P.
1
Section A Mechanics
SECTION A : MECHANICS
Chapter 2
:
DYNAMTICS
2.1
Uniformly acceleration motion
a)
Definitions of some general terms
i)
distance
–
it is the length of the actual path
ii)
displacement (
)
–
it is
the
distance moved in a const
ant direction
iii)
average speed = total distance moved
time taken
iv)
average velocity =
= total displacement moved
time taken
v)
instantaneous velocity =
vi)
average
acceleration =
=
vii)
instantaneous acceleration =
** N.B.
Instantaneous speed is equal to the magnitude of the instantaneous
velocity but average speed is no need
to have the same value as the
magnitude of the average velocity.
b)
Graphical interpretation
i)
displacement
–
time graph

magnitude of the instantaneous velocity at a certain point is equal to
the
gradient there
ii)
velocity
–
time graph

magnitude of the
instantaneous acceleration at a certain point is equal to
the
gradient there.
P.
2
s = total displacement moved in time
interval t
1
t
2

Hence, displacement is equal to the area under the graph.
Area above time
axis is positive while area below time axis is negative.
c)
Equation of uniformly accelerated
linear
motion
t
=
time taken during the motion
s
=
displacement moved by a body in this time interval
u
=
interval velocity of the body
v
=
final velocity of the body
(ii)
(iii)
s = average velocity x time taken
(iv)
e.g.
An elevator ascends with an upward acceleration of 1ms

2
. At the instant its
upward speed is 2ms

1
, a loose bolt drops from the coiling of
the
elevator 3m from
the floor.
Calculate
a)
the time of flight of the bolt fr
om coiling to floor and
b)
the
distance it has fallen relative to
the
elevator shaft
P.
3
(Take
g=10ms

2
)

s
1
+ s
2
= 3
t
2
)+2t+
2
= 3
t = 0.7
4
s
** ALTERNATI
VE METHOD
choose the lift as the reference frame,
u
=
initial velocity of the bolt w.r.t the lift
=
0
a
=
acceleration of the bolt w.r.t the lift
=

11ms

2
s
=

3m
since
t
=
0.74s
b) u = 2ms

1
,
a =

10ms

2
,
t = 0.74s
s
1
= ut +
at
2
=

1.26m
2.2)
Motion under gravity

The acceleration towards the surface of the earth under free fall in equal to g.
G = the acceleration due to gravity = 9.8ms

2
apprex.
2.3)
Projectiles
a)
property

the vertical
motion of a projected ball (a const. Acc
= g) is unaffected by
its horizontal motion (a constant velocity).
The
two motions are quite
independent
of each other.
b)
Consider a body projected obliquely from O with velocity u at an angle
θ
to
the horizontal.
(Choose sign convention as)
H
=
max. height of the projectile
R
=
range of the projectile
t
’
=
time of flight
t
=
time used from O to A
P.
4
Vertical motion
horizontal motion
consider OA,
from(2)
(v = u + at)
Hence R is max
imum at
θ
= 45
˚
.
time of flight = 2t
’
c)
Consider the velocity v at point B,
velocity at B,
(it can also easil
y be obtained by using
conservation of energy)
(

ve means the angle is clockwise)
d)
The trajectory of a particle is parabolic
Let the body be at point C(x, y) at time t after projection from 0.
sub. (1) into (2),
This
is of the form y = ax + bx
2
which is
the
equation of a parabola. In
practice air resistance causes slight d
eviation from a parabolic path.
P.
5
e.g.
A rifle with a muzzle velocity of 50ms

1
sheets a bullet at a small target 50m
away.
a)
How high above the target must the gun be aimed so that the bullet will
hit the target, and
b)
Find the time of flight of the bullet.
(take g = 10ms

