{"id":2472,"date":"2023-05-27T14:10:55","date_gmt":"2023-05-27T14:10:55","guid":{"rendered":"https:\/\/vartmaaninstitutesirsa.com\/?page_id=2472"},"modified":"2024-03-14T12:32:55","modified_gmt":"2024-03-14T12:32:55","slug":"class-11-physics-notes-chapter-5-laws-of-motion","status":"publish","type":"page","link":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/class-11-physics-notes-chapter-5-laws-of-motion\/","title":{"rendered":"class 11 physics notes chapter 5 laws of motion"},"content":{"rendered":"

class 11 physics notes chapter 5 laws of motion\u00a0<\/em><\/strong><\/h1>\n

Chapter 5 Laws of Motion :\u00a0 Intuitive concept of force, Inertia, Newton’s first law of motion; momentum
\nand Newton’s second law of motion; impulse; Newton’s third law of motion. Law of conservation of linear momentum and its applications. Equilibrium of concurrent forces, Static and kinetic friction, laws of friction, rolling friction, lubrication. Dynamics of uniform circular motion: Centripetal force, examples of circular motion (vehicle on a level circular road, vehicle on a banked road).<\/p>\n

\n

Unit-3 Laws of Motion:-<\/span><\/strong><\/p>\n

1. Force:<\/span><\/strong><\/p>\n

Force may be defined as a pull or push which changes or tends to changes the state of rest or state of motion of a body.<\/span><\/em><\/p>\n

\uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>If force is applied directly on a body then force is called contact force and if force is applied without actual contact from a distance then it is called action at a distance.<\/span><\/p>\n

Effect Produced by the force:<\/span><\/strong><\/p>\n

i.<\/span><\/em><\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/em><\/strong>Force can change speed of an object:-\u00a0<\/span><\/em><\/strong><\/p>\n

e.g.:- A ball thrown-thrown with a larger force will have larger speed and force applied on breaks of a car makes car slow down.<\/span><\/p>\n

ii.<\/span><\/em><\/strong>\u00a0<\/span><\/em><\/strong>Force can change direction of motion of an object:-<\/span><\/em><\/strong><\/p>\n

e.g.:- Force applied on the steering wheel of a car changes the direction of motion of car.<\/span><\/p>\n

\u00a0<\/span><\/em><\/strong>iii.<\/span><\/em><\/strong>\u00a0<\/span><\/em><\/strong>Force can change the shape of an object:-<\/span><\/em><\/strong><\/p>\n

e.g.:- By pushing the rubber ball we can change its shape.<\/span><\/p>\n

\u00a0<\/span><\/strong>Question: 1 Give the magnitude and direction of the net force acting on\u00a0<\/span><\/strong>
\n(a) a drop of rain falling down with a constant speed,<\/span><\/strong>
\n(b) a cork of mass 10 g floating on water,<\/span><\/strong>
\n(c) a kite skillfully held stationary in the sky,<\/span><\/strong>
\n(d) a car moving with a constant velocity of 30 km\/h on a rough road,<\/span><\/strong>
\n(e) a high-speed electron in space far from all material objects, and free of electric and magnetic fields.<\/span><\/strong>
\nAnswer:<\/span><\/strong>\u00a0<\/span>(a) As the drop of rain is falling with constant speed, in accordance with first law of motion, the net force on the drop of rain is zero.<\/span>
\n(b) As the cork is floating on water, its weight is being balanced by the up thrust (equal to. Weight of water displaced). Hence net force on the cork is zero.<\/span>
\n(c) Net force on a kite skillfully held stationary in sky is zero because it is at rest.<\/span>
\n(d) Since car is moving with a constant velocity, the net force on the car is zero.<\/span>
\n(e) Since electron is far away from all material agencies producing electromagnetic and gravitational forces, the net force on electron is zero.<\/span><\/p>\n

2. Aristotle\u2019s Fallacy:<\/span><\/strong>\u00a0<\/span><\/em><\/strong><\/p>\n

According to Aristotle\u2019s Law of motion, an external force is necessary to keep a body moving with uniform velocity\u00a0<\/span><\/p>\n

i.e. according to this law a body will not remains in motion without a external force. This was proved wrong by Galileo law of inertia.<\/span><\/p>\n

3. Galileo\u2019s Law of Inertia<\/span><\/strong>:<\/span><\/em><\/strong>\u00a0<\/span><\/p>\n

On the basis of a series of experiment performed by Galileo, he gave the law of inertia. According to this law, it is inability of a body to change the state of rest or uniform motion of a body without the help of any external force.<\/span><\/p>\n

    \n
  • Inertia:<\/span><\/strong>\u00a0<\/span><\/li>\n<\/ul>\n

    The inherent (inability) property of a material body by virtue of which it cannot change its state of rest or uniform motion is called inertia. Inertia means resistance to change.<\/span><\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/strong>Different Types of Inertia:<\/span><\/strong><\/p>\n

    a. Inertia of rest:<\/span><\/strong>\u00a0<\/span><\/em><\/strong><\/p>\n

    The inability of a body changes its state of rest called inertia of rest:-<\/span><\/p>\n

    \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Why a person falls backward when bus suddenly starts moving forward? when bus starts moving forward then the lower part of the body starts moving forward while the upper part is in rest due to this the person falls backward.<\/span><\/p>\n

    b. Inertia of motion<\/span><\/strong>:\u00a0<\/span><\/em><\/strong><\/p>\n

    The inability of a body changes its state of rest motion is called inertia of motion:-<\/span><\/p>\n

    \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Why a person falls backward when bus suddenly stops. This is due reason that when bus, suddenly stops then lower part of the body comes into rest while<\/span>\u00a0\u00a0<\/span>the upper part is still in the motion due to this person falls forward.\u00a0<\/span><\/p>\n

    c. Inertia of direction:<\/span><\/strong>\u00a0<\/span><\/em><\/strong><\/p>\n

    The inability of a body changes itself its direction of motion is called inertia of direction:-\u00a0<\/span><\/p>\n

    \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Why a person sitting inside the bus experience force acting away from the center of curved path when a bus takes a sharp turn. This is due to reason that the body of person does not like to change its direction of motion.<\/span><\/p>\n

    Question: 2 how is the inertia of a body measured?<\/span><\/strong><\/p>\n

    Inertia of a body is measured by the mass of the body; greater is the amount of mass greater is the inertia of the body.\u00a0<\/span><\/p>\n

    e.g.:- It is easy to change the position of football than a stone of same size because the mass of football is smaller than the mass of the stone. It does not depend on the shape of the body.<\/span><\/p>\n

    4. Momentum:<\/span><\/strong><\/p>\n

    The quantity of motion possessed by the body is called the momentum. It is equal to product of mass and velocity of the body.<\/span><\/p>\n

    \"\"<\/p>\n

    Or\u00a0<\/span>\"\"<\/p>\n

    Momentum is vector quantity. Its direction is same as that of velocity. The\u00a0<\/span>S.I<\/span><\/em>. unit of momentum =\u00a0<\/span>\"\"\u00a0<\/span>\"\".<\/span><\/p>\n

      \n
    • Different cases of Momentum:<\/span><\/em><\/strong><\/li>\n<\/ul>\n

      \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>When two object each of mass\u00a0<\/span>\"\"\u00a0moves with velocity\u00a0<\/span>\"\"and\u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

