UNIT 2 FORCE - MOTION.pdf

SUBJECT (Applied Physics) LECTURE COMPANION SEMESTER: I PREPARED BY: Dhaval Desani CHAPTER 2 FORCE & MOTION...

SUBJECT (Applied Physics) LECTURE COMPANION SEMESTER: I PREPARED BY: Dhaval Desani CHAPTER 2 FORCE & MOTION ⮚ NEWTON’S FIRST LAW OF MOTION Every body continues to be in its state of rest or of uniform motion in a straight line unless compelled by some external force to act otherwise. The state of rest or uniform linear motion both imply zero acceleration. The first law of motion can, therefore, be simply expressed as: If the net external force on a body is zero, its acceleration is zero. Acceleration can be non zero only if there is a net external force on the body. Two kinds of situations are encountered in the application of this law in practice. In some examples, we know that the net external force on the object is zero. In that case we can conclude that the acceleration of the object is zero. For example, a spaceship out in interstellar space, far from all other objects and with all its rockets turned off, has no net external force acting on it. Its acceleration, according to the first law, must be zero. If it is in motion, it must continue to move with a uniform velocity. More often, however, we do not know all the forces to begin with. In that case, if we know that an object is unaccelerated (i.e. it is either at rest or in uniform linear motion), we can infer from the first law that the net external force on the object must be zero. Gravity is everywhere. For terrestrial phenomena, in particular, every object experiences gravitational force due to the earth. Also objects in motion generally experience friction, viscous drag, etc. If then, on earth, an object is at rest or in uniform linear motion, it is not because there are no forces acting on it, but because the various external forces cancel out i.e. add up to zero net external force. Consider a book at rest on a horizontal surface Fig (a). It is subject to two external forces : the force due to gravity (i.e. its weight W) acting downward and the upward force on the book by the table, the normal force R. R is a self-adjusting force. This is an example of the kind of situation mentioned above. The forces are not quite known fully but the state of motion is known. We observe the book to be at rest. Therefore, we conclude from the first law that the magnitude of R equals that of W. A statement often encountered is : “Since W = R, forces cancel and, therefore, the book is at rest”. This is incorrect reasoning. The correct statement DEPARTMENT OF PHYSICS Page | 1 *Proprietary material of SILVER OAK UNIVERSITY  SUBJECT (Applied Physics) LECTURE COMPANION SEMESTER: I PREPARED BY: Dhaval Desani is : “Since the book is observed to be at rest, the net external force on it must be zero, according to the first law. This implies that the normal force R must be equal and opposite to the weight W ”. Consider the motion of a car starting from rest, picking up speed and then moving on a smooth straight road with uniform speed (Fig. (b)). When the car is stationary, there is no net force acting on it. During pick-up, it accelerates. This must happen due to a net external force. Note, it has to be an external force. The acceleration of the car cannot be accounted for by any internal force. This might sound surprising, but it is true. The only conceivable external force along the road is the force of friction. It is the frictional force that accelerates the car as a whole. (You will learn about friction in section 5.9). When the car moves with constant velocity, there is no net external force. ⮚ Inertia We can call it as the unwillingness of particles to change their state of rest or uniform motion → The more the mass, the bigger the inertia → There are 3 types of inertia (a) Inertia of rest (b)Inertia in motion (c) Inertia of direction DEPARTMENT OF PHYSICS Page | 2 *Proprietary material of SILVER OAK UNIVERSITY  SUBJECT (Applied Physics) LECTURE COMPANION SEMESTER: I PREPARED BY: Dhaval Desani (a) Inertia of rest → If there is no external force applied, the object which is steady, will remain steady → For e.g. you will have experienced the backward push when the car moves ahead. It is because car has gone from steady to moving ahead but your body will not move ahead until unless a force from behind be experienced that will push you forward. In this case that force is provided by seats on our backs. (b) Inertia in motion → If there is no external force applied, the body in motion will remain in motion → For e.g. when you apply breaks, the