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Physics Beyond 2000 Chapter 2 Newtons Laws of Motion.

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Presentation on theme: "Physics Beyond 2000 Chapter 2 Newtons Laws of Motion."— Presentation transcript:

1 Physics Beyond 2000 Chapter 2 Newtons Laws of Motion

2 Dynamics Concerned with the motion of bodies under the action of forces. Bodies are assumed to have inertia. Momentum and force.

3 Momentum = linear momentum of a body m = mass of the body = velocity of the body

4 Momentum = linear momentum of a body m = mass of the body = velocity of the body

5 Momentum = linear momentum of a body m = mass of the body = velocity of the body Note that is a vector quantity. Unit of is

6 Momentum = linear momentum of a body m = mass of the body = velocity of the body Note that is a vector quantity. Unit of is kg m s or N s.

7 Newtons First Law of Motion A body continues in a state of rest or uniform motion in a straight line unless it is acted upon by external forces

8 Newtons First Law of Motion A body continues in a state of rest or uniform motion in a straight line unless it is acted upon by external forces The body changes its state of motion under an external force.

9 Newtons First Law of Motion A body continues in a state of rest or uniform motion in a straight line unless it is acted upon by external forces The body does not change its state of motion when there is not any external force.

10 Newtons First Law of Motion A body continues in a state of rest or uniform motion in a straight line unless it is acted upon by external forces The body changes its state of motion under an external force.

11 Newtons First Law of Motion A body continues in a state of rest or uniform motion in a straight line unless it is acted upon by external forces. Linear air track –Vehicle without external force – Vehicle under constant force

12 Inertia and Mass Inertia is a property of matter that causes it to resist any change in its motion or to keep its state of motion. Mass of a body is a quantitative measure of its inertia. SI unit of mass: kg.

13 Newtons Second Law of Motion The rate of change of momentum of a body is proportional to and in the same direction as the resultant force (net force) that acts on it.

14 Newtons Second Law of Motion If the mass is a constant,

15 Newtons Second Law of Motion If the mass is a constant, where k is a proportional constant

16 Newtons Second Law of Motion If the mass is a constant, where k is a proportional constant In SI units, define 1 newton of force as the net force acting on the mass of 1 kg and producing an acceleration of 1 m s k = 1 in SI units. -2

17 Newtons Second Law of Motion If the mass is a constant, where k = 1 in SI units Note that the above equation is correct on condition that SI units are used. a F net m

18 Newtons Second Law of Motion No matter the mass is a constant or not, Note that the above equation is correct on condition that SI units are used.

19 Newtons Second Law of Motion If F = 0, mv = constant This is the case of Newtons First Law of Motion.

20 The origin of force Gravitational force –Attraction between two massive particles. Electromagnetic force –Electrostatic force: Force between two charged particles. –Electromagnetic force: Force on a moving charged particle in a magnetic field. Nuclear force –Force between two the particles of the nucleus.

21 Weight W = m.g It is a gravitational force. Use spring balance to measure the weight. g varies on earth. The measured value of g may be affected by the rotation of the planet.

22 Normal Contact Force The box is at rest on the ground. W N N = W

23 Normal Contact Force The box is at rest on the ground. W N N = W + F F

24 Normal Contact Force The box is at rest on the ground. W N N = W - F F

25 Newtons 3 rd Law of Motion If one body exerts a force on another, there is an equal and opposite force, called a reaction, exerted on the first body by the second. A B Exerted on A by B. Exerted on B by A.

26 Newtons 3 rd Law of Motion A B Exerted on A by B. Exerted on B by A. are action and reaction pair

27 Newtons 3 rd Law of Motion Are they action and reaction pair?

28 Feeling of Ones Weight A man is standing on a balance inside a lift.

29 Feeling of Ones Weight What are the forces acting on the man? weight W = mg Normal contact force N. This is the reading on the balance.

30 Feeling of Ones Weight The net force and the acceleration. Newtons 2 nd law of motion. N W net force F = N – W = m.a m = mass of the man a

31 Feeling of Ones Weight The lift is moving up/down at constant speed or at rest. a = 0 N – W = 0 N = W N W The reading on the balance is the weight of the man.

32 Feeling of Ones Weight Moving up with acceleration or Moving down with retardation. N – W = m.a N = W + m.a N W The reading on the balance is bigger than the weight of the man. a

33 Feeling of Ones Weight Moving up with retardation or Moving down with acceleration. W – N = m.a N = W - m.a N W The reading on the balance is less than the weight of the man. a

34 Feeling of Ones Weight Free falling. W – N = m.g N = W - m.g = 0 N W The reading on the balance is zero. weightless. g

35 Momentum and Impulse Impulse J = F.t –F is the force –t is the time for the force to act

36 Momentum and Impulse If a force F acts on an object of mass m for a time t and changes its velocity from u to v, prove that J = mv – mu (i.e. mv). m F u m F m F m F m F v

37 Momentum and Impulse If a force F acts on an object of mass m for a time t and changes its velocity from u to v, prove that J = mv – mu (i.e. mv). m F v

38 Impulse and F - t graph The area under F-t graph gives the impulse as long as the mass does not change. F t 0

39 Examples of Impulse Catching a baseball –Momentum of the baseball decreases to zero. –Increase the time of action and reduce the force.