      Such that\u00a0<\/span>\"\"\u00a0then<\/span>\u00a0\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

      \n
      \n
      \n

      Or<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

      \n
      \n
      \n

      As<\/span>\u00a0 \u00a0<\/span>\"\"<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

      Hence equal massive bodies having larger velocity will have larger momentum.<\/span><\/p>\n

        \n
      • \u00a0 \u00a0 \u00a0When two objects of masses <\/span>\"\"\u00a0such that\u00a0<\/span>\"\"\u00a0moves with same velocity then<\/span><\/li>\n<\/ul>\n

        \"\"<\/p>\n

        \u00a0\u00a0<\/span>\"\"\u00a0\u00a0<\/span>\"\"<\/p>\n

        As<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

        \"\"Hence the two bodies having same velocities will have larger momentum which have larger mass.<\/span><\/p>\n

        3. When two bodies have equal linear momentum then\u00a0<\/span><\/p>\n

        \"\"\u00a0\u00a0<\/span><\/p>\n

        Or<\/span>\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

        Or<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

        If<\/span>\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

        \"\"<\/p>\n

        Thus velocities of the bodies are inversely proportional to masses of the bodies i.e. heavier body will have smaller velocity and smaller body will have larger velocity.<\/span><\/p>\n

        5. Newton\u2019s Laws of Motion:\u00a0<\/span><\/strong>M.Imp.<\/sup><\/span><\/strong><\/p>\n

          \n
        1. Newton\u2019s First Law of Motion<\/strong>:<\/em><\/strong><\/li>\n<\/ol>\n

          A body in rest will remain in rest or a body is in motion will remain in motion till any external force is applied on the body.<\/span><\/p>\n

          \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>How Newton\u2019s first law defines force?<\/span><\/em><\/p>\n

          According to Newton\u2019s first law, force is external agency which changes the state of rest or uniform motion of the body. Hence this law gives the qualitative definition of force.<\/span><\/p>\n

          \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Newton\u2019s first law sometimes called law of inertia.<\/span><\/p>\n

            \n
          1. \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <\/span>Newton\u2019s Second Law of Motion:<\/span><\/li>\n<\/ol>\n

            According to Newton\u2019s second law of motion, the rate of change of linear momentum is directly proportional to applied force.<\/span><\/p>\n

            Mathematically<\/span><\/p>\n

            \"\"<\/p>\n

            Or\u00a0<\/span>\u00a0<\/span>\"\"\u00a0\u00a0<\/span>\u00a0<\/span>\u00a0(<\/span>\"\")<\/span><\/p>\n

            Or\u00a0<\/span>\u00a0<\/span>\"\"<\/p>\n

            Or\u00a0<\/span>\u00a0<\/span>\"\"<\/p>\n

            Or\u00a0<\/span>\u00a0<\/span>\"\"<\/p>\n

            Where<\/span>\u00a0<\/span>\"\"<\/p>\n

            \"\"\u00a0<\/span>\u00a0<\/span>\"\"<\/p>\n

            Force is a vector quantity.\u00a0<\/span><\/p>\n

            Its dimensional formula is\u00a0<\/span>\"\"<\/p>\n

            Unit of Force:<\/span><\/strong>\u00a0<\/span><\/em><\/strong>There are two types of unit of Force<\/span><\/em><\/p>\n

              \n
            • Absolute Unit of Force:<\/span><\/strong><\/li>\n<\/ul>\n

              \u00a0<\/span>An absolute unit of force is defined as the force which produces unit acceleration in unit mass of a body. In S.I. absolute unit of force is Newton (<\/span>\"\")<\/span><\/p>\n

              \u00a0<\/span>In\u00a0<\/span>\"\"absolute unit of force is dyne (1 dyne<\/span>\"\")<\/span><\/p>\n

                \n
              • Gravitational Units of Force<\/span><\/strong>:<\/span><\/em><\/strong><\/li>\n<\/ul>\n

                A gravitational unit of force is defined as the force which produces\u00a0<\/span>\"\"\u00a0\u00a0<\/span>(acceleration on due to gravity) acceleration in a body of unit mass.<\/span><\/p>\n

                In S.I. the gravitational unit of force is Kilogram weight\u00a0<\/span>\"\"\u00a0or Kilogram force\u00a0<\/span>\"\"<\/p>\n

                \"\"<\/p>\n

                In\u00a0<\/span>\"\"\u00a0the gravitational unit of force is gram weight\u00a0<\/span>\"\"\u00a0or gram force\u00a0<\/span>\"\"<\/p>\n

                \"\"<\/p>\n

                Relation between Newton and dyne<\/span><\/p>\n

                \"\"<\/p>\n

                \"\"<\/p>\n

                \"\"<\/p>\n

                Or\u00a0<\/span>\u00a0<\/span>\"\"<\/p>\n

                Difference between Gravitational unit and absolute unit of force:<\/span><\/strong><\/p>\n

                \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>The absolute units of force remains the same throughout the universe but the Gravitational unit depends upon\u00a0<\/span>\"\"\u00a0so changes from place to place.<\/span><\/p>\n

                \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>A gravitational unit is\u00a0<\/span>\"\"\u00a0times the corresponding absolute unit. Gravitational unit is also called weight or practical unit of force.<\/span><\/p>\n

                Question: 3 A stone of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the stone,<\/span><\/strong><\/p>\n

                (a) during its upward motion, .<\/span><\/strong><\/p>\n

                (b) During its downward motion,<\/span><\/strong>
                \n(c) at the highest point where it is momentarily at rest.\u00a0<\/span><\/strong><\/p>\n

                Do your answers change if the stone was thrown at an angle of 45\u00b0 with the horizontal direction?\u00a0<\/span><\/strong><\/p>\n

                Answer:<\/span><\/strong>\u00a0<\/span><\/strong>(a) When the stone is moving upward, the acceleration g is acting downward, so the force is acting downward is equal to<\/span>\u00a0<\/span>F = mg = 0.05 kg x 10 ms<\/span>-2<\/sup><\/span>\u00a0<\/span>= 0.5 N.<\/span>
                \n(b) In this case also F = mg = 0.05 x 10 = 0.5 N. (downwards).<\/span>
                \n(c) The stone is not at rest at highest point but has horizontal component of velocity. The direction and magnitude of the net force on the stone will not alter even if it is thrown at 45\u00b0 because no other acceleration except \u2018g\u2019 is acting on stone.<\/span><\/p>\n

                Question: 4 give the magnitude and direction of the net force acting on a stone of mass 0.1 kg,<\/span><\/strong><\/p>\n

