{"id":2475,"date":"2023-05-27T14:15:44","date_gmt":"2023-05-27T14:15:44","guid":{"rendered":"https:\/\/vartmaaninstitutesirsa.com\/?page_id=2475"},"modified":"2024-06-30T04:05:07","modified_gmt":"2024-06-30T04:05:07","slug":"chapter-6-work-energy-and-power-physics-notes","status":"publish","type":"page","link":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/chapter-6-work-energy-and-power-physics-notes\/","title":{"rendered":"Chapter 6 Work, Energy and Power physics notes"},"content":{"rendered":"

Chapter 6 Work, Energy and Power physics notes<\/h1>\n

To download pdf of class 11 Chapter 6 Work, Energy and Power physics notes click on the link given below<\/p>\n

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\n

\n<\/div>\n

UNIT 4 WORK- POWER- ENERGY<\/span><\/strong><\/p>\n

1. WORK<\/span><\/strong><\/p>\n

\"Work\"Work is said to be done when a force applied on a body displace the body through a distance in the direction of force.<\/span><\/p>\n

Let us consider that force F displace the body from point A to point B through a distance S. than work done to displace the body is<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Suppose a force F is applied on a body at an angle \u03b8 with the horizontal and displace the body from point A to B through a distance S.<\/span><\/p>\n

Now resolving F into components<\/span><\/p>\n

\"\"\u00a0In direction of displacement<\/span><\/p>\n

\"\"\u00a0In perpendicular direction of displacement<\/span><\/p>\n

\"\"Clearly here, the motion is due to\u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

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

\"\"<\/p>\n

Thus, work done by a force is equal to dot product of force and displacement of the body.<\/span><\/p>\n

Dimensions and Unit of Work:-<\/span><\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/strong><\/p>\n

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

So [W] =[MLT<\/span>-2<\/sup><\/span>][L]<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

Or [W]= [ML<\/span><\/strong>2<\/sup><\/span><\/strong>T<\/span><\/strong>-2<\/sup><\/span><\/strong>]<\/span><\/strong><\/p>\n

Unit of work<\/span><\/strong><\/p>\n

There are of two types of unit of work\u00a0<\/span><\/p>\n

(i) Absolute unit of work:-<\/span><\/strong><\/p>\n

    \n
  • In SI the absolute unit of work is Joule.<\/span><\/strong><\/li>\n<\/ul>\n

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

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

    Thus, work done is said to be one joule if one Newton force displaces the body through one meter in its own direction.<\/span><\/p>\n

      \n
    • I<\/span><\/strong>n c,g,s system the absolute unit of work is erg.<\/span><\/strong><\/li>\n<\/ul>\n

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

      \"\"<\/p>\n

      Hence, work done is said to be one erg if one-dyne force displace a body to one-metre distance in its own direction. The relation between one joule and one erg is<\/span><\/p>\n

      1 J=10<\/span><\/strong>7<\/sup><\/span><\/strong>\u00a0erg<\/span><\/strong><\/p>\n

      Gravitational unit of work:-<\/span><\/strong><\/p>\n

      In SI the gravitational unit of work is kilogram metre<\/span><\/p>\n

      \"\"<\/p>\n

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

      Thus, work done is said to be one kg m if one kg wt force displaces the body through one-meter distance in its own direction.\u00a0<\/span><\/p>\n

      In cgs the gravitational unit of work is Gram centimeter\u00a0<\/span><\/p>\n

      \"\"<\/p>\n

      1g cm=980 erg\u00a0<\/span><\/strong><\/p>\n

      Thus, work done is said to be one g cm if one g wt force displaces the body through one-centimeter distance in its own direction.<\/span><\/p>\n

      Question 1 A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1.\u00a0<\/span><\/strong><\/p>\n

      Compute the\u00a0<\/span><\/strong>
      \n(a) Work done by the applied force in 10 s<\/span><\/strong>
      \n(b) Work done by friction in 10 s<\/span><\/strong>
      \n(c) Work done by the net force on the body in 10 s<\/span><\/strong>
      \n(d) Change in kinetic energy of the body in 10 s and interpret your results.<\/span><\/strong>
      \nAnswer:<\/span><\/strong>\u00a0<\/span><\/strong>(a) We know that\u00a0<\/span>\"\"<\/p>\n

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

      \"\"<\/p>\n

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

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

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

      \"\"<\/p>\n

      \"\"and distance traveled<\/span>\u00a0\u00a0<\/span>,\u00a0<\/span>\"\"<\/p>\n

      \"\"<\/p>\n

      \"\"So work done by applied force\u00a0<\/span>\"\"<\/p>\n

      \"\"<\/p>\n

      \"\"
      \n(b) Work done by friction\u00a0<\/span><\/p>\n

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

      \"\"(c) Work done by net force<\/span><\/p>\n

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

      \"\"(d) Change in the kinetic energy of the body = work done by the net force in 10 seconds = 635.04 J\u00a0<\/span><\/p>\n

      Question 2 A trolleys of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km\/h on a friction less track. After a while, sand starts leaking out of a hole on the trolley\u2019s floor at the rate of 0.05 kg s<\/span><\/strong>-1<\/sup><\/span><\/strong>.\u00a0<\/span><\/strong><\/p>\n