car tends to deaccelerate but we feel some sort of push ahead. It is because our body is in motion and tries to remain in motion. (C) Inertia of direction → If no external force is applied, the object will not change the direction of it’s motion → For e.g. moon is circulating around the earth constantly because gravitational pull of earth is constantly changing the direction of the moon → For e.g. tie a stone at the end of a thread and move it in circular path. If the thread breaks, the stone will go in tangential direction ⮚ Basic (Fundamental) forces in nature → There are various types of forces around us, which affects our motion in daily life. → Examples of the different types of forces are (a) Gravitational force (b) Electromagnetic force (c) Strong Nuclear Force (d) Weak Nuclear Force DEPARTMENT OF PHYSICS Page | 3 *Proprietary material of SILVER OAK UNIVERSITY  SUBJECT (Applied Physics) LECTURE COMPANION SEMESTER: I PREPARED BY: Dhaval Desani (a) Gravitational force The gravitational force is the force of mutual attraction between any two objects by virtue of their masses. It is a universal force. Every object experiences this force due to every other object in the universe. All objects on the earth, for example, experience the force of gravity due to the earth. In particular, gravity governs the motion of the moon and artificial satellites around the earth, motion of the earth and planets around the sun, and, of course, the motion of bodies falling to the earth. It plays a key role in the large-scale phenomena of the universe, such as formation and evolution of stars, galaxies and galactic clusters. (b) Electromagnetic force Electromagnetic force is the force between charged particles. In the simpler case when charges are at rest, the force is given by Coulomb’s law : attractive for unlike charges and repulsive for like charges. Charges in motion produce magnetic effects and a magnetic field gives rise to a force on a moving charge. Electric and magnetic effects are, in general, inseparable – hence the name electromagnetic force. Like the gravitational force, electromagnetic force acts over large distances and does not need any intervening medium. It is enormously strong compared to gravity. The electric force between two protons, for example, is 1036 times the gravitational force between them, for any fixed distance. Matter, as we know, consists of elementary charged constituents like electrons and protons. Since the electromagnetic force is so much stronger than the gravitational force, it dominates all phenomena at atomic and molecular scales. (The other two forces, as we shall see, operate only at nuclear scales.) Thus it is mainly the electromagnetic force that governs the structure of atoms and molecules, the dynamics of chemical reactions and the mechanical, thermal and other properties of materials. It underlies the macroscopic forces like ‘tension’, ‘friction’, ‘normal force’, ‘spring force’, etc. Gravity is always attractive, while electromagnetic force can be attractive or repulsive. Another way of putting it is that mass comes only in one variety (there is no negative mass), but charge comes in two varieties : positive and negative charge. This is what makes all the difference. Matter is mostly electrically neutral (net charge is zero). Thus, electric force is DEPARTMENT OF PHYSICS Page | 4 *Proprietary material of SILVER OAK UNIVERSITY  SUBJECT (Applied Physics) LECTURE COMPANION SEMESTER: I PREPARED BY: Dhaval Desani largely zero and gravitational force dominates terrestrial phenomena. Electric force manifests itself in atmosphere where the atoms are ionised and that leads to lightning. If we reflect a little, the enormous strength of the electromagnetic force compared to gravity is evident in our daily life. When we hold a book in our hand, we are balancing the gravitational force on the book due to the huge mass of the earth by the ‘normal force’ provided by our hand. The latter is nothing but the net electromagnetic force between the charged constituents of our hand and the book, at the surface in contact. If electromagnetic force were not intrinsically so much stronger than gravity, the hand of the strongest man would crumble under the weight of a feather ! Indeed, to be consistent, in that circumstance, we ourselves would crumble under our own weight (c) Strong Nuclear Force The strong nuclear force binds protons and neutrons in a nucleus. It is evident that without some attractive force, a nucleus will be unstable due to the electric repulsion between its protons. This attractive