                (a) just after it is dropped from the window of a stationary train,<\/span><\/strong>
                \n(b) just after it is dropped from the window of a train running at a constant velocity of 36km\/h,<\/span><\/strong>
                \n(c) just after it is dropped from the window of a train accelerating with 1 ms<\/span><\/strong>-2<\/sup><\/span><\/strong>,<\/span><\/strong>
                \n(d) lying on the floor of a train which is accelerating with 1 m s~2, the stone being at rest relative to the train.\u00a0<\/span><\/strong>
                \nAnswer:<\/span><\/strong>\u00a0<\/span><\/strong>(a) Mass of stone = 0.1 kg<\/span>
                \nNet force,\u00a0<\/span>\"\"\u00a0= 0.1 x 10 = 1.0 N. (vertically downwards).<\/span>
                \n(b) When the train is running at a constant velocity, its acceleration is zero. No force acts on the stone due to this motion. Therefore, the force on the stone is the same (1.0 N.).<\/span>
                \n(c) The stone will experience an additional force F\u2019 (along horizontal)\u00a0<\/span><\/p>\n

                i.e<\/span>\u00a0\u00a0<\/span>\"\"\u00a0= 0.1 x l = 0.1 N<\/span>
                \nAs the stone is dropped, the force F\u2019 no longer acts and the net force acting on the stone\u00a0<\/span>\"\"\u00a0= 0.1 x 10 = 1.0 N. (vertically downwards).<\/span>
                \n(d) As the stone is lying on the floor of the train, its acceleration is same as that of the train. Force acting on the stone,\u00a0<\/span>\"\"= 0.1 x 1 = 0.1 N. It acts along the direction of motion of the train.<\/span><\/p>\n

                Question: 5 One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is:<\/span><\/strong>
                \n(i) T,<\/span><\/strong>\u00a0\u00a0\u00a0<\/span><\/strong>(ii) T \u2013 mv<\/span><\/strong>2<\/sup><\/span><\/strong>\/l,<\/span><\/strong>\u00a0\u00a0\u00a0<\/span><\/strong>(iii) T +mv<\/span><\/strong>2<\/sup><\/span><\/strong>\/l,<\/span><\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/strong>(iv) 0<\/span><\/strong>
                \nT is the tension in the string. [Choose the correct alternative].<\/span><\/strong>
                \nAnswer:<\/span><\/strong>\u00a0<\/span>(i) T<\/span>
                \nThe net force T on the particle is directed towards the centre. It provides the centripetal force required by the particle to move along a circle.<\/span><\/p>\n

                Question: 6 A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15<\/span><\/strong>\u00a0<\/span><\/strong>ms<\/span><\/strong>-3<\/sup><\/span><\/strong>. How long does the body take to stop?<\/span><\/strong>
                \nAnswer:<\/span><\/strong>\u00a0<\/span><\/strong>\u00a0<\/span>Here m = 20 kg, F = \u2013 50 N (retardation force)\u00a0<\/span><\/p>\n

                As F = ma<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                So\u00a0<\/span>\"\"<\/p>\n

                As\u00a0<\/span>\"\"<\/p>\n

                Using equation<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                We get<\/span>\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

                \"\"
                \nQuestion: 7 constant forces acting on a body of mass 3.0 kg changes its speed from 2.0 ms<\/span><\/strong>-1<\/sup><\/span><\/strong>\u00a0<\/span>to 3.5 ms<\/span><\/strong>-1<\/sup><\/span><\/strong>\u00a0<\/span><\/strong>in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?<\/span><\/strong>
                \nAnswer:<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>given\u00a0<\/span>\u00a0<\/span>\"\"\u00a0,<\/span>\u00a0\u00a0<\/span>\"\",<\/span>\u00a0\u00a0<\/span>t=25s<\/span>\u00a0\u00a0<\/span><\/p>\n

                As we know<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

                \"\"<\/p>\n

                \"\"<\/p>\n

                \"\"\u00a0<\/span><\/p>\n

                here the force is in direction of motion.<\/span><\/p>\n

                Question: 8 A bodies of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.<\/span><\/strong>
                \nAnswer:\u00a0<\/span><\/strong>here\u00a0<\/span>\"\"<\/p>\n

                \"ncert-class-11-solutions-physics-5-laws-motion-3\"The resultant force on the body<\/span><\/p>\n

                \u00a0<\/span>\"\"=10N<\/span><\/p>\n

                So<\/span>\u00a0\u00a0<\/span>\"\"\u00a0\u00a0<\/span>in direction of force<\/span><\/p>\n

                The direction of acceleration may be given by<\/span><\/p>\n

                \u00a0\u00a0<\/span>\"\"\u00a0\u00a0\u00a0<\/span><\/p>\n

                So<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

                Question: 9 the driver of a three-wheeler moving with a speed of 36 km\/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. <\/span><\/strong>What is the average retarding force on the vehicle? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg.<\/span><\/strong>
                \nAnswer:<\/span><\/strong>\u00a0<\/span>Here mass of three-wheeler m<\/span>i<\/sub><\/span>\u00a0= 400 kg,\u00a0<\/span><\/p>\n

                Mass of driver = m<\/span>2<\/sub><\/span>\u00a0<\/span>= 65 kg,\u00a0<\/span><\/p>\n

                Initial speed of auto, u = 36 km\/h = 36 x<\/span>\"\"\u00a0= 10 ms<\/span>-1<\/sup><\/span>,\u00a0<\/span><\/p>\n

                Final speed, v = 0 and t = 4s.<\/span><\/p>\n

                As acceleration\u00a0<\/span>\"\"<\/p>\n

                \"\"<\/p>\n

                \"\"\u00a0\u00a0<\/span><\/p>\n

                Now\u00a0<\/span>\"\"<\/p>\n

                \"\"<\/p>\n

                \"\"<\/p>\n

                \"\"<\/p>\n

                Here \u2013ve sign shows that force is of retarding nature.<\/span><\/p>\n

                3.\u00a0 Newton\u2019s Third Law of Motion:<\/strong><\/p>\n

                It states that to every action, there is always an equal and opposite reaction.<\/span><\/p>\n

                \"\"Suppose <\/span>\"\"\u00a0is the amount of force exerted by body<\/span>\u00a0\u00a0<\/span>\"\"\u00a0on body<\/span><\/p>\n

                \u00a0<\/span>\"\"\u00a0Is the amount of force on\u00a0<\/span>\"\"\u00a0\u00a0<\/span>due to\u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

                Then from Newton\u2019s Third law of motion<\/span>\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

                I.e. Force on\u00a0<\/span>\"\"<\/p>\n

                e.g.: During swimming a person pushes water back word then the water pushes person in forward direction.<\/span><\/p>\n

                \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Here Newton\u2019s third law shows that a single force cannot exist in nature, the force exists in pairs.<\/span><\/p>\n

                \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Newton\u2019s third law of motion is applicable irrespective of the nature of the forces.<\/span><\/p>\n

                \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Action and reaction always acts on the different bodies.<\/span><\/p>\n

                \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>The force of action and reaction cannot cancel each other because act on different bodies.<\/span><\/p>\n

                \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>No action act occurs in the absence of reaction.<\/span><\/p>\n

                Illustrations of Newton\u2019s Third Law of Motion:<\/span><\/strong><\/p>\n

                \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>A book kept on table exert a force\u00a0<\/span>\"\"\u00a0due to weight on the table in down ward direction due to which a equal and opposite force\u00a0<\/span>\"\"\u00a0is Act on book due to table in upward direction. Hence both remain in equilibrium.<\/span><\/p>\n

                \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>While walking, we press the ground in backward direction, them the ground exerts an equal and opposite force on us due to which we walks.<\/span><\/p>\n

                \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>It is difficult to walk on slippery ground or sand because we are unable to exert force due to which reaction is not sufficient to walk.<\/span><\/p>\n