40 Examples of Impulse Catching a baseball –Momentum of the baseball decreases to zero. –Increase the time of action and reduce the force. time of contact t momentum of the base ball

41 Examples of Impulse Catching a baseball time of action t momentum of the base ball t mu

42 Examples of Impulse Catching a baseball –Momentum of the baseball decreases to zero. –Increase the time of action and reduce the force. F F F

43 Examples of Impulse Striking a tennis ball Go to the search engine and type the word tennis science.

44 Examples of Impulse Use of a seat belt –Reduce the force on the passenger by prolonging the time of stopping the passenger on crash. Without seat belt With seat belt F t

45 Raindrops versus Hailstones Try to study the passage by yourself. Discuss with your classmates.

46 Example 1 Write the symbol for 0.14 kg. Write the symbol for 30 Write the symbol for 50 Write the symbol for the impulse. What is the formula connecting the above quantities? Note the directions!

47 Example 1 What is the formula connecting the force and the action time?

48 Example 1 Note the directions! Momentum of the ball time time of action 0

49 Collisions Studying different kinds of collisions. –Perfectly elastic collision. –Completely inelastic collision. –Collision in-between

50 Principle of conservation of momentum When bodies in a system interact, the total momentum remains constant, provided no external force acts on the system.

51 Principle of conservation of momentum When bodies in a system interact, the total momentum remains constant, provided no external force acts on the system. Before collision, –Total momentum = m1.u1 + m2.u2 m1 m2 u1 u2

52 Principle of conservation of momentum When bodies in a system interact, the total momentum remains constant, provided no external force acts on the system. Before collision, –Total momentum = m1.u1 + m2.u2 m1 u1 m2 u2 m1 u1 m2 u2 m1 u1 m2 u2 m1 u1 m2 u2 m1 u1 m2 u2 m1 u1 m2 u2

53 Principle of conservation of momentum When bodies in a system interact, the total momentum remains constant, provided no external force acts on the system. After collision, –Total momentum = m1.v1 + m2.v2 m1 v1 m2 v2 m1 v1 m2 v2 m1 v1 m2 v2 m1 v1 m2 v2

54 Principle of conservation of momentum When bodies in a system interact, the total momentum remains constant, provided no external force acts on the system. Without external force, m1.u1 + m2.u2 = m1.v1 + m2.v2 m1 v1 m2 v2 m1 u1 m2 u2

55 Principle of conservation of momentum The momentum of m1 m1 v1 m2 v2 m1 u1 m2 u2 m1.u1 m1v1 time time of action

56 Principle of conservation of momentum The momentum of m2 m1 v1 m2 v2 m1 u1 m2 u2 m2.v2 m2.u2 time time of action

57 Principle of conservation of momentum For N bodies in collision. Without external force, sum of momenta before = sum of momenta after

58 Collisions in 2-dimension The collision is not head-on. The collision is oblique. Resolve each momentum into 2 perpendicular components (x- and y- components). Without external force, the momenta is conserved along x-direction. Without external force, the momenta is conserved along y-direction.

59 Collisions in 2-dimension Right-angled fork Two equal masses in oblique and elastic collision. Before collision, one mass is stationary and the other mass is moving. After collision, they move out in directions perpendicular to each other.

60 Collisions in 2-dimension Right-angled fork Two equal masses in oblique and elastic collision. Before collision, one mass is stationary and the other mass is moving. After collision, they move out in directions perpendicular to each other.

61 Collisions in 2-dimension Right-angled fork Two equal masses in oblique and elastic collision. Before collision, one mass is stationary and the other mass is moving. After collision, they move out in directions perpendicular to each other.

62 Collisions in 2-dimension Right-angled fork Two equal masses in oblique and elastic collision. Before collision, one mass is stationary and the other mass is moving. After collision, they move out in directions perpendicular to each other.

63 Collisions in 2-dimension Right-angled fork Prove : After collision, they move out in directions perpendicular to each other. θ+φ= 90 0 θ ψ Hint: Use conservation of momentum. Use conservation of kinetic energy. mu mv 1 mv 2

64 Collisions in 2-dimension Right-angled fork Example: α-particle colliding with helium atom. θ+φ= 90 0 θ ψ mu mv 1 mv 2

65 Collisions in 2-dimension Two unequal masses in oblique and elastic collision. Before collision, one mass is stationary and the other mass is moving. θ ψ m1um1u m1v1m1v1 m2v2m2v2 If m 1 >m 2, then θ+ψ<90 o

66 Collisions in 2-dimension Two unequal masses in oblique and elastic collision. Before collision, one mass is stationary and the other mass is moving. θ ψ m1um1u m1v1m1v1 m2v2m2v2 If m 1 90 o

67 Friction To act along the common surface between two bodies in contact. To resist the relative motion (or tendency of relative motion) of two bodies. direction of motion direction of friction f

68 Friction To act along the common surface between two bodies in contact. To resist the relative motion (or tendency of relative motion) of two bodies. direction of motion direction of friction f

69 Friction Static friction : –The object is stationary. Kinetic friction : –The object is moving.

70 Static Friction To resist the tendency of relative motion of two bodies. Static friction has a maximum value, limiting frictional force f L. tendency of motion in this direction direction of static friction f F F = f. The object is stationary.

