Transcript No Slide Title

Momentum and Conservation
momentum
m
p = mv
v
Law of conservation of momentum
* isolated system, sum of external forces
acting on system is zero S F = 0
* collision or explosion
* S(initial momentum) = S(final momentum)
* S(initial momentum)k = S(final momentum)k
subscript k represents x, y or z
component of momentum
before
(initial)
during
After (final)
CENTRE OF MASS
Centre of Mass (CM) - point
where the total mass of a
collection of objects can be
regarded as being located.
Diver jumping into a swimming
pool: The motion of the CM can
only change with the application
of an external force, and no new
force is applied, the CM of the
diver moves in a parabolic path.
VECTORS
Ax = A cosq
Y
Ay = A sinq
Ay
A
A = Ax2 + Ay2
tanq = Ay / Ax
q
Ax
X
WORK W (SI unit: joules J) - measure of the amount of energy
transferred into or out of a system by the action of a single applied
(external) force acting through a distance:
F = constant
W = F d cosq
F(r) = variable
WP1-->P2 =  F(r) . dr =  F(r) cosq dr (integration limits r1 to r2)
The total work done by a number of forces is Wtotal = SWi
When work is done on a system to overcome inertia, it goes into
kinetic energy K. The total work done Wtotal on the system produces
a change in its kinetic energy DK.
F
Wtotal = DK
K = ½ m v²
q
r
Newton’s First Law of Motion
* Defines an inertial frame of reference observer’s acceleration is zero
* Force F - agent of change (SI unit: netwon, N)
SF=0  a=0
aobserver = 0
v=0
v = constant
Newton’s Second Law of Motion
* Force F - agent of change (SI unit: netwon, N)
a = SF/m
external environment
interaction
system boundary
SYSTEM
internal environment,
mass m
aobserver = 0
SYSTEM
Disturbance SF
mass m
response a
Newton’s Third Law of Motion
Forces always act in pairs, interacting between
two systems such that the two forces have the
same magnitude but are opposite in direction.
aobserver = 0
Interaction between systems A and B
FBA = - FAB
SYSTEM A
SYSTEM B
FBA = FAB
B acting on A: FAB
A acting on B: FBA
Impluse J (SI Unit N.s or kg.m.s-1)
Impulse  Change in Momentum
 t2 


J   F dt  mv2  mv1
t1
An impulse exerted by a racquet changes a ball's
mometum. The greater the contact time between
the racquet and ball, the longer the force of the
racquet acts upon the ball and hence the greater the
ball's change in momentum and hence greater
speed of ball flying away from the racquet.
Contact J > 0, balls momentum (speed) increases
NO contact J = 0 0, balls momentum (speed) is constant
SUPERPOISION PRINCIPLE When two or more
disturbances of the same kind overlap, the resultant amplitude at
any point in the region is the algebraic sum of the amplitudes of
each contributing wave.
The Principle of Superposition leads to the phenomena known as
interference. For example, assume that there are two
monochromatic and coherent light sources (waves of a single
frequency which are always "in-step" with each other). The waves
from each source reaching arbitray points within a region will have
traveled different path lengths and therefore will have different
phases. At some points the waves will be in phase (in step difference in pathlengths Dd= ml m = 0, 1, 2, ...) and reinforce
each other giving maximum disturbance at that point constructive interference. At other points, the two waves will be
out of phase (out of step - difference in path lengths = (m+1/2)l m
= 0, 1, 2, ...) and cancel each other - destructive interference.
This region is characterized by bright and dark areas called
interference fringes.
Superposition and Interference for light
Waves out of phase  destructive interference  dark fringe
Dd = (m+½)l
Two
monochromatic
& coherent light
sources
+
Dd = m l
Waves in phase  constructive interference  bright fringe
ARCHIMEDES PRINCIPLE AND BUOYANCY
An object immersed in a fluid will be "lighter", that is,
buoyed up by an amount equal to the weight of the fluid
it displaces.
Object: mass m, weight FG,, volume V, density r
r = m / V m = V r FG = m g
FB
Buoyant
force
Fluid
density rF
Weight
FG
Volume of water displaced Vd = Volume of object submerged Vs
Newton's Second Law: FB + FG = m a  a = FB / m - g
FB = weight fluid displaced = Vd rF g = Vs rF g 
a = (Vs rfluid g/ V r } - g
= g { (Vs / V) . (rF / r) - 1}
Object partially submerged and floating ---> a = 0
Vs = ( r / rF )V  greater the density of the object compared to
fluid then the greater the volume of the object submerged.
Object fully submerged:
a = 0 r = rF object floats under water
a > 0 r < rF object rises to surface
a < 0 r > rF object sinks to the bottom
OBJECT MOVING WITH A CONSTANT ACCELERATION
(one dimensional motion)
a = constant
displacement, s
+X
time t = 0
initial velocity, u
v=u+at
s = u t + ½ a t2
v2 = u 2 + 2 a s
average velocity <v> = s / t = (u + v) /2
time t
final velocity, v
OSCILLATORY MOTION
Mass / spring systems, vibrations, waves, sound, light, colour, ...
Period, T  time for a complete cycle or oscillation (SI unit: second, s)
Frequency, f  number of cycles (oscillations) in one second (SI unit:
hertz Hz)
Angular frequency  “angle swept out” in a time interval (SI unit:
radian/second rad.s-1)
amplitude  max deviation of disturbance from equilibrium position
(for simple harmonic motion, amplitude is independent of period or
frequency)
T=1/f
f=1/T
w=2pf=2p/T
Simple Harmonic Motion SHM
Hooke’s Law and mass / spring systems
Hooke’s Law F = - k x
spring constant, k
F restroing force acting on mass by spring when the
extension of the spring from its natural length is x
Potential energy of system U = ½ k x2
SHM
ma=-kx
mass, m
Amplitude of
oscillation ,A
a = - w2 x
a = - (k / m) x
w2 = k / m  T = 2 p (m / k)
x = A cos( w t + f )
v = A w sin( w t + f )
a = - A w2cos( w t + f ) = - w2 x
STANDING WAVES ON A STRING (WIRE)
t time, T period, A antinode, N node
t = 0, T , 2T , 3T , ...
t = T /2, 3T /2, 5T /2,
...
A
t = T /4, 3T /4, 5T /4,
...
m mass of string
L length of string
T string tension
v speed of wave along string
l resonance wavelength of standing wave on
string
f resonance frequency of vibration of string
 linear density of string
N mode number (integer)
 = m /L
l = 2L / N
N
A
N
A
N
N = 1, 2, 3, ....
v = f l =  (T / )
f = (N /2L). (T / m )