                6. Horse and Cart Problems:<\/span><\/strong><\/p>\n

                Suppose a cart connected to a horse by a string. Suppose horse while pull the cart produce T tension in the string, same tension is produce by cart in opposite direction.<\/span><\/p>\n

                \"E:\\Pradeep\\Untitled.png\"Now again suppose that the horse pushes the ground with a force F and R is normal reaction. Resolving R into rectangular components we get:-<\/span><\/p>\n

                \"\"<\/p>\n

                \"\"<\/p>\n

                Hence if\u00a0<\/span>\"\"is greater then\u00a0<\/span>\"\"\u00a0then horse will pull cart & if\u00a0<\/span>\"\"\u00a0is greater then\u00a0<\/span>\"\"\u00a0than cart will pull horse.<\/span><\/p>\n

                \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>On an inclined plane cart may pull horse.<\/span><\/p>\n

                Question :10<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>Explain why<\/span><\/strong>
                \n(a) a horse cannot pull a cart and run in empty space,<\/span><\/strong>
                \n(b) passengers are thrown forward from their seats when a speeding bus stops suddenly,<\/span><\/strong>
                \n(c) it is easier to pull a lawn mower than to push it,<\/span><\/strong>
                \n(d) a cricketer moves his hands backwards while holding a catch.<\/span><\/strong>
                \nAnswer:<\/span><\/strong> (a) A horse by itself cannot move in space due to law of inertia and so cannot pull a cart in space.
                \n(b) The passengers in a speeding bus have inertia of motion. When the bus is suddenly stopped the passengers are thrown forward due to this inertia of motion.
                \n(c) In the case of pull, the effective weight is reduced due to the vertical component of the pull. In the case of push, the vertical component increases the effective weight.
                \n(d) The ball comes with large momentum after being hit by the batsman. When the player takes catch it causes large impulse on his palms which may hurt the cricketer. When he moves his hands backward the time of contact of ball and hand is increased so the force is reduced.<\/p>\n

                7. Show that Newton\u2019s second law is real law of motion?<\/span><\/strong>\u00a0M.Imp<\/sup><\/span><\/strong><\/p>\n

                  \n
                1. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>First law is contained in second law:-<\/li>\n<\/ol>\n

                  According to Newton\u2019s second law<\/span>\u00a0\u00a0\u00a0<\/span><\/p>\n

                  \"\"\u00a0<\/span><\/p>\n

                  In the absence of external force<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                  \"\"<\/p>\n

                  \u00a0\u00a0<\/span>Here<\/span>\u00a0<\/span>\u00a0<\/span>\"\"<\/p>\n

                  Or\u00a0<\/span>\"\"\u00a0means\u00a0<\/span>\"\"<\/p>\n

                  Hence if no external force on the body then if body is rest will remain is rest or if body is moving with uniformly then will remain in motion which is first law.\u00a0<\/span><\/p>\n

                    \n
                  1. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Third law is contained in second law:-<\/li>\n<\/ol>\n

                    Consider an isolated system of two bodies<\/span>\"\". Let\u00a0<\/span>\"\"\u00a0is force on\u00a0<\/span>\"\"\u00a0and\u00a0<\/span>\"\"\u00a0is force on<\/span>\"\".<\/span><\/p>\n

                    Then from second law<\/span>\u00a0\u00a0\u00a0<\/span><\/p>\n

                    \"\"<\/p>\n

                    \"\"<\/p>\n

                    \"\"<\/p>\n

                    \"\"\u00a0\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

                    \"\"\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

                    \"\"<\/p>\n

                    \"\"\u00a0\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

                    \"\"\u00a0\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

                    This is Newton\u2019s third law of motion.<\/span><\/p>\n

                    Hence both Newton\u2019s first law and third law can be explained by second law, so we can say Newton\u2019s second law is real of motion or fundamental law of motion.<\/span><\/p>\n

                    Question: 11 a body of mass 0.40 kg moving initially with a constant speed of 10 ms<\/span><\/strong>-1<\/sup><\/span><\/strong>\u00a0<\/sup><\/span><\/strong>to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its position at t = -5 s, 25 s, 100 s.\u00a0<\/span><\/strong>
                    \nAnswer: given that m= 0.40 kg,<\/span><\/p>\n

                    \u00a0<\/span>\"\"<\/p>\n

                    \"\"<\/p>\n

                    \"\"<\/p>\n

                    \u00a0<\/span>As\u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

                    Also<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

                    \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Position at t=-5s\u00a0 is<\/p>\n

                    \u00a0\"\"<\/p>\n

                    \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Position at t=-25s\u00a0 is<\/p>\n

                    \u00a0\"\"<\/p>\n

                    \"\"<\/p>\n

                    \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Position at t=-100s\u00a0 is<\/p>\n

                    \"\"<\/p>\n

                    \"\"<\/p>\n

                    At t=30s, \"\"<\/p>\n

                    so\u00a0\u00a0 v= 10+ (-20X30) =-590m\/s<\/p>\n

                    Now motion from 30s to 100s<\/p>\n

                    \u00a0\"\"<\/p>\n

                    \"\"<\/p>\n

                    So total distance \"\"<\/p>\n

                    \"\"<\/p>\n

                    Question: 12<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>\u00a0<\/span><\/strong>A truck starts from rest and accelerates uniformly at 2.0 ms<\/span><\/strong>-2<\/sup><\/span><\/strong>. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11s? (Neglect air resistance.)<\/span><\/strong>
                    \nAnswer:\u00a0<\/span><\/strong>here<\/span>\u00a0\u00a0<\/span> u = 0, a = 2 ms-2<\/sup><\/span>, t= 10 s
                    \nUsing equation, \"\",<\/p>\n

                    we get\u00a0 \"\"<\/p>\n

                    \"\"
                    \n(a) Initial velocity along x axis\u00a0\u00a0 \"\"ms-1<\/sup><\/span>\"ncert-class-11-solutions-physics-5-laws-motion-6\"Let us now consider vertical motion which is controlled by force of gravity.
                    \n\"\"=0, a = g = 10 ms-2<\/sup><\/span>, t = (11 \u2014 10) s = 1 s<\/p>\n

                    Using\u00a0\u00a0 \"\"<\/p>\n

                    we get \"\"<\/p>\n

                    Resultant velocity \"\"<\/p>\n

                    \"\"<\/p>\n

                    \"\"<\/p>\n

                    The direction of resultant velocity can be given by<\/p>\n

                    \u00a0\"\"<\/p>\n

                    \"\"\u00a0along horizontal direction.<\/p>\n

                    (b) The moment the stone is dropped from the car, horizontal force on the stone is zero. The only acceleration of the stone is that due to gravity.<\/p>\n

                    This gives a vertically downward acceleration of 10 ms-2<\/sup><\/span>. This is also the net acceleration of the stone.<\/p>\n

                    8. Impulse:<\/span><\/strong>\u00a0imp<\/sup><\/span><\/strong><\/p>\n

                    It is the total effect of a large force which act on a body for a very short time and is equal to the product of force and time. Some time it is called action of force.<\/span><\/p>\n

                    Mathematically impulse may be defined as the product of force and time.<\/span><\/p>\n