      What is the speed of the trolley after the entire sand bag is empty?<\/span><\/strong>
      \nAnswer:<\/span><\/strong>\u00a0<\/span><\/strong>\u00a0<\/span>The system of trolley and sandbag is moving with a uniform speed. Clearly, an external force is not acting upon the system. If the sand leaks out, even then no external force acts. Therefore, there shall be no change in the speed of the trolley.<\/span><\/p>\n

      2. Nature of work done:-<\/span><\/strong><\/p>\n

      i.\u00a0<\/span><\/strong>If\u00a0<\/span>\"\"<\/p>\n

      \"\"<\/p>\n

      When a body is lifted, the work done by the lifting force is positive.<\/span><\/p>\n

      When a spring is stretched, work done by the stretching force is positive.<\/span><\/p>\n

      \"\"ii.\u00a0<\/span><\/strong>If\u00a0<\/span>\"\"<\/p>\n

      \"\"<\/p>\n

      A person holding a heavy stone at rest is doing no work. I.e. S=0\u00a0<\/span><\/p>\n

      A coolie carrying a suitcase on his head is doing no work. I.e. \u03b8=90<\/span>0<\/sup><\/span><\/p>\n

      iii.\u00a0<\/span><\/strong>If\u00a0<\/span>\"\"<\/p>\n

      (a)When break are applied on a moving vehicle, the work done by the breaking force is negative.<\/span><\/p>\n

      (b) Work done by the frictional force is negative.<\/span><\/p>\n

      Q. 3 the sign of work done by a force on a body is important to understand, State carefully if the following quantities are positive or negative :<\/span><\/strong>
      \n(a) Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket,<\/span><\/strong>
      \n(b) Work done by gravitational force in the above case.<\/span><\/strong>
      \n(c) Work done by friction on a body sliding down an inclined plane.<\/span><\/strong>
      \n(d) Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity.<\/span><\/strong>
      \n(e) Work done by the resistive force of air on a vibrating pendulum in bringing it to rest.<\/span><\/strong>
      \nAnswer:<\/strong> Work done\"\"
      \n(a) Work done \u2018positive\u2019, because force is acting in the direction of displacement<\/p>\n

      i.e., \u03b8 = 0\u00b0.
      \n(b) Work done is negative, because force is acting against the displacement<\/p>\n

      i.e., \u03b8 = 180\u00b0.
      \n(c) Work done is negative, because force of friction is acting against the displacement<\/p>\n

      i.e., \u03b8=180\u00b0.
      \n(d) Work done is positive, because body moves in the direction of applied force<\/p>\n

      i.e., \u03b8= 0\u00b0.
      \n(e) Work done is negative, because the resistive force of air opposes the motion<\/p>\n

      i.e., \u03b8 = 180\u00b0.<\/p>\n

      \"ncert-solutions-class-11-physics-chapter-6-work-energy-power-4\"Question 4 : In Fig.(i), the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. (ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end.\u00a0<\/span><\/strong><\/p>\n

      In which case is the work done greater?<\/span><\/strong>
      \nAnswer:<\/span><\/strong>\u00a0In Fig. (i) Force is applied on the mass, by the man in vertically upward direction, but distance is moved along the horizontal.<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

      \"\"<\/p>\n

      \"\"<\/p>\n

      \"\"In Fig. (ii), force is applied along the horizontal and the distance moved is also along the horizontal. Therefore, \u03b8 = 0\u00b0.<\/span>
      \n\"\"\u00a0<\/span><\/p>\n

      \"\"<\/p>\n

      \"\"
      \nThus, work done in (ii) case is greater.<\/span><\/p>\n

      3. Work Done In Term of Rectangular Components:-<\/span><\/strong><\/p>\n

      In term of rectangular components the force and displacement can be given as\u00a0<\/span><\/p>\n

      \"\"<\/p>\n

      \u2234<\/span>\u00a0Work<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

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

      Question 5. A body constrained to move along the z-axis of a coordinate system is subject to a constant force<\/span><\/strong>\u00a0<\/span><\/strong>F given by<\/span><\/strong>\"\"\u00a0\u00a0<\/span>where i, j, k, are unit vectors along the x- y- and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis?<\/span><\/strong>
      \nAnswer:\u00a0<\/span><\/strong>since the body is displaced 4m along Z-axis only,<\/span><\/p>\n

      So displacement vector<\/span><\/p>\n

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

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

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

      \"\"<\/p>\n

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

      4. CONSERVATIVE AND NON CONSERVATIVE FORCES:-<\/span><\/strong>M.Imp<\/sup><\/span><\/strong><\/p>\n

      (a)<\/span><\/strong>\u00a0\u00a0\u00a0<\/span><\/strong>Conservative force:-<\/span><\/strong>\u00a0\u00a0\u00a0\u00a0<\/span><\/strong><\/p>\n

      A force is said to be conservative if the amount of work done by or against the force does not depend on the path followed by the body but depend only on the initial and final position of the body.\u00a0<\/span><\/p>\n

      E.g. gravitational force, electromagnetic force<\/span><\/p>\n

      (b)<\/span><\/strong>\u00a0\u00a0\u00a0<\/span><\/strong>Non-conservative force:-<\/span><\/strong>\u00a0\u00a0\u00a0\u00a0<\/span><\/strong><\/p>\n