force cannot be gravitational since force of gravity is negligible compared to the electric force. A new basic force must, therefore, be invoked. The strong nuclear force is the strongest of all fundamental forces, about 100 times the electromagnetic force in strength. It is charge-independent and acts equally between a proton and a proton, a neutron and a neutron, and a proton and a neutron. Its range is, however, extremely small, of about nuclear dimensions (10–15m). It is responsible for the stability of nuclei. The electron, it must be noted, does not experience this force. Recent developments have, however, indicated that protons and neutrons are built out of still more elementary constituents called quarks. (d) Weak Nuclear Force The weak nuclear force appears only in certain nuclear processes such as the β-decay of a nucleus. In β-decay, the nucleus emits an electron and an uncharged particle called neutrino. The weak nuclear force is not as weak as the gravitational force, but much weaker than the strong nuclear and electromagnetic forces. The range of weak nuclear force is exceedingly small, of the order of 10-16 m. DEPARTMENT OF PHYSICS Page | 5 *Proprietary material of SILVER OAK UNIVERSITY  SUBJECT (Applied Physics) LECTURE COMPANION SEMESTER: I PREPARED BY: Dhaval Desani ⮚ Momentum Momentum of a body is defined to be the product of its mass m and velocity v, and is denoted by p: p=mv Momentum is clearly a vector quantity. The following common experiences indicate the importance of this quantity for considering the effect of force on motion. Suppose a light-weight vehicle (say a small car) and a heavy weight vehicle (say a loaded truck) are parked on a horizontal road. We all know that a much greater force is needed to push the truck than the car to bring them to the same speed in same time. Similarly, a greater opposing force is needed to stop a heavy body than a light body in the same time, if they are moving with the same speed. If two stones, one light and the other heavy, are dropped from the top of a building, a person on the ground will find it easier to catch the light stone than the heavy stone. The mass of a body is thus an important parameter that determines the effect of force on its motion. Speed is another important parameter to consider. A bullet fired by a gun can easily pierce human tissue before it stops, resulting in casualty. The same bullet fired with moderate speed will not cause much damage. Thus for a given mass, the greater the speed, the greater is the opposing force needed to stop the body in a certain time. Taken together, the product of mass and velocity, that is momentum, is evidently a relevant variable of motion. The greater the change in the momentum in a given time, the greater is the force that needs to be applied. A seasoned cricketer catches a cricket ball coming in with great speed far more easily than a novice, who can hurt his hands in the act. One reason is that the cricketer allows a longer time for his hands to stop the ball. As you may have noticed, he draws in the hands backward in the act of catching the ball (Fig. 5.3). The novice, on the other hand, keeps his hands fixed and tries to catch the ball almost instantly. He needs to provide a much greater force to stop the ball instantly, and this hurts The conclusion is clear: force not. only depends on the DEPARTMENT OF PHYSICS Page | 6 *Proprietary material of SILVER OAK UNIVERSITY  SUBJECT (Applied Physics) LECTURE COMPANION SEMESTER: I PREPARED BY: Dhaval Desani change in momentum, but also on how fast the change is brought about. The same change in momentum brought about in a shorter time needs a greater applied force. In short, the greater the rate of change of momentum, the greater is the force. Thus the same force for the same time causes the same change in momentum for different bodies. ⮚ NEWTON’S SECOND LAW OF MOTION These qualitative observations lead to the second law of motion expressed by Newton as follows : The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts. Thus, if under the action of a force F for time interval ∆t, the velocity of a body of mass m changes from v to v + ∆v i.e. its initial momentum p = m v changes by ∆p = m∆v. According to the Second Law, DEPARTMENT OF PHYSICS Page | 7 *Proprietary material of SILVER OAK UNIVERSITY  SUBJECT (Applied Physics) LECTURE COMPANION SEMESTER: I PREPARED BY: Dhaval Desani where k is a constant of proportionality. Taking the limit ∆t → 0, the term ∆p/∆t becomes the derivative or differential co-efficient of p with respect to t, denoted by