                    From Newton\u2019s second law<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                    \"\"\u00a0<\/span>\u00a0<\/span>\u00a0<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                    \"\"\u00a0<\/span>\"\"<\/p>\n

                    Integrating both side<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                    \u00a0<\/span>\"\"<\/p>\n

                    \"\"\u00a0<\/span>\"\"<\/p>\n

                    \"\"\u00a0<\/span>\"\"<\/p>\n

                    Here<\/span>\u00a0\u00a0\u00a0<\/span>Ft\u00a0<\/span><\/em>is called impulse, so impulse may be defined as the change in momentum of a body.<\/span><\/p>\n

                    Unit:-<\/span><\/em><\/strong>\u00a0<\/span>in S.I. system the unit of impulse is NS or Kgm\/s & in c,g,s system dyne\/sec or gcm\/s.<\/span><\/em><\/p>\n

                    Usefulness of concept of Impulse:<\/span><\/strong><\/p>\n

                    By the knowledge of impulse we can find out the time for which a force is acted on the body.<\/span><\/p>\n

                    9. Application of concept of Impulse:<\/span><\/strong>\u00a0imp<\/sup><\/span><\/strong><\/p>\n

                      \n
                    1. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>A cricket player lower his hand while catching a ball because by doing this he increase the time of catch due to which he have to apply smaller force to catch and hence in impulse\u00a0 felled by person decrease.<\/li>\n
                    2. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>A person falling from a height on a cemented floor get more injuries as compared when he fall on a help of sand because the time of action of force on sandy surface is large as compared to cement surface.<\/li>\n
                    3. automobiles are provided with shockers because shockers increase the time of action of force due to which we feel small Jerk.<\/li>\n
                    4. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Buffers are provided between the bogies of a train because they increase the time of Jerk due to which action of force decreases and hence we feel small Jerk.<\/li>\n
                    5. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Chinaware\u2019s are packed in straw paper because they increase the time of experience the Jerk during transportation. Hence they strike each other with a small force and get loss damaged.<\/li>\n<\/ol>\n

                      Question: 13 two billiard balls, each of mass 0.05 kg, moving in opposite directions with speed 6 ms<\/span><\/strong>-1<\/sup><\/span><\/strong>collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?<\/span><\/strong>
                      \nAnswer:<\/span><\/strong>\u00a0<\/span><\/strong>Initial momentum of each ball before collision<\/p>\n

                      = 0.05 x 6 kg ms-1<\/sup><\/span>\u00a0<\/span>= 0.3 kg ms-1<\/sup><\/span>
                      \nFinal momentum of each ball after collision<\/p>\n

                      = \u2013 0.05 x 6 kg ms-1<\/sup><\/span>\u00a0<\/span>= \u2013 0.3 kg\u00a0ms-1<\/sup><\/span>\u00a0<\/span><\/p>\n

                      Impulse imparted to each ball due to the other<\/p>\n

                      = final momentum \u2013 initial momentum<\/p>\n

                      = 0.3 kg m s-1 \u2013 0.3 kg ms-1<\/sup><\/span><\/p>\n

                      = \u2013 0.6 kg ms-1<\/sup><\/span>\u00a0<\/span>= 0.6 kg ms-1<\/sup><\/span>\u00a0<\/sup><\/span>(in magnitude)
                      \nThe two impulses are opposite in direction.<\/p>\n

                      Question: 14 Figure shows the position-time graph of a particle of mass 0.04 kg. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the particle? What is the magnitude of each impulse?<\/span><\/strong>
                      \nAnswer:<\/span><\/strong>\u00a0<\/span><\/strong>This graph can be of a ball rebounding between two walls situated at position 0 cm and 2 cm.<\/p>\n

                      The ball is rebounding from one wall to another, time and again every 2 s with uniform velocity.
                      \nHere \"\"<\/p>\n

                      So\u00a0\u00a0\u00a0 \"\"<\/p>\n

                      \"\"<\/p>\n

                      \"\"<\/p>\n

                      \"ncert-class-11-solutions-physics-5-laws-motion-17\"<\/p>\n

                      And\u00a0 \"\"<\/p>\n

                      \"\"<\/p>\n

                      \"\"<\/p>\n

                      \"\"<\/p>\n

                      The time between two impulse is 2 sec . the ball recieves an impulse every 2 second<\/p>\n

                      Question: 15 A batsmen deflect a ball by an angle of 45\u00b0 without changing its initial speed which is equal to 54 km\/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)<\/span><\/strong>
                      \nAnswer:<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>suppose o is the position of bat. AO is the line shows the path along which the ball strike the bat with velocity u and OB is the path showing deflection such that angle AOB =45\u00a0<\/span><\/p>\n

                      So initial momentum of the ball<\/span><\/p>\n

                      \u00a0<\/span>\"\"<\/p>\n

                      \"\"<\/p>\n

                      \"ncert-class-11-solutions-physics-5-laws-motion-14\"<\/p>\n

                      Final momentum of the ball\u00a0<\/span><\/p>\n

                      =<\/span>\"\"\u00a0<\/span><\/strong>along ON\u00a0<\/span><\/p>\n

                      So Impulse = change in momentum<\/span><\/p>\n

                      \"\"<\/p>\n

                      10. Law of Conservation of Linear Momentum:<\/span><\/strong>\u00a0imp<\/sup><\/span><\/strong><\/p>\n

                      According to law of conservation of linear momentum if there is no any external force is acting on an isolated system then the linear momentum of the system remains constant.<\/span><\/p>\n

                        \n
                      • \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Derivation of Law of Conservation of Linear Momentum form Second law:<\/strong><\/li>\n<\/ul>\n

                        Suppose a system of\u00a0<\/span>n<\/span><\/em>\u00a0particles having masses\u00a0<\/span>\"\"\u00a0and velocity\u00a0<\/span>\"\"\u00a0then total linear momentum of\u00a0<\/span>n<\/span><\/em>\u00a0particles will be\u00a0<\/span><\/p>\n

                        \"\"<\/p>\n

                        \"\"\u00a0<\/span>\"\"<\/p>\n

                        If\u00a0<\/span>\"\"\u00a0is the external force acting on the system then<\/span>\u00a0\u00a0\u00a0<\/span>\"\"\u00a0=<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

                        For isolated system\u00a0<\/span>F = 0<\/span><\/em>\u00a0\u00a0<\/span><\/em>\"\"\u00a0<\/span>\"\"<\/p>\n

                        \"\"<\/p>\n

                        Thus in absence of external force the linear momentum of isolated system remains constant.<\/span><\/p>\n

                          \n
                        • \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong><\/span>Derivation of Law of Conservation of Linear Momentum form Third law:<\/strong><\/li>\n<\/ul>\n

                          Suppose a isolated system of two bodies such that\u00a0<\/span>\"\"\u00a0is the force on body\u00a0<\/span>A<\/span><\/em>\u00a0due to\u00a0<\/span>B<\/span><\/em>\u00a0and\u00a0<\/span>\"\"\u00a0is the force on body\u00a0<\/span>B<\/span><\/em>\u00a0due to\u00a0<\/span>A<\/span><\/em>. Then from third law<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