      A force is said to be non-conservative if the amount of work done by or against the force depend on the path followed by the body.\u00a0<\/span><\/p>\n

      E.g. frictional force<\/span><\/p>\n

      5. Show that Gravitational Force is conservative in nature:-<\/span><\/strong><\/p>\n

      Suppose a body of mass M is raised to a height h by the following four steps.\u00a0<\/span><\/p>\n

      If equal work is done to move the body to same height than work done by the gravitational force is conservative in nature.<\/span><\/p>\n

      (i) As shown in fig. (1)\u00a0<\/span><\/p>\n

      As the body is raised vertically upward to a height h by the force\u00a0<\/span><\/p>\n

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

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

      \"\"<\/p>\n

      \"\"<\/p>\n

      \u2234\u00a0<\/span>\u00a0<\/span>\"\"<\/p>\n

      \"\"\u00a0\u00a0\u00a0<\/span>(1)<\/span><\/p>\n

      (ii) Again consider Fig (2)\u00a0<\/span><\/p>\n

      Here body moves upward stepwise through steps I, II &III at a height\u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

      Here when body moves horizontal then work done is zero and the total work done is equal to the total vertical distance covered by the body,\u00a0<\/span><\/p>\n

      Hence\u00a0<\/span>\"\"\u00a0\u00a0\u00a0<\/span>(2)\u00a0<\/span><\/p>\n

      Now as shown in fig (3).here body moves at an inclined plane inclined at an angle<\/span>\u00a0\u00a0<\/span>\u03b8<\/span><\/p>\n

      Here clearly\u00a0<\/span>\"\"along AB<\/span>\u00a0<\/span><\/p>\n

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

      \"\"<\/p>\n

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

      \"\"<\/p>\n

      Now as shown in fig (4) here body moves in zig- zag path .this path can be assumed to form from many infinite small horizontal and vertical segments.<\/span><\/p>\n

      Here also total work done is equal to vertical segments\u00a0<\/span><\/p>\n

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

      Hence, we can say that work done by the gravitational force on a body is independent of path, so it is conservative in nature.<\/span><\/em><\/p>\n

      Question 6. A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 ms<\/span><\/strong>-1<\/sup><\/span><\/strong>\u00a0<\/span><\/strong>?<\/span><\/strong>
      \nAnswer:<\/span><\/strong>\u00a0<\/span><\/strong>Here,.<\/span>\"\"
      \nDistance moved in each half of the journey,<\/span><\/p>\n

      \"\"
      \nDensity of water,\u00a0<\/span>\"\"\u00a0\u00a0\u00a0<\/span><\/p>\n

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

      \"\"<\/p>\n

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

      \"\"<\/p>\n

      \"\"
      \nNote: Whether the drop moves with decreasing acceleration or with uniform speed work done by the gravitational force on the drop remains the same. If there is no resistive forces, energy of drop on reaching the ground.<\/span>
      \n\"\"\u00a0<\/span><\/p>\n

      \"\"<\/p>\n

      \"\"
      \nActual energy,<\/span>\"\"<\/p>\n

      \"\"<\/p>\n

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

      Work done by the resistive forces,<\/span><\/p>\n

      \"\"<\/p>\n

      \"\"<\/p>\n

      \"\"<\/p>\n

      \u00a0<\/span><\/strong>6. ENERGY<\/span><\/strong><\/p>\n

      Energy is defined as ability of a body to do work. Energy can be measured by the total amount of work that a body can do.<\/span><\/p>\n

      Unit of energy is same as that of work i.e. joule\u00a0<\/span><\/p>\n

      7. TYPES OF ENERGY:-<\/span><\/strong><\/p>\n

      Energy of is of many type i.e. solar energy, wind energy, electrical energy, mechanical energy, in this chapter we will discuss only mechanical energy.<\/span><\/p>\n

      8. MECHENICAL ENERGY<\/span><\/strong><\/p>\n

      Mechanical energy is defined as sum of potential energy and kinetic energy<\/span><\/p>\n

      (<\/span><\/strong>a)KINETIC ENERGY<\/span><\/strong><\/p>\n

      The energy possess by a body due to its motion is called kinetic energy.<\/span><\/p>\n

      E.g. flowing water, moving vehicle possess kinetic energy\u00a0<\/span><\/p>\n

      Expression for kinetic energy<\/span><\/strong>\u00a0:-\u00a0<\/span><\/strong><\/p>\n

      let us consider a body is in rest initially, now a force F is applied on the body due to which it moves a distance S with v velocity then<\/span><\/p>\n

      As v<\/span>2<\/sup><\/span>-u<\/span>2<\/sup><\/span>=2aS<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

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

      So<\/span>\u00a0\u00a0\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0<\/span>(1)<\/span><\/p>\n

      Also<\/span>\u00a0\u00a0<\/span>\"\"\u00a0\u00a0\u00a0<\/span>(2)<\/span>\u00a0\u00a0<\/span><\/p>\n

      From equation (1) and (2) as<\/span>\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

      And this work done is stored as kinetic energy of the body.\u00a0<\/span><\/p>\n

      So kinetic energy =<\/span>\"\"\u00a0<\/span><\/p>\n

      Relation between K.E. and linear momentum:-\u00a0<\/span><\/strong><\/p>\n

      As we know, the kinetic energy of the body is<\/span>\u00a0\u00a0<\/span><\/p>\n

      \"\"<\/p>\n

      Now dividing and multiplying the above equation by m in RHS, we get<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