71 Static Friction Static friction has a maximum value, limiting frictional force f L. f L depends on the nature of the surface and normal reaction. f L = μ s.R where μ s is the coefficient of static friction. direction of static friction f F R= normal reaction

72 Kinetic Friction Kinetic frictional force f k is almost a constant. f k depends on the nature of the surface and normal reaction. f k = μ k.R where μ k is the coefficient of kinetic friction. f k < f L direction of kinetic friction fkfk F R= normal reaction direction of motion

73 Chang of friction Gradually increase the applied force F F f static kinetic f F 0 f = F fLfL fkfk

74 Example 3 Given: m = 2.0 kg, μ s = 1.2, F = W. Stationary. Find f. F W R f

75 Example 4 Given: m = 3 kg, F = 30N, a = 2 ms -2. Find f k and μ k. F W R fkfk a

76 Example 5 Gradually increase the angle of inclination.

77 Example 5 Express f in terms of W and θ when the object is still stationary. What is the maximum angle θ if coefficient of static friction is μ s ? W R f θ

78 Friction Cause of friction Reducing friction Role of friction Friction in a car

79 Spring and force constant Hookes law –The extension or compression of a spring is proportional to the force acting on it, provided it does not exceed the elastic limit. F = k.e –where k is the force constant of the spring –and e is the extension F e In equilibrium

80 Spring and force constant Example 6

81 Combination of springs In series In parallel k1k1 k2k2 k1k1 k2k2 F F

82 Combination of springs In series k1k1 k2k2 F e

83 Combination of springs In parallel k1k1 k2k2 F e F = (k 1 + k 2 ).e

84 Work and Energy Work is the transfer of energy.

85 Work W = F.s.cos F s

86 Work W = F.s.cos F s F F FFF

87 Work W = F.s.cos F s Note that F.cos is the component of F in the direction of s.

88 Work If F varies with s, F s

89 Sign of Work done Free falling –F is the gravitational force to do work. –Work done is positive. –The ball gains K.E. F s FFF

90 Sign of Work done Moving up an object at steady speed –Apply an upward force F to do work. –Work done by F is positive. –The object gains gravitational P.E. F s FFF

91 Sign of Work done Lowering down an object at steady speed –Apply an upward force F to do work. –Work done by F is negative. –The object loses gravitational P.E. F s FFF

92 Sign of Work done Holding an object. –Apply an upward force F. –Displacement s = 0 –Work done by F is zero. –The object neither gains nor loses energy. s = 0 F

93 Sign of Work done Stretching a spring unstretched

94 Sign of Work done Stretching a spring –Apply a force F to extend the spring. –Work done by F is positive. –The spring gains elastic P.E. unstretched F stretched s

95 Sign of Work done Releasing a stretched spring –A restoring force F acts on the spring. –Work done by F is negative. –The spring loses elastic P.E. stretched F

96 Sign of Work done Releasing a stretched spring –A force F acts on the mass. –Work done by F is positive. –The mass gains K.E. stretched unstretched s F

97 Work Example 7 Example 8

98 Forms of Energy Kinetic energy (K.E. or E k ) Potential energy –Gravitational –Elastic –Electrostatic Thermal and internal energy

99 Forms of Energy Radiant energy Chemical energy Nuclear energy Mass equivalent In microscopic scale, molecules possesses kinetic energy and electrostatic potential energy only.

100 Kinetic Energy m v

101 Kinetic energy Prove that the work done F.s is equal to the gain of kinetic energy s m F u = 0 m v

102 Gravitational potential energy

103 Prove that the work done F.s is equal to the gain of gravitational potential energy E p =mgh. Note that the motion is a steady one. F F s

104 Energy Example 9

105 Elastic potential energy F = k.e Find the work done to extend a spring for a length e. F unstretched

106 Elastic potential energy F = k.e Find the work done to extend a spring for a length e. F unstretched stretched e

107 Elastic potential energy F = k.e Find the work done to extend a spring for a length e. e F x F=k.x 0

108 Elastic potential energy F = k.e Find the work done to extend a spring for a length e. e F x F=k.x 0

109 Elastic potential energy F = k.e Find the work done to extend a spring for a length e. e F x F=k.x 0 This is also the area below the graph.

110 Elastic potential energy Example 10.


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