2
)
2.4 Newton
’
s laws of Motion
a)
Newton
’
s first law
If a body has no net forces acting on it, it will not
undergo
a change in its state
of motion. That is, a body at rest will remain at rest; a body in motion will
continue in mo
tion with constant velocity.
b)
Mass

the first law introduces the concept of inertia, that is, every body tends to
remain its original state of motion.
(e.g.
passengers are caused to lurch forwards when a vehicle is suddenly
stopped and jerked backwards whe
n it is started.)

mass of a body is a measure of its inertia which shows reluctance in changing
its state of rest or motion.
Mass under such consideration is called inertial mass.
c)
Newton
’
s second law
It states that the rate of change of momentum of a
body is proportional to the
resultant
force acting on the body and occurs in the direction of the force.
F
=
force acting at the
instant t
m
=
mass at time t
In S.I. unit, k=1
= accel. At time t
F =
v
=
vel. At time t
F =
= rate of change of
mass
(if mass of the system is unchanged)
P.
6
**
From this, 1 newton is de
fined as the force which gives a mass of 1kg as acceleration
of 1ms

2
.
e.g.
A hose eject water at a speed of 20cms

1
through a hole of area 100cm
2
. If the
water strikes a wall normally, calculate force on the call in newton
s
, assuming
the velocity o
f the water normal to the wall is zero after collision. (density of
water = 1gcm

3
)
Solution:
Volume of water striking = 100
×
20 = 2000cm
3
s

1
the wall per second
=
mass of striking the = 2000gs

1
= 2kgs

1
v
–
u = velocit
y change of water
= 0
–
20 =

20cms

1
=

0.2ms

1
F =
force acting on the water
=

0.4N
force acting on the wall
=

F (by Newton
’
s third law action a reaction)
= 0.4N
e.g.
Sand drops vertica
lly at the rate of 2kgs

1
onto a conveyor belt moving
horizontally with a velocity of 0.1ms

1
.
Calculate
(
i
)
the extra power needed to keep the belt moving, and
(ii)
the rate of K.E. of the sand.
Why is the power twice as great as the rate of K.E.?
Solution:
(i)
(by Newton
’
s second law),
F =
force to
= 0 + 0.1
×
2
keep the belt
= 0.2N
moving
Power = W/t = F
∙
s/t = Fv
= 0.2
×
0.1
= 0.02W
(ii)
K.E. per unit time
Thus the power supplied is twice as great as the rate of change K.E. The
extra is due to the fact that the sand does not immediately have the velocity
of the belt., so that the belt has first moved relative to the sand. The ex
tra
power is needed to overcome the friction between the sand and belt.)
P.
7
e.g.
Rocket Propulsion
Consider a rocket of mass moving with velocity v at a certain instant.
The fuel is burning and ejected as gases with velocity u.
△
M fuel is used
to i
ncrease its velocity by
△
v after a time
△
t. Find the force acting on the
rocket.
Solution:
F
r
=
force acting
on the
rocket
F
g
=
force acting on the ejected gases
(Newton
’
s third law)
(Newton
’
s seco
nd law)
acceleration of the rocket can be found to be
If
M
=
mass of a rocket at certain instant = 10000kg
=
exhausted rate of the gases =

5kgs

1
(u

v
)
=
velocity if the ejected gases relative to the rocket =

2000ms

1
then,
acc. at this instant = F
r
/ M = 1ms

2
d)
Weight

the weight W of a body is the f
orce of gravity acting on it towards the centre of
the earth.
(g = acc. due to gravity)

since weight
mass, we can find the mass of a body by comparing its weight
with standard mass on a beam balance.

