stores energy in the spring, this energy is called strain energy or elastic potential energy. Exercise: 2.12 Solve the following questions from past papers May/June 2010, Paper 1, question 19, 20 Oct/Nov 2010, Paper 22, question 4 Oct/Nov 2009, Paper 12, question 20 Oct/Nov 2009, Paper 22, question 4 May/June 2008, Paper 1, question 23 Kinetic and potential energies: Kinetic Energy: The energy which a body of mass m possess only because it is moving is called kinetic energy E k or the amount of work that must have been done on an object to increase its velocity from zero to the velocity v is. Derivation of E k If a body of mass m moves a distance s under the action of a constant force F, the work done W by the force is given by W = Fs F m v m s and if the constant acceleration of the body is a, then according to Newton s second law of motion F=ma but work done is represented by if the body has acceleration from rest to some velocity v, then initial velocity u is zero 2 2 v u 2as or Replacing this value of as in the equation of work 2 v W m or 2 as 2 v 2 1 W mv 2 due to this work done the object gain motion and hence kinetic energy therefore this work done can be expressed as E k Ek 1 mv 2 2 2

22 Prepared by Faisal Jaffer, revised on Nov 2011 Gravitational potential energy: The energy stored in a body due to its change of position or change of shape is called potential energy. The gravitational potential energy of a body is the amount of work done in lifting an object of mass m to the height h above the surface of earth. m But this work done is changing the position of mass from ground to height h then E p mgh h Relationship between gravitational potential energy and m height: Consider a sky diver jumping from the aeroplane. He possesses a potential energy due to his height above the ground. If no air resistance is considered then the graph between the gravitational potential energy and height is given by straight line. Hence this concludes that gravitational energy is directly proportional to the height. Gravitational potential energy is directly proportional height Law of conservation of mechanical energy: Mechanical energy of a body is the sum of kinetic and potential energies of the body. E k E p Constant Loss of E Gain of 1 2 mv mgh 2 k E p or v 2gh Exercise: 2.13 Solve the following questions from past papers. May/June 2010, Paper 23, question 3 Oct/Nov 2009, Paper 22, question 3(c) Oct/Nov 2009, Paper 11, question 14 Oct/Nov 2009, Paper 11, question 15 May/June 2008, Paper 2, question 2(b) Oct/Nov 2007, Paper 1, question 14 Oct/Nov 2008, Paper 1, question 17 May/June 2002, Paper 1, question 16 Electrical potential energy (E p ): The electric potential energy due to charge Q (capital Q) at a point in an electric field is defined as being equal to the work done in bringing a unit positive charge q (small q) from infinity to that point. It follows that the electrical potential energy of a charge Q at a point where the potential is V is given by

23 AS Physics 9702 unit 2: Motion, Force and Energy 23 Internal energy The internal energy of a system is the sum of the kinetic and electric potential energies of the molecules of the system. It follows from the equation Q U w The heat energy ( Q) supplied to a system is equal to the increase in the internal energy ( U) of the system plus the work done ( w) by the system on its surrounding. Temperature is a measure of the average random kinetic energy or internal energy of the molecules of an object. When an object is moving it possess the ordered kinetic energy. Power: Power of a machine is the rate at which it does work or the rate at which it supplies energy. The unit W of power is joules per second or watt. P t But W Fs or Fs P t P Fv Efficiency of machines: The ratio of useful work done, or energy output, to the work or energy input, in an energy transfer system such as a machine or an engine. Exercise: 2.14 Solve the following questions from past paper. Oct/Nov 2008, Paper 1, question 18 Oct/Nov 2009, Paper 12, question 15 May/June 2002, Paper 1, question 19 May/June 2008, Paper 1, question 19 Oct/Nov 2006, Paper 1, question 17, 18