                          Now impulse of\u00a0<\/span>\"\"<\/p>\n

                          And impulse of\u00a0<\/span>\"\"<\/p>\n

                          As impulse of\u00a0<\/span>\"\"\u00a0= impulse of\u00a0<\/span>\"\"<\/p>\n

                          \"\"\u00a0<\/span>\"\"<\/p>\n

                          \"\"\u00a0<\/span>\"\"<\/p>\n

                          \"\"\u00a0<\/span><\/p>\n

                          11. Application of Law of Conservation of Linear Momentum:<\/span><\/strong>\u00a0imp<\/sup><\/span><\/strong><\/p>\n

                          \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>While firing a bullet, the gun should be hold tight to the shoulder because as we know the recoiling gun can\u2019t hear the shoulder.\u00a0<\/span><\/p>\n

                          If the gun is hold tight to the body then gun and body becomes a isolated system of large mass and recoil velocity becomes small.<\/span><\/p>\n

                          \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>When a men jumps out of a boat to the shore, the boat moves slightly away from the shore becomes initially total momentum of both is zero, but during jumping man accrue a momentum in forward direction due to which boat accrue momentum<\/span>\u00a0\u00a0<\/span>in backward direction so it happens.<\/span><\/p>\n

                          \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Rocket and Jet planes work on the principle of conservation of linear momentum:<\/span><\/p>\n

                          In rockets and Jet planes the fuels impart momentum in downward direction due to which rocket get a momentum in upward direction.<\/span><\/p>\n

                          \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Recoil of gun: suppose a bullet of mass m is fired from a gun of mass M. initially both are in rest but after firing suppose v is the velocity of bullet and V is the velocity of gun. Then from conservation of linear momentum<\/span><\/p>\n

                          \"\"<\/p>\n

                          Or<\/span>\u00a0<\/span>\"\"<\/p>\n

                          Or<\/span>\u00a0<\/span>\"\"<\/p>\n

                          Here\u00a0<\/span>\"\"\u00a0\u00a0<\/span>sign show that both bullet and gun moves in opposite direction and clearly<\/span>\"\".\u00a0<\/span><\/p>\n

                          So bullet moves faster from heavy gun.<\/span><\/p>\n

                          \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Explosion of a bomb is also based on conservation of linear momentum.<\/span><\/p>\n

                          Question: 16 a nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two<\/span><\/strong>\u00a0<\/span><\/strong>smaller nuclei the products must move in opposite directions.<\/span><\/strong>
                          \nAnswer:<\/span><\/strong>\u00a0<\/span><\/strong> Let \"\"be the masses of products and \"\"\u00a0<\/span>be their respective velocities.<\/p>\n

                          Therefore, total linear momentum after disintegration = m1<\/sub><\/span>v1<\/sub><\/span>\u00a0<\/span>+m2<\/sub><\/span>\u00a0<\/span>v2<\/sub><\/span>.<\/p>\n

                          Before disintegration, the nucleus is at rest.
                          \nTherefore, its linear momentum before disintegration is zero.<\/p>\n

                          \u00a0According to the principle of conservation of linear momentum,<\/p>\n

                          \u00a0\"\"<\/p>\n

                          \"\"
                          \nNegative sign shows that v1<\/sub><\/span>\u00a0<\/span>and v2<\/sub><\/span>\u00a0<\/span>are in opposite directions.<\/p>\n

                          Question: 17 A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 ms<\/span><\/strong>-1<\/sup><\/span><\/strong>\u00a0<\/span>what is the recoil speed of the gun?<\/span><\/strong>
                          \nAnswer:<\/span><\/strong>\u00a0<\/span><\/strong> m = 0.02 kg, M = 100 kg, v = 80 ms-1<\/sup><\/span>, V =?<\/p>\n

                          So\u00a0\u00a0 \"\"<\/p>\n

                          \"\"<\/p>\n

                          \"\"<\/p>\n

                          \"\"
                          \nhere negative sign indicates that the gun moves in a direction opposite to the direction of motion of bullet.<\/p>\n

                          12.<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>Apparent Weight of a Men in An Elevator\/Lift:<\/span><\/strong>\u00a0M.Imp<\/sup><\/span><\/strong><\/p>\n

                          Suppose a man of mass\u00a0<\/span>\"\"\u00a0standing on a weighting machine placed in a lift. Then various cases associated with the weight of the person are shown below:<\/span><\/p>\n

                          \"E:\\Pradeep\\Untitled1.png\"(1) When the lift is in rest or moving with uniform velocity\u00a0<\/span>\"\"\u00a0downward or upward. In this case<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>\"\"\u00a0in this case the equation becomes<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

                          Or\u00a0<\/span>\"\"<\/p>\n

                          In this case the apparent weight of the person is equal to his real weight.<\/span><\/p>\n

                          (2) When the lift moves upwards with acceleration\"\". In this case the net force act in upward direction<\/p>\n

                          So<\/span>\u00a0\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                          Or\u00a0<\/span>\"\"<\/p>\n

                          Or<\/span>\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

                          Hence apparent weight of a man increase when lift moves in upward direction.<\/span><\/p>\n

                          (3)\u00a0 <\/span>When the lift moves down downward with acceleration\"\".<\/p>\n

                          In this case the net force act in downward direction<\/span><\/p>\n

                          So<\/span>\u00a0\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                          Or<\/span>\u00a0\u00a0<\/span>\"\"\u00a0\u00a0<\/span><\/p>\n

                          Or<\/span>\u00a0<\/span>\"\"<\/p>\n

                          Hence apparent weight of man decreases when moves in downward direction.<\/span><\/p>\n

                          (4)\u00a0 <\/span>When lift falls freely. In this case the acceleration of the body becomes equal to acceleration due to gravity<\/p>\n

                          i.e.\u00a0<\/span>\u00a0<\/span>\"\"\u00a0\u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

                          Question: 18 A man of mass 70 kg, stands on a weighing machine in a lift,<\/span><\/strong>\u00a0<\/span><\/strong>which is moving\u00a0<\/span><\/strong>
                          \n(a) upwards with a uniform speed of 10 ms<\/span><\/strong>-1<\/sup><\/span><\/strong>.<\/span><\/strong>
                          \n(b) Downwards with a uniform acceleration of 5 ms<\/span><\/strong>-2<\/sup><\/span><\/strong>.<\/span><\/strong>
                          \n(c) Upwards with a uniform acceleration of 5 ms<\/span><\/strong>-2<\/sup><\/span><\/strong>.<\/span><\/strong>
                          \nWhat would be the readings on the scale in each case?<\/span><\/strong>
                          \n(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?<\/span><\/strong>
                          \nAnswer:<\/span><\/strong>\u00a0<\/span><\/strong>Here, m = 70 kg, g = 10 m\/s2<\/sup><\/span>
                          \n(a) When the lift moves upwards with a uniform speed, its acceleration is zero.<\/p>\n

                          \u00a0\"\"<\/p>\n

                          \"\"
                          \n(b) When the lift moves downwards with a = 5 ms-2<\/sup><\/span><\/p>\n

                          Than \"\"<\/p>\n

                          \"\"(c) When the lift moves upwards with\u00a0a = 5 ms-2<\/sup><\/span><\/p>\n

                          Than \"\"<\/p>\n

                          \"\"<\/p>\n

                          \"\"(d) If the lift were to come down freely under gravity, downward acc. A = g<\/p>\n