d / dt. Thus For a body of fixed mass m, i.e the Second Law can also be written as which shows that force is proportional to the product of mass m and acceleration a. The unit of force has not been defined so far. In fact, we use Eq. (5.4) to define the unit of force. We, therefore, have the liberty to choose any constant value for k. For simplicity, we choose k = 1. The second law then is In SI unit force is one that causes an acceleration of 1 m s-2 to a mass of 1 kg. This unit is known as newton : 1 N = 1 kg m s-2. DEPARTMENT OF PHYSICS Page | 8 *Proprietary material of SILVER OAK UNIVERSITY  SUBJECT (Applied Physics) LECTURE COMPANION SEMESTER: I PREPARED BY: Dhaval Desani ⮚ KINEMATIC EQUATIONS FOR UNIFORMLY ACCELERATED MOTION For uniformly accelerated motion, we can derive some simple equations that relate displacement (x), time taken (t), initial velocity (v0 ), final velocity (v) and acceleration (a). Equation (1) already obtained gives a relation between final and initial velocities v and v0 of an object moving with uniform acceleration a : …….(1) This relation is graphically represented in Fig. The area under this curve is : Area between instants 0 and t = Area of triangle ABC + Area of rectangle OACD the area under v-t curve represents the displacement. Therefore, the displacement x of the object is : DEPARTMENT OF PHYSICS Page | 9 *Proprietary material of SILVER OAK UNIVERSITY  SUBJECT (Applied Physics) LECTURE COMPANION SEMESTER: I PREPARED BY: Dhaval Desani ……..(2) can be written as ……..(3) From Eq. (1), t = (v – v0 )/a. Substituting this in Eq. (3), we get ………..(4) we have obtained three important equations : DEPARTMENT OF PHYSICS Page | 10 *Proprietary material of SILVER OAK UNIVERSITY  SUBJECT (Applied Physics) LECTURE COMPANION SEMESTER: I PREPARED BY: Dhaval Desani Example 1: A bullet of mass 0.04 kg moving with a speed of 90 m s-1 enters a heavy wooden block and is stopped after a distance of 60 cm. What is the average resistive force exerted by the block on the bullet? Solution: The retardation ‘a’ of the bullet (assumed constant) is given by The retarding force, by the second law of motion, is Example 2: A car moving along a straight highway with speed of 126 km h–1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop ? Solution: DEPARTMENT OF PHYSICS Page | 11 *Proprietary material of SILVER OAK UNIVERSITY  SUBJECT (Applied Physics) LECTURE COMPANION SEMESTER: I PREPARED BY: Dhaval Desani ⮚ Impulse We sometimes encounter examples where a large force acts for a very short duration producing a finite change in momentum of the body. For example, when a ball hits a wall and bounces back, the force on the ball by the wall acts for a very short time when the two are in contact, yet the force is large enough to reverse the momentum of the ball. Often, in these situations, the force and the time duration are difficult to ascertain separately. However, the product of force and time, which is the change in momentum of the body remains a measurable quantity. This product is called impulse: DEPARTMENT OF PHYSICS Page | 12 *Proprietary material of SILVER OAK UNIVERSITY  SUBJECT (Applied Physics) LECTURE COMPANION SEMESTER: I PREPARED BY: Dhaval Desani A large force acting for a short time to produce a finite change in momentum is called an impulsive force. In the history of science, impulsive forces were put in a conceptually different category from ordinary forces. Newtonian mechanics has no such distinction. Impulsive force is like any other force – except that it is large and acts for a short time. Example 3: A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12 m s–1. If the mass of the ball is 0.15 kg, determine the impulse imparted to the ball. (Assume linear motion of the ball) ⮚ NEWTON’S THIRD LAW OF MOTION The second law relates the external force on a body to its acceleration. What is the origin of the external force on the body ? What agency provides the external force ? The simple answer in Newtonian mechanics is that the external force on a body always arises due to some other body. Consider a pair of bodies A and B. B gives rise to an external force on A. A natural question is: Does A in turn give rise to an external force on B ? In some examples, the answer seems clear. If you press a coiled spring, the spring is compressed by the force of your hand. The compressed spring in turn exerts a force on your hand and you can feel it. But what if the bodies are not in contact ? The earth pulls a stone downwards due to gravity. Does the stone exert a force on the earth ? The answer is not