# Chapter 2 Newton’s Laws of Motion

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Chapter 2 Newton’s Laws of Motion
Physics Beyond 2000 Chapter 2 Newton’s Laws of Motion

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

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

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

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

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. -1

Newton’s 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

Newton’s 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.

Newton’s 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.

Newton’s 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.

Newton’s 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

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.

Newton’s 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.

Newton’s Second Law of Motion
If the mass is a constant,

Newton’s Second Law of Motion
If the mass is a constant, where k is a proportional constant

Newton’s 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

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

Newton’s 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.

Newton’s Second Law of Motion
If F = 0, mv = constant This is the case of Newton’s First Law of Motion.

The origin of force Gravitational force Electromagnetic 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.

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.

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

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

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

Newton’s 3rd 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.

Newton’s 3rd Law of Motion
are action and reaction pair A B Exerted on A by B. Exerted on B by A.

Newton’s 3rd Law of Motion
Are they action and reaction pair?

Feeling of One’s Weight
A man is standing on a balance inside a lift.

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

Feeling of One’s Weight
The net force and the acceleration. Newton’s 2nd law of motion. N a net force F = N – W = m.a m = mass of the man W

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

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

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

Feeling of One’s Weight
Free falling. W – N = m.g N = W - m.g = 0 N g The reading on the balance is zero.  weightless. W

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

Momentum and Impulse u m F v m F m F m F m F
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) . u m F v m F m F m F m F

Momentum and Impulse v F m
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) . v m F

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

Examples of Impulse http://www.exploratorium.com/sports/
Catching a baseball Momentum of the baseball decreases to zero. Increase the time of action and reduce the force.

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

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

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

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

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

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

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

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

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

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

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

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 u2 u1 m1 m2

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

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

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 u1 m2 u2 m1 v1 m2 v2

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

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

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

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.

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.

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.

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.

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.

Collisions in 2-dimension Right-angled fork
Prove : After collision, they move out in directions perpendicular to each other. θ+φ= 900 mv2 mu θ ψ mv1 Hint: Use conservation of momentum. Use conservation of kinetic energy.

Collisions in 2-dimension Right-angled fork
Example: α-particle colliding with helium atom. θ+φ= 900 mv2 mu θ ψ mv1

Collisions in 2-dimension
Two unequal masses in oblique and elastic collision. Before collision, one mass is stationary and the other mass is moving. m2v2 m1u θ ψ m1v1 If m1>m2, then θ+ψ<90o

Collisions in 2-dimension
Two unequal masses in oblique and elastic collision. Before collision, one mass is stationary and the other mass is moving. m2v2 m1u θ ψ m1v1 If m1<m2, then θ+ψ>90o

Friction f direction of motion direction of 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

Friction f direction of motion direction of 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 f direction of friction

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

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

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

Kinetic Friction R= normal reaction direction of motion F fk
Kinetic frictional force fk is almost a constant. fk depends on the nature of the surface and normal reaction. fk = μk .R where μk is the coefficient of kinetic friction. fk < fL direction of kinetic friction fk F R= normal reaction direction of motion

Chang of friction Gradually increase the applied force F
kinetic static fL fk f = F F f

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

Example 4 R a F fk W Given: F = 30N, a = 2 ms-2 . Find fk and μk.
m = 3 kg, F = 30N, a = 2 ms-2 . Find fk and μk. F W R fk a

Example 5 Gradually increase the angle of inclination.

Example 5 W R f θ 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 μs?

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

Spring and force constant
Hooke’s 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 In equilibrium F e

Spring and force constant
Example 6

Combination of springs
In series In parallel k1 k2 F k1 F k2

Combination of springs
In series k1 k2 F e

Combination of springs
k1 k2 F In parallel e F = (k1 + k2).e

Work and Energy Work is the transfer of energy.

Work W = F.s.cos F s

Work W = F.s.cos F F F F F F s

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

Work If F varies with s, F s

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

Sign of Work done F F F s F 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 F F s

Sign of Work done F F F s F 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 F F s

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

Sign of Work done Stretching a spring unstretched

Sign of Work done unstretched F s stretched 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 s stretched

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

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

Work Example 7 Example 8

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

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

Kinetic Energy m v

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

Gravitational potential energy

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

Energy Example 9

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

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

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

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

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

Elastic potential energy
Example 10.