                          \"\"<\/p>\n

                          \"\"<\/p>\n

                          \"\"<\/p>\n

                          13. Motion of the connected bodies:<\/span><\/strong>\u00a0imp<\/sup><\/span><\/strong><\/p>\n

                          \"E:\\Pradeep\\Untitled5.png\"Suppose two masses\u00a0<\/span>\"\"\u00a0such that\u00a0<\/span>\"\"\u00a0move on a pully. Suppose\u00a0<\/span>\"\"\u00a0is the tension in the string. The mass\u00a0<\/span>\"\"\u00a0moves downward and\u00a0<\/span>\"\"\u00a0upward, the resultant downward force on mass\u00a0<\/span>\"\"\u00a0is<\/span><\/p>\n

                          \"\"<\/p>\n

                          And resultant upward force on mass\u00a0<\/span>\"\"\u00a0is\u00a0<\/span><\/p>\n

                          \"\"<\/p>\n

                          Adding eq. (i) and (ii) we get<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                          \u00a0<\/span>\"\"<\/p>\n

                          Or<\/span>\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                          Clearly\u00a0<\/span>\"\"<\/p>\n

                          Hence acceleration of the two connected bodies is smaller than the acceleration due to gravity.<\/span><\/p>\n

                          Also dividing eq. (i) by (ii)<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                          \u00a0\u00a0<\/span>\"\"<\/p>\n

                          Or\u00a0<\/span>\"\"<\/p>\n

                          \"\"\u00a0<\/span>\"\"<\/p>\n

                          \"\"\u00a0\u00a0<\/span>\"\"<\/p>\n

                          \"\"\u00a0<\/span>\"\"<\/p>\n

                          \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>A diagram for each body of the system showing all the forces exerted on the body by the remaining part of the body is called free body diagram.<\/span><\/p>\n

                          \"ncert-class-11-solutions-physics-5-laws-motion-11\"Question: 19 two masses 8 kg and 12 kg are connected at the two ends of a light in extensible string that goes over a friction less pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.<\/span><\/strong>
                          \nAnswer:<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>For block\u00a0<\/span>\"\"<\/p>\n

                          \"\"\u2026\u2026(1)<\/span><\/p>\n

                          And For block\u00a0<\/span>\"\"<\/p>\n

                          \"\"\u2026.(2)<\/span><\/p>\n

                          Adding equation<\/p>\n

                          We get \"\"<\/p>\n

                          \u00a0Or\u00a0\u00a0 \"\"g<\/p>\n

                          =\"\"10<\/p>\n

                          =2 \"\"<\/p>\n

                          Substituting value of a into equation (ii) we get<\/p>\n

                          \u00a0 \"\"<\/p>\n

                          \"\"<\/p>\n

                          \"ncert-class-11-solutions-physics-5-laws-motion-10\"Question: 20 Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a tight string. A horizontal force F = 600 N is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case?<\/span><\/strong>
                          \nAnswer:<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

                          1. When force is applied on 10 kg mass<\/p>\n

                          Than \"\"<\/p>\n

                          \"\"<\/p>\n

                          2. When force is applied on 20 kg mass<\/p>\n

                          than \"\"<\/p>\n

                          Question: 21 Ten one rupee coins are put on top of one another on a table. Each coin has a mass m kg. Give the magnitude and direction of<\/span><\/strong>
                          \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/strong>(a)<\/span><\/strong>\u00a0\u00a0\u00a0\u00a0<\/span><\/strong>the force on the 7th coin (counted from the bottom) due to all coins above it.<\/span><\/strong>
                          \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/strong>(b)<\/span><\/strong>\u00a0\u00a0\u00a0\u00a0<\/span><\/strong>The force on the 7th coins by the eighth coin and<\/span><\/strong>
                          \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/strong>(c)<\/span><\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/strong>the reaction of the sixth coin on the seventh coin.<\/span><\/strong>
                          \nAnswer:<\/span><\/strong>\u00a0\u00a0\u00a0<\/span><\/strong>\u00a0<\/span>(a) The force on 7th coin is due to weight of the three coins lying above it<\/span><\/p>\n

                          Therefore,<\/span>\u00a0\u00a0<\/span>\"\"\u00a0\u00a0\u00a0<\/span><\/p>\n

                          This force acts vertically downwards.<\/span>
                          \n\u00a0(b) The eighth coin is already under the weight of two coins above it and it has its own weight too. Hence force on 7th coin due to 8th coin is sum of the two forces\u00a0<\/span><\/p>\n

                          i.e.<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

                          \"\"<\/p>\n

                          \"\"\u00a0\u00a0<\/span><\/p>\n

                          the force acts vertically downwards.<\/span>
                          \n\u00a0(c) The sixth coin is under the weight of four coins above it.<\/span>
                          \n\u00a0Reaction,\u00a0<\/span>\"\"<\/p>\n

                          \"\"<\/p>\n

                          \"\"\u00a0<\/span><\/p>\n

                          Minus sign indicates that the reaction acts vertically upwards, opposite to the weight.<\/span><\/p>\n

                          \"ncert-class-11-solutions-physics-5-laws-motion-25\"Question: 22 A monkeys of mass 40 kg climb on a rope (Fig.) which can stand a maximum tension of 600 N.\u00a0<\/span><\/strong><\/p>\n

                          In which of the following cases will the rope break: the monkey<\/span><\/strong>
                          \n(a) climbs up with an acceleration of 6 ms<\/span><\/strong>-2<\/sup><\/span><\/strong>
                          \n(b) climbs down with an acceleration of 4 ms<\/span><\/strong>-2<\/sup><\/span><\/strong>
                          \n(c) climbs up with a uniform speed of 5 ms<\/span><\/strong>-1<\/sup><\/span><\/strong>
                          \n(d) falls down the rope nearly freely under gravity?<\/span><\/strong>
                          \nAnswer:<\/span><\/strong>\u00a0<\/span><\/strong>(a) When the monkey climbs up with acceleration a, <\/span><\/p>\n

                          then\u00a0<\/span>\"\"\u00a0so\u00a0<\/span>\"\"<\/p>\n

                          \"\"<\/p>\n

                          \"\"<\/p>\n

                          \"\"<\/p>\n

                          But rope can withstand with a force of 600N so will break.<\/span><\/p>\n

                          (b) When monkey is climbing down with acceleration, than\u00a0<\/span><\/p>\n

                          \"\"<\/p>\n

                          \"\"<\/p>\n

                          \"\"<\/p>\n

                          \"\"<\/p>\n

                          \"\"\u00a0The rope will not break.<\/span><\/p>\n

                          c. When monkey climb up with uniform speed\u00a0<\/span><\/p>\n

                          than<\/span>\"\".\u00a0<\/span><\/p>\n

                          The rope will not break.<\/span><\/p>\n

                          D, When the monkey is falling freely, it would be a state of weightlessness.\u00a0<\/span><\/p>\n