The equation W=mg is slightly affected by the rotation of the earth.
(dis
cussed later, ref.: Duncan MM P.265)
P.
8
e)
Difference between weight and mass
(i)
mass of a body is a scalar quantity which depends only on its intrinsic
properties (e.g. density, volume)
and
is independent of external environment.
(ii)
Weight of a body is a vector quan
tity
which
depends not only on its intrinsic
properties but also depends on the external environment (e.g. gravitational
field.)
f)
Newton
’
s third law

it states that if body A exerts a force (action) on body B, then body is exerts an
equal but opposite force
(reaction) on body A.
e.g. A man M pulls two block A and B along a horizontal smooth surface.
Draw the free body diagrams of M, A and B
Write down all the action

reaction Pairs.
free body
diagrams
The
action

reaction pair are (r
B
, m
Bg
); (R
A
, m
Ag
); (R
M
, m
Mg
); (F
A
, F
B
); (F
M
,F
A
)
g)
Limitations of
Newtonian
mechanics
1)
The size of
the
body must not be too small.(e.g. atomic size), otherwise
quantum mechanics must to be used.
It is because Newtonian mec
hanics assumes that we can determine the velocity
and position of the object at the same instant but actuall we cannot.
2)
The velocity must not to be too large.(e.g. speed of light), otherwise special
relativity have to be used.
It is because the measured m
ass appears to change under high velocity.
P.
9
h)
Other comments on Newton
’
s Law
(i)
1
st
law is a special case of 2
nd
law.
(ii)
2
nd
law provides a method to measure the inertial mass of a body.
(same force)
The inertial mass m can be
found by comparing its acceleration with t
h
at of the
standard mass.
*** EXAMPLES (Duncan MM P.224)
2.5
Friction
a)
Introduction

frictional forces act along the surface between 2 bodies whenever one moves
or tries to move over the other.
The direction o
f the frictional force is always oppose the motion.
b)
Laws of friction
1.
The frictional force between 2 surface opposes their relative motion.
2.
The frictional force does not depend
on the
area of contact of the surface
i
f the
normal reaction is constant.
3.
a)
Wh
en the surface are at rest,
the
limiting frictional force F is directly
proportional to the normal force N.
i.e.
(
μ
= coefficient of static friction)
b)
When motion occurs, the dynamic frictional force F1 is directly
proportional to the normal force N.
ie.
(
μ’
= coefficient of dynamic / sliding
f
riction)
P.
10
c)
Nature of friction

Close examination of the flattest and most highly polished surface show that
they
have hollows and humps more atoms high.

When one solid is placed on another, contact therefore only occurs at a few
places of small area. The c
ontact area is increased by deformation of the
humps until
the
upper solid is supported.

It is thought that at the points of contact,
‘
joints
’
are formed by the strong
adhesive forces between molecules which are very close together. These
have to b
e broken before one surface can move over the other.
Accounting for law 1

Since changing the apparent area of contact of the bodies has little effect on the
actual area.
Accounting for law 2

Normal reaction between the surfaces
actual area of contact
actual area of contact
hence,
accounting for law 3
d)
(OPTIONAL) Rolling friction

Ideally there should be on rolling friction, because there should be no relative
motion betw
een the surfaces in contact. N reality, a wheel or ball is slightly
flattened when it rests on a surface, which itself is slightly dented. A resistive
force arises when the wheel or ball rolls, partly because it and the surface must
be continually deform
ed and partly because there is some relative motion
between them owing to the deformation. Coefficients of rolling friction are
nevertheless much smaller than those of sliding friction.
e.g. An object with mass 2kg is just able to be pushed upwards by a h
orizontal
force 20N along an inclined plane which is 30
º
to the horizontal. Find the
coefficient of sliding friction
μ
between the object and the plane.
Solution:
(f = friction between the plane and the object
N = n
ormal reaction between the plane and the object
)
From (1), (2) and (3),
P.
11
2.6
Principle of conservation of momentum

It states that the total linear momentum of colliding bodies ids urcharge before
and after the collision if there is n
o external force acting on them.
Proof: Consider the following case,
By Newton
’
s second law
F
1
= (m
1
v
1
–
m
1
u
1
) / t