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

      Therefore, we obtain that\u00a0<\/span><\/p>\n

      K.E<\/span>\u00a0\u00a0<\/span>\u221d<\/span>\u00a0square of linear momentum<\/span><\/p>\n

      Special cases\u00a0<\/span><\/strong><\/p>\n

      (1) If P =constant, K.E\u00a0<\/span>\"\"<\/p>\n

      (ii)<\/span>\u00a0\u00a0\u00a0<\/span>If K.E =constant, P<\/span>2<\/sup><\/span>\u00a0\u00a0\u00a0<\/span>\u221d<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

      (iii)<\/span>\u00a0\u00a0<\/span>If m=constant,<\/span>\u00a0\u00a0\u00a0<\/span>P\u00a0<\/span>\u221d<\/span>\u00a0<\/span>\"\"\u00a0as shown in fig respectively.<\/span><\/p>\n

      Question 7. An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton? Obtain the ratio of their speeds, (electron mass = 9.11 x 10<\/span><\/strong>-31<\/sup><\/span><\/strong>\u00a0<\/span><\/strong>kg, proton mass = 1.67 x 10<\/span><\/strong>-27<\/sup><\/span><\/strong>\u00a0<\/span><\/strong>kg, 1 eV= 1.60 x 10<\/span><\/strong>19<\/sup><\/span><\/strong>J).<\/span><\/strong>
      \nAnswer:<\/span>\"\"<\/p>\n

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

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

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

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

      \"\"<\/p>\n

      \"\"<\/p>\n

      \"\"<\/p>\n

      Thus, electron is travelling faster.<\/span><\/p>\n

      (b) POTENTIAL ENERGY<\/span><\/strong><\/p>\n

      The energy possess by a body due to its position is called potential energy\u00a0<\/span><\/p>\n

      E.g. the water stored in a dam possesses potential energy which can be converted in to kinetic energy. A stretched spring possess potential energy.<\/span><\/p>\n

      Gravitational potential energy<\/span><\/strong><\/p>\n

      \u00a0<\/span><\/strong>It<\/span>\u00a0<\/span><\/strong>is due to position above the surface of earth<\/span>.<\/span><\/strong><\/p>\n

      Elastic potential energy\u00a0<\/span><\/strong><\/p>\n

      It is due to compression and stretching of body (I,e-spring)\u00a0<\/span><\/p>\n

      (i) Expression for gravitational potential energy:-\u00a0<\/span><\/strong><\/p>\n

      Let us consider a body of mass m is raised to a height h, by applying a force F=mg in upward direction, then work done by the force is<\/span><\/p>\n

      \"\"<\/p>\n

      \"\"<\/p>\n

      \"\"<\/p>\n

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

      \"\"<\/p>\n

      Work done by the gravitational force is\u00a0<\/span><\/p>\n

      \"\"<\/p>\n

      Potential energy is of two types:-<\/span><\/strong><\/p>\n

      (ii)POTENTIAL ENERGY OF A SPRING (elastic potential energy)<\/span><\/strong><\/p>\n

      When a spring is stretched or compressed from its normal position (x=0) by a small distance x, then a restoring force is produced in the spring to bring it to normal position.\u00a0<\/span><\/p>\n

      \"\"The restoring force so produced is proportional to the displacement x in opposite direction i.e.<\/span><\/p>\n

      \"\"<\/p>\n

      \"\"<\/p>\n

      Where k is called spring constant and the above relation is called Hooks Law.<\/span><\/p>\n

      Now work done to move the small distance\u00a0<\/span>\"\"\u00a0is<\/span>\u00a0\u00a0<\/span><\/p>\n

      \"\"<\/p>\n

      total work done to move the spring x against restoring force is distance is<\/span>\u00a0\u00a0<\/span><\/p>\n

      \"\"<\/p>\n

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

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

      This work is stored in the body in form of potential energy.\u00a0<\/span><\/p>\n

      So the potential energy of the spring is\u00a0<\/span><\/p>\n

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

      External force is just equal and opposite of restoring force\u00a0<\/span><\/p>\n

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

      hence potential energy of a spring is defined as the energy possessed due to its compression and expansion.<\/span><\/p>\n

      Spring constant:-\u00a0<\/span><\/strong><\/p>\n

      Spring constant is defined as restoring force per unit displacement of the spring<\/span>\u00a0\u00a0<\/span><\/p>\n

      \"\"<\/p>\n

      Question<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>8. A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with a uniform speed of 7 ms<\/span><\/strong>-1<\/sup><\/span><\/strong>. It hits the floor of the elevator (length of elevator = 3 m) and does not rebound. What is the heat produced by the impact? Would your answer be different if the elevator were stationary ?<\/span><\/strong>
      \nAnswer:<\/span><\/strong>\u00a0<\/span><\/strong>\u00a0<\/span><\/p>\n