obvious since we hardly see the effect of the stone on the earth. The answer according to Newton is: Yes, the stone does DEPARTMENT OF PHYSICS Page | 13 *Proprietary material of SILVER OAK UNIVERSITY  SUBJECT (Applied Physics) LECTURE COMPANION SEMESTER: I PREPARED BY: Dhaval Desani exert an equal and opposite force on the earth. We do not notice it since the earth is very massive and the effect of a small force on its motion is negligible. Thus, according to Newtonian mechanics, force never occurs singly in nature. Force is the mutual interaction between two bodies. Forces always occur in pairs. Further, the mutual forces between two bodies are always equal and opposite. This idea was expressed by Newton in the form of the third law of motion. To every action, there is always an equal and opposite reaction. Newton’s wording of the third law is so crisp and beautiful that it has become a part of common language. For the same reason perhaps, misconceptions about the third law abound. Let us note some important points about the third law, particularly in regard to the usage of the terms : action and reaction. 1. The terms action and reaction in the third law mean nothing else but ‘force’. Using different terms for the same physical concept can sometimes be confusing. A simple and clear way of stating the third law is as follows : Forces always occur in pairs. Force on a body A by B is equal and opposite to the force on the body B by A. 2. The terms action and reaction in the third law may give a wrong impression that action comes before reaction i.e action is the cause and reaction the effect. There is no cause effect relation implied in the third law. The force on A by B and the force on B by A act at the same instant. By the same reasoning, any one of them may be called action and the other reaction. 3. Action and reaction forces act on different bodies, not on the same body. Consider a pair of bodies A and B. According to the third law, DEPARTMENT OF PHYSICS Page | 14 *Proprietary material of SILVER OAK UNIVERSITY  SUBJECT (Applied Physics) LECTURE COMPANION SEMESTER: I PREPARED BY: Dhaval Desani ⮚ CONSERVATION OF MOMENTUM The second and third laws of motion lead to an important consequence: the law of conservation of momentum. Take a familiar example. A bullet is fired from a gun. If the force on the bullet by the gun is F, the force on the gun by the bullet is – F, according to the third law. The two forces act for a common interval of time ∆t. According to the second law, F ∆t is the change in momentum of the bullet and – F ∆t is the change in momentum of the gun. Since initially, both are at rest, the change in momentum equals the final momentum for each. Thus if Pb is the momentum of the bullet after firing and Pg is the recoil momentum of the gun, Pg = – Pb i.e. Pb + Pg = 0. That is, the total momentum of the (bullet + gun) system is conserved. Thus in an isolated system (i.e. a system with no external force), mutual forces between pairs of particles in the system can cause momentum change in individual particles, but since the mutual forces for each pair are equal and opposite, the momentum changes cancel in pairs and the total momentum remains unchanged. This fact is known as the law of conservation of momentum : The total momentum of an isolated system of interacting particles is conserved. which shows that the total final momentum of the isolated system equals its initial momentum. Notice that this is true whether the collision is elastic or inelastic. In elastic collisions, there is a second condition that the total initial kinetic energy of the system equals the total final kinetic energy. Example 4: Find the recoil velocity of a gun having mass equal to 5 kg, if a bullet of 25gm acquires the velocity of 500m/s after firing from the gun. DEPARTMENT OF PHYSICS Page | 15 *Proprietary material of SILVER OAK UNIVERSITY  SUBJECT (Applied Physics) LECTURE COMPANION SEMESTER: I PREPARED BY: Dhaval Desani Here given, Mass of bullet (m1) = 25 gm = 0.025 kg Velocity of bullet before firing (u1) = 0 Velocity of bullet after firing (v1) = 500 m/s Mass of gun (m2) = 5 kg Velocity of gun before firing, (u2) = 0 Velocity of gun after firing = ? Example 5: A bullet of 5 gm is fired from a pistol of 1.5 kg. If the recoil velocity of pistol is 1.5 m/s, find the velocity of bullet. DEPARTMENT OF PHYSICS Page | 16 *Proprietary material of SILVER OAK UNIVERSITY  SUBJECT (Applied Physics) LECTURE COMPANION SEMESTER: I PREPARED BY: Dhaval Desani DEPARTMENT OF PHYSICS Page | 17 *Proprietary material of SILVER OAK UNIVERSITY