(1)
F
2
= (m
2
v
2
–
m
2
u
2
) / t

(2)
By Newton
’
s
third
law
F
1
=

F
2
→
m
1
u
1
+ m
2
u
2
= m
1
v
1
+m
2
v
2
initial total momentum = final total momentum
2.7
Work, energy and power
a)
Work
(i)
Definition
–
work is done when a force moves its point of
application along
the direction of its line of action. Work is
actually a process of
tra
nsferring energy.
W =
W = FS cos
θ
(
θ
is the angle between F and S)
e.g.
e.g
W = 5
×
3
×
cos90
º
W = 3
×
4
×
cos60
º
= 0J
= 6J

work done = positive (support the motion)

work done = negative (against the motion)
e.
g.
e.g.
W = 3
×
(

2)
W = 3
×
2
=

6J
= 6J
(ii)
Force
–
displacement graph
**(work done by a force during displacement
S is represented by area OABC)
P.
12
b)
Power
–
Power of a machine is the
note at which it does work.
=
c)
Energy
–
it is the capacity to be work.

It is the capacity to do work.
i)
Kinetic energy
–
it is the energy that a mass possesses by virtue of its motion.
K.E. =
2
(v = velocity of the mass)
Proof:
Consider a body of mass m at rest. Let a constant force F act on it and
bring its velocity to v in a distance s.
from (1),
K.E. gained = work
done on the body by the force
ii)
Potential
energy
–
it is the energy that a body Possesses by virtue of its position
P.E. = mgh
( h vertical displacement)
Proof:
Let the P.E. is zero at the surface of the earth.
A force, eq
ual and opposite to mg, has to be exerted on the body if the
mass m is moved from A to B.
P.E. of the mass at height h = work done by the force on
the mass
P.E.
= Fs
= mgh
2.8
Principle of conservation of energy

It states that the total energy w
hich the bodies in an isolated system posses is
constant . i.e. energy can neither be created nor destroyed.

The total amount of mechanical energy (P.E + K.E>) is also
constant
if there is
no frictional force i.e. no mechanical energy is
changed
into int
ernal energy.
P.
13
e.g. A body of mass m accelerates
uniformly
from rest to a speed v
1
in time t
1
.
a)
show that the work done on the body as a function of time t, in terms of v
1
and
t
1
is
b)
As a function of time t, what is the instantan
eous power delivered to the body?
c)
What is the instantaneous power at the end of 10sec. delivered to a 1000kg
body which accelerates from rest to 30ms

1
on 10sec? Hat is the average
power delivered in this 10sec?
Solution:
a)
u = 0 , a = constant accelerat
ion
At time =
At time =
From (1),
Work done = K.E. gained
from (2),
b)
instan
taneous power =
c)
(i)
instantaneous power =
= 90kW
(ii)
average power
= total work done / time taken
= total K.E. gained / time taken
2.9
Collisions
a)
Coefficient of resitution (e)
Before collision (u
1
>u
2
)
after collision (v
2
<v
1
)
Relative velocity of separation = v2
–
v1
Relative velocity of approaching = u1
–
u2
P.
14
b)
Different types of collisions

momentum is alwa
ys conserved in a collision if there is no external force.

There is usually a change of some K.E. into internal energy or sound energy
except the elastic case.
(i)
Inelastic collisions (perfectly inelastic)

the collision mass stick together after collision
i.e. e = 0

K.E. is dissipated
e.g.
collision of 2 vehicles in air track with plasticine at their collision ends.
(ii)
Partly inelastic collisions

the relative velocity after the interaction is reversed in direction but is
smaller than the relative velocit
y initially.
i.e. 0<e<1

K.E. is dissipated
e.g. rubber ball bouncing on the wall
(iii)
Elastic collisions (perfectly elastic)

the relative velocity after the interaction is reversed in direction and is equal
in magnitude to the relative velocity initially.
i.
e. e = 1

K.E. is conserved.