      P.E. of bolt\u00a0<\/span><\/p>\n

      = mgh\u00a0<\/span><\/p>\n

      = 0.3\u00a0<\/span>\"\"9.8\u00a0<\/span>\"\"\u00a03\u00a0<\/span><\/p>\n

      = 8.82 J<\/span>
      \nThe bolt does not rebound.\u00a0<\/span><\/p>\n

      So the whole of the energy is converted into heat.<\/span><\/p>\n

      Since the value of acceleration due to gravity is the same in all-inertial system, therefore the answer will not change even if the elevator is stationary.<\/span><\/p>\n

      Question 9 A 1 kg block situated on a rough incline is connected to a spring with spring constant 100 Nm<\/span><\/strong>-1<\/sup><\/span><\/strong>\u00a0<\/span><\/strong>as shown in Figure. The block is released from rest with the spring in the unstretched position. The block moves 10 cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline.\u00a0<\/span><\/strong><\/p>\n

      Assume that the spring has negligible mass and the pulley is friction less.<\/span><\/strong><\/p>\n

      Answer: from the above figure\u00a0<\/span><\/p>\n

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

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

      \"\"<\/p>\n

      \"ncert-solutions-class-11-physics-chapter-6-work-energy-power-18\"Net force on the block down the incline<\/span><\/p>\n

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

      \"\"<\/p>\n

      \"\"<\/p>\n

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

      In equilibrium work done = potential energy of stretched spring<\/span><\/p>\n

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

      =<\/span>\u00a0\u00a0<\/span>2<\/span>\"\"<\/p>\n

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

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

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

      8. CONGERVATIVE NATURE OF ENERGY<\/span><\/strong>\u00a0M.Imp<\/sup><\/span><\/strong><\/p>\n

      Law of conservation of energy:-<\/span><\/strong><\/p>\n

      Energy can neither be created nor be destroyed but it can be converted from one form to another form.\u00a0<\/span><\/p>\n

      Proof conservation of energy of a freely falling body:-<\/span><\/strong><\/p>\n

      The mechanical energy (K.E.+P.E.) is constant in case of freely falling body<\/span>.\u00a0<\/span><\/strong><\/p>\n

      Let us consider the body is initially at rest at a height h from ground\u00a0<\/span><\/p>\n

      Then at point A\u00a0<\/span><\/p>\n

      Kinetic Energy =<\/span>\"\"=0<\/span><\/em>\u00a0\u00a0\u00a0<\/span><\/p>\n

      And<\/span>\u00a0\u00a0\u00a0<\/span>Potential Energy = mgh<\/span><\/p>\n

      So the mechanical energy at\u00a0<\/span>\"\"<\/p>\n

      \"\"<\/p>\n

      \"\"<\/p>\n

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

      \"\"Now let body fall freely from A to B with velocity v<\/span>1<\/sub><\/span><\/p>\n

      such that A B = x.\u00a0<\/span><\/p>\n

      Therefore, the height of B from the ground is (h-x)<\/span><\/p>\n

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

      \"\"<\/p>\n

      \"\"<\/p>\n

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

      \"\"<\/p>\n

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

      Hence the mechanical energy at point B\u00a0<\/span><\/p>\n

      \"\"<\/p>\n

      Mechanical Energy<\/span><\/p>\n

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

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

      Now let the body just reaches the ground point C so\u00a0<\/span>\"\"<\/p>\n

      Here potential Energy<\/span><\/p>\n

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

      Let v<\/span>2<\/sub><\/span>\u00a0is the velocity of the body at just reaching the point C<\/span><\/p>\n

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

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

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

      Hence mechanical energy at point C\u00a0<\/span><\/p>\n

      \"\"<\/p>\n

      \"\"<\/p>\n

      \"\"\u2026 (3)<\/span><\/p>\n

      From equation (1), (2) and (3), it is clear that total mechanical energy of a body during the free fall of a body under gravity remain constant<\/span><\/p>\n

      When body reaches the ground, the kinetic energy is converted into heat energy and sound energy<\/span><\/p>\n

      Question 10 : An artificial satellite orbiting around<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>earth loses its energy gradually due to atmospheric resistance. Why then does its speed increase progressively as it comes closer and closer to the earth?<\/span><\/strong>
      \nAnswer: As an artificial satellite gradually loses its energy due to dissipation against atmospheric resistance, its potential decreases rapidly. As a result, kinetic energy of satellite slightly increases i.e., its speed increases progressively.<\/span><\/p>\n

      Question 11. A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated,\u00a0<\/span><\/strong><\/p>\n

      (a) How much work does she do against the gravitational force?<\/span><\/strong><\/p>\n

      (b) Fat supplies 3.8 x 10<\/span><\/strong>7<\/sup><\/span><\/strong>J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?<\/span><\/strong>
      \nAnswer:<\/span><\/strong>\u00a0<\/span><\/strong>Here<\/span>\"\"
      \n(a) work done against gravitational force.\u00a0<\/span><\/p>\n

      \"\"<\/p>\n

      \"\"<\/p>\n

      \"\"
      \n(b) Mechanical energy supplied by 1 kg of fat\u00a0<\/span><\/p>\n

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

      \"\"
      \n.-. Fat used up by the dieter<\/span><\/p>\n

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

      \"\"<\/p>\n

      Question 12. A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 ms<\/span><\/strong>-1<\/sup><\/span><\/strong>\u00a0<\/span><\/strong>?<\/span><\/strong>
      \nAnswer:<\/span><\/strong>\u00a0<\/span><\/strong>Here,.<\/span>\"\"
      \nDistance moved in each half of the journey,.<\/span><\/p>\n