                          So, tension will be zero and the rope will not break.<\/span><\/p>\n

                          14. Example of Variable mass situation: Rocket Propulsion:<\/span><\/strong>\u00a0imp<\/sup><\/span><\/strong><\/p>\n

                          \"E:\\Pradeep\\Untitled6.png\"<\/p>\n

                           <\/p>\n

                           <\/p>\n

                           <\/p>\n

                           <\/p>\n

                           <\/p>\n

                           <\/p>\n

                           <\/p>\n

                          Suppose at\u00a0<\/span>\"\"\u00a0is mass of rocket including fuel and\u00a0<\/span>\"\"\u00a0is initial velocity of the rocket and at\u00a0<\/span>\"\"\u00a0is mass of rocket left and\u00a0<\/span>\"\"\u00a0is velocity accursed by the rocket.<\/span><\/p>\n

                          \u00a0<\/span>As<\/span>\"\"<\/p>\n

                          Now again suppose that in a small interval of time\u00a0<\/span>\"\"is small decrease in mass of rocket and\u00a0<\/span>\"\"\u00a0is corresponding increase in small increase in velocity and\u00a0<\/span>\"\"\u00a0is the velocity of exhaust gasses w.r.t. the earth.<\/span><\/p>\n

                          Now from conservation of linear momentum<\/span><\/p>\n

                          \"\"<\/p>\n

                          Here\u00a0<\/span>\"\"\u00a0sign indicate that gasses escape downward.<\/span><\/p>\n

                          \"\"\u00a0\u00a0<\/span>\"\"<\/p>\n

                          As\u00a0<\/span>\"\"\u00a0is small so can be neglect<\/span><\/p>\n

                          \"\"\u00a0\u00a0<\/span>\"\"<\/p>\n

                          \"\"<\/p>\n

                          Now suppose\u00a0<\/span>\"\"\u00a0is the relative velocity of burnt gas w.r.t. rocket then<\/span><\/p>\n

                          \"\"<\/p>\n

                          i.e.\u00a0<\/span>\"\"<\/p>\n

                          \"\"<\/p>\n

                          Using in (i)<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

                          Or<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

                          Integrating both sides we get<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                          \u00a0<\/span>\"\"<\/p>\n

                          \"\"<\/p>\n

                          \"\"<\/p>\n

                          \"\"<\/p>\n

                          Or<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

                          This is the velocity of rocket at any time\u00a0<\/span>\"\"<\/p>\n

                          Now at<\/span>\u00a0\u00a0<\/span>\"\"\u00a0\u00a0\u00a0<\/span><\/p>\n

                          \u00a0<\/span>\"\"\u00a0<\/span>\u00a0<\/span>\"\"<\/p>\n

                          \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Thus in the absence of any external force, the instantaneous velocity of the rocket depends on exhaust speed of burnt gas and mass of the rocket and burnt fuel.<\/span><\/p>\n

                          Burnt out speed of the rocket:-<\/span><\/strong><\/p>\n

                          The speed acquired by the rocket when whole of its fuel gets burnt is called brunt of speed of the rocket<\/span>.<\/span><\/p>\n

                          Let\u00a0<\/span>\"\"\u00a0is the residual mass of the rocket then burnt out speed\u00a0<\/span>\"\"\u00a0can be given as<\/span><\/p>\n

                          \"\"<\/p>\n

                          Burnt out speed of rocket is its maximum speed.<\/span><\/p>\n

                          Thrust on the rocket:-<\/span><\/strong><\/p>\n

                          The force with which a rocket moves upward is called thrust of the rocket.<\/span><\/p>\n

                          As we know<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

                          Dividing both sides by\u00a0<\/span>\"\"\u00a0we get\u00a0<\/span><\/p>\n

                          \"\"<\/p>\n

                          Or<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

                          Or\u00a0<\/span>\"\"<\/p>\n

                          Thus upward thrust of the rocket is equal to the product of the exhaust speed of the burnt gas and the rate of consumption of fuel at that instant.<\/span><\/p>\n

                          Question: 23 A rockets with a lift-off mass 20,000 kg are blasted upwards with an initial acceleration of 5.0 ms<\/span><\/strong>-2<\/sup><\/span><\/strong>. Calculate the initial thrust (force) of the blast.<\/span><\/strong>
                          \nAnswer:<\/span><\/strong> Here, \"\"<\/p>\n

                          Initial acceleration = 5 ms-2<\/sup><\/span>
                          \nclearly, the thrust should be such that it overcomes the force of gravity besides giving it an upward acceleration of 5 ms-2<\/sup><\/span>.<\/p>\n

                          Thus the force should produce a net acceleration of<\/p>\n

                          \"\".
                          \nSince, thrust = force = mass x acceleration<\/p>\n

                          so \"\"<\/p>\n

                          15. Equilibrium of Concurrent forces:-<\/span><\/strong><\/p>\n

                          \"C:\\Users\\SURENDER\\Desktop\\Untitled.png\"If a number of forces act on a body at a same point and the net unbalanced force is zero, the body will continue in its state of rest or of uniform motion along straight line then concurrent forces (forces acting on single point) are said to be in equilibrium.<\/span><\/p>\n

                          Suppose three concurrent forces\u00a0<\/span>\"\"\u00a0are acting on a body shown in fig.<\/span><\/p>\n

                          Here resultant of\u00a0<\/span>\"\"\u00a0and\u00a0<\/span>\"\"\u00a0is<\/span>\"\".\u00a0<\/span><\/p>\n

                          If the third force having equal magnitude as that of\u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

                          And act in opposite to<\/span>\u00a0\u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

                          Then\u00a0<\/span>\"\"<\/p>\n

                          Or<\/span>\u00a0<\/span>\"\"<\/p>\n

                          This is the required condition for equilibrium of the concurrent forces. If there are\u00a0<\/span>\"\"\u00a0forces then<\/span><\/p>\n

                          \"\"<\/p>\n

                          16. Lami’s Theorem:-<\/span><\/strong><\/p>\n

                          \"E:\\Pradeep\\Untitled7.png\"Suppose there concurrent forces\u00a0<\/span>\"\"\u00a0are acting on the body. Such that\u00a0<\/span>\u03b1<\/span>\u00a0is angle between<\/span>\"\",\u00a0<\/span>\u03b2<\/span>\u00a0is angle between\u00a0<\/span>\"\"\u03b3<\/span>\u00a0is angle between\u00a0<\/span>\"\"\u00a0as shown in fig.<\/span><\/p>\n

                          Then according to<\/span>\u00a0\u00a0<\/span>Lami’s theorem.<\/span><\/p>\n

                          At equilibrium<\/span>\u00a0\u00a0\u00a0<\/span><\/p>\n

                          \u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

                          Thus according to Lami’s theorem if three forces are acting on a particle in equilibrium then each force is proportional to the sine of angle between the other two<\/span>\u00a0\u00a0<\/span>forces.\u00a0<\/span><\/p>\n

                          Unit-3(B) Friction:<\/span><\/strong><\/p>\n

                          17. Friction:<\/span><\/strong><\/p>\n

                          An opposing force which comes into play when a body moves or tends to moves on the surface of another body is called frictional force.<\/span><\/p>\n

                          Explanation of Origin of Frictional Force:<\/span><\/strong><\/p>\n

                          The force of friction is due to the force of attraction between atoms and molecules of two surfaces at the point of actual contact.<\/span><\/p>\n

                            \n
                          1. Types of Friction:<\/span><\/li>\n<\/ol>\n

                            There are three types of friction<\/span><\/p>\n