Post on 27-Mar-2015
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
• = linear momentum of a body
• m = mass of the body
• = velocity of the body
vmp
.p
v
Momentum
• = linear momentum of a body
• m = mass of the body
• = velocity of the body
vmp
.p
v
Note that is a vector quantity.Unit of is
p
p
Momentum
• = linear momentum of a body
• m = mass of the body
• = velocity of the body
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
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.
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 motionwhen 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 underan 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.
)( vmdt
dFnet
Newton’s Second Law of Motion
• If the mass is a constant,
)( vmdt
dFnet
amdt
vdmFnet
.
Newton’s Second Law of Motion
• If the mass is a constant,
amdt
vdmFnet
.
akmFnet
where k is a proportional constant
Newton’s Second Law of Motion
• If the mass is a 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
Newton’s Second Law of Motion
• If the mass is a constant,
amFnet
where k = 1 in SI units
Note that the above equation is correcton condition that SI units are used.
a
Fnet
m
Newton’s Second Law of Motion
• No matter the mass is a constant or not,
Note that the above equation is correcton condition that SI units are used.
)( vmdt
dFnet
Newton’s Second Law of Motion
• 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
Normal Contact Force
The box is at rest on the ground.
W
N
N = W + F
F
Normal Contact Force
The box is at rest on the ground.
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.
Newton’s 3rd Law of Motion
A B
Exerted on A by B.
Exerted on B by A.
are action and reaction pair
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?
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
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) .
mF
u
mF
mF
mF
mF
v
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) .
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/
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
momentumof the baseball
Examples of Impulse
• Catching a baseball
time of actiont
momentumof the baseball
t
t
muF
0
mu
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
Examples of Impulse
• Striking a tennis ball
• Go to the search engine http://www.askjeeves.com
• and type the word “ tennis science “.
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.
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
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
v1
m2
v2
m1
u1
m2
u2
Principle of conservation of momentum
The momentum of m1
m1
v1
m2
v2
m1
u1
m2
u2
m1.u1
m1v1
time
time of action
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
Principle of conservation of momentum
• 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.
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.
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.
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.
Collisions in 2-dimensionRight-angled fork
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
Collisions in 2-dimensionRight-angled fork
Example: α-particle colliding with helium atom.
θ+φ= 900
θ
ψ
mu
mv1
mv2
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
m1u
m1v1
m2v2
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.
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
• 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
Combination of springs
• In series k1 k2
F
ekk
kkF .
21
21
e
Combination of springs
• 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
Elastic potential energy
• F = k.e• Find the work done to extend a spring for a
length e.
F
unstretched
stretchede
Elastic potential energy
• F = k.e• Find the work done to extend a spring for a
length e.
edxFW
0.
e
F
x
F=k.x
0
Elastic potential energy
• F = k.e• Find the work done to extend a spring for a
length e.
edxkxW
0.
e
F
x
F=k.x
0
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
2
2
1keW
This is also the area below the graph.
Elastic potential energy
• Example 10.