      \"\"
      \nDensity of water,\u00a0<\/span>\"\"\u00a0\u00a0\u00a0<\/span><\/p>\n

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

      \"\"<\/p>\n

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

      \"\"<\/p>\n

      \"\"<\/p>\n

      Note: Whether the drop moves with decreasing acceleration or with uniform speed, work done by the gravitational force on the drop remains the same\u00a0<\/span><\/p>\n

      If there was no resistive forces, energy of drop on reaching the ground<\/span>
      \n\"\"\u00a0<\/span><\/p>\n

      \"\"<\/p>\n

      \"\"
      \nActual energy,<\/span>\"\"<\/p>\n

      \"\"<\/p>\n

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

      Work done by the resistive forces,<\/span><\/p>\n

      \"\"<\/p>\n

      \"\"<\/p>\n

      \"\"<\/p>\n

      9. VIBRATION OF A SIMPLE PENDULUM:-<\/span><\/strong><\/p>\n

      \"C:\\Users\\vartmaan\\Downloads\\main-qimg-cc7d21ae3004d2e6f7e4ef8003a14274.jpg\"Let us consider initially bob of pendulum is released from B.<\/span> \u00a0<\/span>At point B, initially the P.E. is maximum now on releasing from B its potential energy decrease and K.E increases.<\/span><\/p>\n

      At point A its K.E becomes maximum. Now when bob moves upward towards points C again its potential energy increases and Kinetic energy decreases now at point C again its potential energy becomes maximum.\u00a0<\/span><\/p>\n

      Hence during the complete cycle of pendulum there is conservation of energy.<\/span><\/p>\n

      Question 13 The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance?<\/span><\/strong>
      \nAnswer:<\/span><\/strong>\u00a0<\/span><\/strong>\u00a0<\/span>On releasing the bob of pendulum from horizontal position, it falls vertically downward by a distance equal to length of pendulum\u00a0<\/span><\/p>\n

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

      As 5% of loss in P.E. is dissipated against air resistance, the balance 95% energy is transformed into K.E.\u00a0<\/span><\/p>\n

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

      \"\"<\/p>\n

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

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

      10. Different forms of energy:-<\/span><\/strong><\/p>\n

      Heat energy<\/span><\/strong>:-<\/span><\/p>\n

      The energy possessed by the molecules due to their random motion is called heat energy.<\/span><\/p>\n

      Internal energy: –<\/span><\/strong>\u00a0<\/span><\/p>\n

      The energy possessed by the body due to temperature and intermolecular forces is called internal energy<\/span><\/p>\n

      Electrical energy: –<\/span><\/strong>\u00a0<\/span><\/p>\n

      The energy possessed by the body due to motion of charge particles called electrical energy.<\/span><\/p>\n

      Nuclear energy: –<\/span><\/strong>\u00a0<\/span><\/p>\n

      The energy required to hold the nucleons inside a nucleus is called nuclear energy.\u00a0<\/span><\/p>\n

      Transformation of Energy:-<\/span><\/strong><\/p>\n

      The phenomenon of change of one form of energy into another form is called transformation of energy.\u00a0<\/span><\/p>\n

      For example In a dam the potential energy of the water due to its height is converted into kinetic energy then into electrical energy with the help of a turbine.<\/span><\/p>\n

      In an electric bulb, the electric energy is converted into light energy and heat energy.<\/span><\/p>\n

      In a thermal power plant, the chemical energy stored in coal is converted into electric energy with the help of a turbine.<\/span><\/p>\n

      Hence, one form of energy can be converted into another form.<\/span><\/p>\n

      11. VARIATION OF MASS WITH VELOCITY:-<\/span><\/strong><\/p>\n

      According to Einstein, the variation of mass of a body with velocity is given by\u00a0<\/span><\/p>\n

      \"\"<\/p>\n

      Where m<\/span>o<\/sub><\/span>\u00a0is the rest mass of the body v is the velocity of the body and c=3\u00d710<\/span>8<\/sup><\/span>\u00a0ms<\/span>-1<\/sup><\/span>\u00a0is the velocity of light. Let the body have velocity v equal to velocity of light\u00a0<\/span><\/p>\n

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

      \"\"<\/p>\n

      \"\"<\/p>\n

      I.e.<\/span>\u00a0\u00a0<\/span>A body having \u221e mass means acceleration produced in the body =0<\/span>\u00a0\u00a0\u00a0<\/span><\/p>\n

      I.e.<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>\"\"=0<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

      Thus, no material particle can achieve velocity greater than velocity of light.<\/span><\/p>\n

      12. EINSTEIN\u2019S MASS-ENERGY RELATIONSHIP:-<\/span><\/strong><\/p>\n

      According to Einstein\u2019s mass and energy are related to each other as\u00a0<\/span><\/p>\n

      E=mc<\/span>2<\/sup><\/span><\/p>\n

      Hence, mass of a body can be converted into energy and energy can be converted into mass.<\/span><\/p>\n

      13. WORK ENERGY THEOREM\u00a0<\/span><\/strong>M.Imp<\/sup><\/span><\/strong><\/p>\n

      We know that small amount of work done to move a body through small distance\u00a0<\/span>\"\"\u00a0may be given as<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

