Physics Beyond 2000 Chapter 2 Newtons 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

vmp

.p

v

Momentum




vmp

.p

v

Note that is a vector quantity.Unit of is

p

p

Momentum




vmp

.p

v

Note that is a vector quantity.Unit of is kg m s or N s.

p

p -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



The body changes its state of motion underan external force.



The body does not change its state of motionwhen there is not any external force.



The body changes its state of motion underan external force.


• 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.

)( vmdt

dFnet


• If the mass is a constant,

)( vmdt

dFnet

amdt

vdmFnet

.



amdt

vdmFnet

.

akmFnet

where k is a proportional constant



akmFnet

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



amFnet

where k = 1 in SI units

Note that the above equation is correcton condition that SI units are used.

a

Fnet

m


• No matter the mass is a constant or not,

Note that the above equation is correcton condition that SI units are used.

)( vmdt

dFnet


• If F = 0,

mv = constant This is the case of Newton’s First Law of Motion.

)( vmdt

dFnet

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.

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.

W

N

N = W



W

N

N = W + F

F



W

N

N = W - F

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.


A B

Exerted on A by B.

Exerted on B by A.

are action and reaction pair


• 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?

weight W = mg

Normal contact force N.This is the reading on the balance.

Feeling of One’s Weight• The net force and the acceleration. Newton’s 2nd

law of motion.

N

W

net forceF = N – W = m.am = mass

of the man

a

Feeling of One’s Weight• The lift is moving up/down at constant speed or at

rest.

a = 0N – W = 0 N = W

N

WThe reading on the balance is the weight of the man.

Feeling of One’s 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

Feeling of One’s 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

Feeling of One’s Weight• Free falling.

W – N = m.g N = W - m.g = 0

N

W

The reading on the balance is zero. weightless.

g

http://www.physlink.com/ae54.cfm





Momentum and Impulse

• Impulse J = F.t– F is the force– t is the time for the force to act


• 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) .

mF

u

mF

mF

mF

mF

v


• 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) .

mF

v

Impulse and F - t graph

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

F

t0

Examples of Impulse

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

http://www.exploratorium.com/sports/

http://www.exploratorium.com/sports/

Examples of Impulse


time of contact

t

momentumof the baseball

Examples of Impulse

• Catching a baseball

time of actiont

momentumof the baseball

t

t

muF

0

mu

Examples of Impulse


F F F

Examples of Impulse

• Striking a tennis ball

• Go to the search engine http://www.askjeeves.com

• and type the word “ tennis science “.

http://www.askjeeves.com/

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

http://www.inel.gov/resources/ep/physics.html






Raindrops versus Hailstones

• Try to study the passage by yourself.

• Discuss with your classmates.

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?

1ms1ms

Note the directions!

Example 1

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

Example 1

Note the directions!

Momentum of the ball

time

time of action

0

Collisions

• Studying different kinds of collisions.

– Perfectly elastic collision.

– Completely inelastic collision.– Collision in-between

• http://webphysics.ph.msstate.edu/javamirror/Default.html

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



• 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



• After collision,– Total momentum = m1.v1 + m2.v2

m1

v1

m2

v2

m1

v1

m2

v2

m1

v1

m2

v2

m1

v1

m2

v2



• Without external force,m1.u1 + m2.u2 = m1.v1 + m2.v2

m1

v1

m2

v2

m1

u1

m2

u2


The momentum of m1

m1

v1

m2

v2

m1

u1

m2

u2

m1.u1

m1v1

time

time of action


The momentum of m2

m1

v1

m2

v2

m1

u1

m2

u2

m2.v2

m2.u2

time

time of action


• For N bodies in collision.

Without external force,sum of momenta before = sum of momenta after

N

iii

N

iii vmum

00

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-dimensionRight-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.


Prove : After collision, they move out in directions perpendicular to each other.

θ+φ= 900

θ

ψ

Hint: Use conservation of momentum. Use conservation of kinetic energy.

mu

mv1

mv2


Example: α-particle colliding with helium atom.

θ+φ= 900

θ

ψ

mu

mv1

mv2


Two unequal masses in oblique and elastic collision.

Before collision, one mass is stationary and the other mass is moving.

θ

ψ

m1u

m1v1

m2v2

If m1>m2, then θ+ψ<90o


Two unequal masses in oblique and elastic collision.

Before collision, one mass is stationary and the other mass is moving.

θ

ψ

m1u

m1v1

m2v2

If m1<m2, then θ+ψ>90o

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 frictionf

Friction

• Static friction : – The object is stationary.

• Kinetic friction : – The object is moving.

Static Friction• To resist the tendency of relative motion of

two bodies.• Static friction has a maximum value, limitin

g frictional force fL .

tendency of motionin this direction

direction of static frictionf

F

F = f. The object is stationary.

Static Friction• Static friction has a maximum value, limiting fricti

onal force fL .• fL depends on the nature of the surface and normal

reaction. • fL = μs.R where μs is the coefficient of static frictio

n.

direction of static frictionf

FR= normal reaction

Kinetic Friction• Kinetic frictional force fk is almost a constant.• fk depends on the nature of the surface and normal r

eaction. • fk = μk .R where μk is the coefficient of kinetic fricti

on.• fk < fL

direction of kinetic frictionfk

FR= normal reaction

direction of motion

Chang of frictionGradually increase the applied force F

Ff

static kineticf

F0f = F

fLfk

Example 3

Given: m = 2.0 kg,

μs = 1.2,

F = W.

Stationary.

• Find f.

F

W

R

f

Example 4

• Given: 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• 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?

W

Rf

θ

Friction

• Cause of friction

• Reducing friction

• Role of friction

• Friction in a car

Spring and force constant

• Hooke’s law– The extension or compre

ssion 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 cons

tant of the spring– and e is the extension

F

e

In equilibrium

Spring and force constant

• Example 6

Combination of springs

• In series

• In parallel

k1 k2

k1

k2

F

F


• In series k1 k2

F

ekk

kkF .

21

21

e


• In parallel k1

k2

Fe

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

s

F F F F F

Work

• W = F.s.cos

F

s

Note that F.cos is the component of F in thedirection of s.

Work

• If F varies with s,

2

1

cos.S

SdsFW

F

sSFW .cos.

Sign of Work done

• Free falling– F is the gravitational force to do work.– Work done is positive.– The ball gains K.E.

F

sF

F

F

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

F

F

F

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

F

F

F

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

Sign of Work done

• Stretching a springunstretched

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

stretcheds

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

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

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

2

2

1mvEk

m

v

Kinetic energy

• Prove that the work done F.s is equal to the gain of kinetic energy

s

mF

u = 0

m

v

2

2

1mvEk

Gravitational potential energy

mghEp

Gravitational potential energy

• 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.

F

F

s

Energy

• Example 9

Elastic potential energy

• F = k.e• Find the work done to extend a spring for a

length e.

F

unstretched



length e.

F

unstretched

stretchede



length e.

edxFW

0.

e

F

x

F=k.x

0



length e.

edxkxW

0.

e

F

x

F=k.x

0



length e.

e

F

x

F=k.x

0

2

2

1keW

This is also the area below the graph.


• Example 10.

Physics Beyond 2000 Chapter 2 Newtons Laws of Motion.

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Transcript of Physics Beyond 2000 Chapter 2 Newtons Laws of Motion.