24 Three forces in equilibrium: Prepared by Faisal Jaffer, revised on Nov 2011

25 AS Physics 9702 unit 2: Motion, Force and Energy 25 PRACTICE QUESTIONS UNIT 2: Laws of motion: 1) A body of mass 7.0 kg rests on the floor of lift. Calculate the force, R, exerted on the body by the floor when the lift a) has upward acceleration of 2.0 m/s 2, b) has downward acceleration of 3.0 m/s 2, c) is moving down with constant velocity. (assuming g = 10 m/s 2 ) 2) A train is moving along a straight horizontal track. A pendulum suspended from the roof of one of the carriages of the train is inclined at 4 o to the vertical. Calculate the acceleration of the train (assuming g = 10 m/s 2 ) 3) Two blocks, A of mass m and B of mass 3m, are side by side in contact with each other. They are pushed along a smooth floor under the action of a constant force F applied to A. Find a) the acceleration of the blocks b) the force exerted by B on A. 4) A body of mass 6.0 kg moves under the influence of two oppositely directed forces whose magnitudes are 60N and 18N. Find the magnitude and direction of the acceleration of the body. 5) Two forces of magnitudes 30N and 40N and which are perpendicular to each other, act on a body of mass 25kg. Find the magnitude and direction of the acceleration of the body. 6) A body of mass 3.0 kg slides down a plane which is inclined at 30 o to the horizontal. Find the acceleration of the body. a) if the place is smooth, b) if there is frictional resistance of 9.0 N. 7) A railway truck of mass 6.0 tones moves with an acceleration of 0,050 m/s 2 down a track which is inclined to the horizontal at an angle where sin = 1/120. Find the resistance to motion, assuming that it is constant. 8) A body hangs from a spring-balance which is suspended from the ceiling of lift. What is the mass of the body if the balance registers a reading of 70N when the lift has an upward acceleration of 4.0 m/s 2. 9) What is the apparent weight during the take-off of an astronaut whose actual weight is 750N if the resultant upward acceleration is 5g. 10) A body of mass 5.0 kg is pulled along smooth horizontal ground by means of 40N acting at 60 o above the horizontal. Find: a) the acceleration of the body; b) the force the body exerts on the ground. 11) A ball is thrown vertically upwards with a velocity 20m/s. Calculate a) the maximum height reached, b) the total time for which the ball is in the air. Projectile Motion: 12) A body is projected with a velocity of 200m/s at an angle of 30 o above the horizontal. Calculate: a) the time taken to reach the maximum height b) its velocity after 16s. 13) A particle is projected with a speed of 25m/s at 30 o above the horizontal. Find: a) the time taken to reach the highest point of the trajectory b) the magnitude and direction of the velocity after 2.0s. 14) A particle is projected with a velocity of 30 m/s at an angle of 40 o above a horizontal plane. Find: a) the time foe which the particle is in the air b) the horizontal distance it travels. 15) A pebble is thrown from the top of a cliff at a speed of 10m/s and at 30 o above the horizontal. It hits the sea below the cliff 6.0s later. Find: a) the height of the cliff, b) the distance from the base of the cliff at which the pebble falls into the sea. 16) An airplane moving horizontally at 150m/s releases a bomb at a height of 500m. The bomb hits the intended target. What was the horizontal distance of the aero plane from the target when the bomb was released?

26 Prepared by Faisal Jaffer, revised on Nov 2011 Moments and equilibrium 17) A uniform plank AB which is 6 cm long and has a weight of 300N is supported horizontally by two vertical ropes at A and B. A weight of 150N rests on the plank at C where AC = 2 cm. Find the tension in each rope. [Ta=250N, Tb=200N] 18) The system of forces is in equilibrium. Find P and Q [P=15.6N, Q=10.2N] 19) A uniform ladder which is 5m long and has mass of 20kg leans with its upper end against a smooth vertical wall and its lower end on rough ground. The bottom of the ladder is 3m from the wall. Calculate the frictional force between the ladder and the ground (g=9.8m/s 2 ). [F=75N] P 60 o 40 o Q 20.0 N Energy 20) Two forces P and Q, acts NW and NE respectively. They are in equilibrium with a force of 50.0N acting due S and a force of 20.0 N acting due E. Find P and Q. 21) A car of mass 800kg and moving at 30m/s. along a horizontal road is brought to rest by a constant retarding force of 5000N. Calculate the distance the car moves whilst coming to rest. [s= 72m] 22) A small block is released from rest at A and slides down a smooth curved track. Calculate the velocity of the block when it reached B, a vertical distance h below A. [v= 2 gh ] 23) A car of mass 1.0 x 10 3 kg increases its speed from 10m/s to 20m/s whilst moving 500 m up a road inclined at an angle to the horizontal where sin = 1/20. There is a constant resistance to motion of 300 N. Find the driving force exerted y the engine, assuming that it is constant. Take g = 10 m/s 2 [F = 1.1 x 10 3 N] B A h