      \"\"<\/p>\n

      Now to calculate total work done to move body from point A to B may be given as\u00a0<\/span><\/p>\n

      \"\"<\/p>\n

      \"\"<\/p>\n

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

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

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

      So using<\/span>\u00a0\u00a0<\/span>in (1)<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

      \"\"<\/p>\n

      \"\"<\/p>\n

      \"\"<\/p>\n

      \"\"<\/p>\n

      \"\"<\/p>\n

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

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

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

      Hence, work done is equal to change in kinetic energy of a body this relationship is called work energy theorem.<\/span><\/p>\n

      14. POWER<\/span><\/strong>\u00a0M.Imp<\/sup><\/span><\/strong><\/p>\n

      Power is defined as the rate at which work is done. It is a scalar quantity\u00a0<\/span><\/p>\n

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

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

      \"\"<\/p>\n

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

      Hence power may also be defined as the dot product of the force and the velocity.<\/span><\/p>\n

      Dimensional formula of power can be given as\u00a0<\/span><\/p>\n

      \"\"<\/p>\n

      Unit of power: –<\/span><\/strong>\u00a0<\/span><\/strong>The S.I. unit of power is Watt\u00a0<\/span><\/p>\n

      \"\"<\/p>\n

      Hence, power is said to be one Watt if one Joule work is done in one second.\u00a0<\/span><\/p>\n

      Power is also measured in\u00a0<\/span><\/p>\n

      1 kilowatt =1000 watt<\/span><\/p>\n

      1 megawatt=10<\/span>6<\/sup><\/span>\u00a0watt<\/span><\/p>\n

      1 horse power =746 watt (this unit is used in engineering)<\/span><\/p>\n

      The cgs unit of power is erg\/sec as 1watt =10<\/span>7<\/sup><\/span>\u00a0erg\/sec\u00a0<\/span><\/p>\n

      Question 14 a bodies is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to<\/span><\/strong>\u00a0<\/span><\/p>\n

      (i)<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>t<\/span><\/strong>1\/2\u00a0<\/sup><\/span><\/strong>\u00a0<\/sup><\/span><\/strong>(ii)<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>t (iii)<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>t<\/span><\/strong>3\/2<\/sup><\/span><\/strong>\u00a0<\/span><\/strong>\u00a0(iv)<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>t<\/span><\/strong>2<\/sup><\/span><\/strong>
      \nAnswer:<\/span><\/strong>\u00a0<\/span><\/strong>From\u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

      We get\u00a0<\/span>\"\"as power,\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

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

      \"\"
      \nSince m and an are constants,\u00a0<\/span><\/p>\n

      therefore, P \u03b1 t<\/span><\/p>\n

      Question 15. A body is moving unidirectional under the influence of a source of constant power.\u00a0<\/span><\/strong><\/p>\n

      Its displacement in time t is proportional to<\/span><\/strong><\/p>\n

      \u00a0<\/span><\/strong>(i)<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>t<\/span><\/strong>1\/2<\/sup><\/span><\/strong>\u00a0<\/sup><\/span><\/strong>\u00a0<\/sup><\/span><\/strong>(ii)<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>t (iii)<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>t<\/span><\/strong>3\/2<\/sup><\/span><\/strong>\u00a0<\/span><\/strong>(iv)<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>t<\/span><\/strong>2<\/sup><\/span><\/strong>
      \nAnswer:<\/span><\/strong>\u00a0<\/span><\/strong>(ii)<\/span>\"\"<\/p>\n

      Dimensional formula of power\u00a0<\/span>\"\"<\/p>\n

      As power is constant so\u00a0<\/span>\"\"<\/p>\n

      \"\"<\/p>\n

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

      Question 16. A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m<\/span><\/strong>3<\/sup><\/span><\/strong>in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump?<\/span><\/strong>
      \nAnswer:<\/span><\/strong>\u00a0<\/span><\/strong>\u00a0<\/span>Here,\u00a0<\/span>\"\"<\/p>\n

      \"\"<\/p>\n

      \"\"As the density of water\u00a0<\/span><\/p>\n

      =\u00a0<\/span>\"\"\u00a0<\/sup><\/span>
      \nMass of water pumped,\u00a0<\/span>\"\"<\/p>\n

      \"\"<\/p>\n

      \"\"
      \nActual power consumed or output power<\/span><\/p>\n

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

      \"\"<\/p>\n

      \u00a0<\/span>\"\"If P<\/span>i<\/sub><\/span>\u00a0is input power (required), then as<\/span><\/p>\n

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

      \"\"<\/p>\n

      15. COLLISION<\/span><\/strong><\/p>\n

      The term collision refers to the interaction between to body or two particles due to which the direction of the magnitude of the colliding particles changes<\/span><\/p>\n

      Types of collision<\/span><\/strong><\/p>\n

      (a)Perfectly elastic collision:-\u00a0<\/span><\/strong><\/p>\n

      A collision between two particles is said to be perfectly elastic if both linear momentum and velocity of the system remain conserved.<\/span><\/p>\n

      e.g. collision between atomic particles and collision between two glass ball are nearly perfectly elastic collision<\/span><\/p>\n

      (b)Inelastic collision:-\u00a0<\/span><\/strong><\/p>\n

      A collision is said to be inelastic if the linear momentum of the system remain conserved but its kinetic energy is not conserved\u00a0<\/span><\/p>\n

      Perfectly inelastic collision:-<\/span><\/strong><\/p>\n

      A collision is said to be perfectly inelastic collision if the two bodies after collision stick together and move as one body. In this case linear momentum remain conserved\u00a0<\/span><\/p>\n

      E.g. when a bullet hit the wooden block and move with the block is<\/span>\u00a0\u00a0<\/span>a example of nearly perfectly inelastic collision\u00a0<\/span><\/p>\n

      16. ELASTIC COLLISION IN ONE DIMENTIONTAL MOTION<\/span><\/strong>\u00a0M.Imp<\/sup><\/span><\/strong><\/p>\n

      Let us consider two bodies of mass m<\/span>1<\/sub><\/span>\u00a0and m<\/span>2<\/sub><\/span>\u00a0collide with each other having initial velocity u<\/span>1<\/sub><\/span>\u00a0and u<\/span>2<\/sub><\/span>\u00a0and final velocity v<\/span>1<\/sub><\/span>\u00a0and v<\/span>2<\/sub><\/span> respectively.<\/span><\/p>\n

      Then total linear momentum before collision<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

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

      \"C:\\Users\\vartmaan\\Downloads\\20190313-164251-orig_orig.jpg\"<\/p>\n

      And total momentum after collision<\/span><\/p>\n

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

      Now according to law of conservation of linear momentum<\/span><\/p>\n

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

      Or\u00a0<\/span>\"\"\u00a0(1)<\/span><\/p>\n

      Also as collision is elastic in nature so kinetic energy is also conserved in nature i.e<\/span><\/p>\n

      \"\"<\/p>\n

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

      Or<\/span>\u00a0\u00a0<\/span>\"\"\u00a0\u00a0\u00a0<\/span>(2)<\/span><\/p>\n

      Now dividing eqns. (2) by (1). We get<\/span><\/p>\n

      \"\"<\/p>\n

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

      Or<\/span>\u00a0\u00a0\u00a0<\/span>\"\"\u00a0\u00a0\u00a0<\/span>(3)<\/span><\/p>\n

      I.e. inelastic one dimension motion relative velocity of approach is equal to relative velocity of separation.<\/span><\/p>\n

      Now velocity after collision:-\u00a0<\/span><\/strong><\/p>\n

      From equation (3)<\/span>\u00a0\u00a0\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0<\/span>(4)<\/span><\/p>\n

      Substituting value of v<\/span>2<\/sub><\/span>\u00a0in equation (1), we get<\/span><\/p>\n

      \"\"<\/p>\n

      \"\"<\/p>\n

      \"\"<\/p>\n

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

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

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

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

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

      Special case\u00a0<\/span><\/p>\n

      When the body have equal mass\u00a0<\/span><\/p>\n

      I,e<\/span>\u00a0\u00a0<\/span>\"\"\u00a0(say)<\/span><\/p>\n

      Than from (5)\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

      Also from (6)\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

      Hence in one dimensional elastic collision two bodies having equal masses changes their velocities after collision.<\/span><\/p>\n

      (ii) When body B is initially at rest\u00a0<\/span><\/p>\n

      I,e<\/span>\u00a0\u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

      Then eqns. (5) and (6) becomes<\/span><\/p>\n

      \"\"<\/p>\n

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

      Here three sub cases arise as given below.<\/span><\/p>\n

      If both the bodies have same masses (m<\/span>1<\/sub><\/span>=m<\/span>2<\/sub><\/span>) then from eqns. (7) & (8)\u00a0<\/span><\/p>\n

      We have v<\/span>1<\/sub><\/span>=0 & v<\/span>2<\/sub><\/span>=u<\/span>1<\/sub><\/span>.\u00a0<\/span><\/p>\n

      Hence, in one-dimensional elastic collision if the bodies have equal masses then after collision the interchange there velocities.<\/span><\/p>\n

      (ii) If body B is lighter than body A\u00a0<\/span><\/p>\n

      I,e<\/span>\u00a0\u00a0<\/span>m<\/span>2<\/sub><\/span>< m<\/span>1<\/sub><\/span>\u00a0then m<\/span>2<\/sub><\/span>\u00a0can be neglected<\/span>\u00a0\u00a0<\/span>eqns. (7) and (8) becomes<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

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

      Hence if a heavy body collides with a lighter body then there is no change in the velocity of heavy body but the velocity of lighter body becomes twice the velocity of heavy body.\u00a0<\/span><\/p>\n

      (iii) When body A is lighter then body B\u00a0<\/span><\/p>\n

      I,e m<\/span>1<\/sub><\/span><<m<\/span>2<\/sub><\/span>. Then eqns. (7) & (8) becomes<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

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

      Hence, when a lighter body collides with a heavy body at rest then there is no change in heavy body and the speed of lighter body reverses.<\/span>\u00a0\u00a0<\/span><\/p>\n

      Coefficient of Restitution:-\u00a0<\/span><\/strong>M.Imp<\/sup><\/span><\/strong><\/p>\n

      It is defined as the ratio of velocity of separation after collision to the velocity of approach before collision; it is represented by e .<\/span><\/p>\n

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