**
If the two masses are identical, they will exchange their velocities after
collision.
e.g. vehicles on the air track hits a fixed rubber band buffer at one end of
the track.
e.g. collision between atoms and molecu
les
(iv)
Hyperelastic collisions

the relative velocity after this interaction is reversed in direction and is
greater than the relative velocity initially.
i.e. e>1

K.E. is gained
e.g. explosion of trolleys or other collisions during which explosions occur.
P.
15
2.10
Lost of K.E. in inelastic collision
Consider a body with mass m
1
strikes on another body with mass m
2
which is
initially at rest. They stick together after the collision with a common velocity v.
Before collision
after collision
since there is no net external force, momentum is conserved
For a given u, when m
1
=m
2
EXERCISE
Calculate the total time of continuous partly elastic collisions by a rub
ber
ball projected at an angle
θ
.
(Hint: only the vertical component
of the
velocity is involved in the
collision)
Find also the total range.
P.
16
2.11
Study of elastic collisions
Consider that a mass m
1
with velocity u
1
strikes o
n another mass m
2
which
is
initially at rest. The velocities of m
1
and m
2
after the collision are v
1
and v
2
respectively.
(before collision)
(after collision)
Since the collision is elastic, K.E. is conserved
By conser
vation of momentum,
sub. (2) into (1), we get
sub. (3) into (2),
from (3), (4)
we have
v
1
= 0
if m
1
= m
2
v
1
> 0
if m
1
> m
2
v
1
< 0
if m
1
< m
2
EXERCISE
Consider the following case, prove that the coefficient of resitution e
is equal to l if elastic collision occurs.
And show that m
1
, m
2
exchange
their
velocity if m
1
= m
2
Elastic collision
(before collision)
(after collision)
P.
17
SOLUTION:
by conservation of momentum:
by conservation of K.E. (elastic collision)
(3) into (1)
2.12
Further applications
a)
Principle of measuring inertial mass
By principle of conservation of momentum,
e.g.
Compare the masses of
two different trolleys X and Y
which
are all initially at
rest. They explode apart and their velocities become 0.16ms

1
and
–
0.96ms

1
respectively.
Hence, the mass of X is 6 times the mass of Y.
b)
Recoil of rifles
By conserv
ation of momentum, there are two factors contribute to the recoil.
(i)
momentum given to the bullet and
(ii)
momentum given to the gases produced by the explosion.
P.
18
e.g.
Let mv
= momentum of bullet after being fired
m
1
v
1
= momentum of the gases after the
explosion
MV
= recoil momentum of the gun.
For a typical rifle,
Find the momentum of the gases.
Solution:
by conservation of momentum,
** It contributes about 34% of the recoil momentum of t
he rifle.
c)
Oblique elastic collision of 2 identical particles with one of them initially at rest
By conservation of kinetic energy (since it is elastic),
Equation (1) satisfy Pythagoras
’
theorem. The only way that
we can draw the
vector diagram for u
1
, v
1
and v
2
so that this equation apples is for u
1
to be the
hypotenuse of a right

angled triangle.
Hence,
OBSERVATION
From the photo of a cloud chamber in which
α
particles collide on helium atoms,
we observe the right

angled fork.
This
implies
α
particles must have the same mass as helium atom and actually it
is He
++
since it was detected to be positively charged by deflection in an electric
field.
By cos of momentum
P.
19
d)
Chadw
ick
’
s estimation of neutron mass
Consider the linear elastic collision between a fast moving neutron and a
stationary atom.
before collision
m
1
=
mass of a neutron in a.m.u.
m
2
=
mass of
the
atom is a.m.u.
after collision
u
1
=
initial velo
city of
the
neutron
v
1
=
final velocity of the neutron
v
2
=
final velocity of the atom
(i)
(ii)
Take experimental observations on the collision of neutrons to hydrogen
and nitrogen respectively
Data:
vel
ocity of hydrogen atom after collision = 3.3
×
10
7
ms

1
Velocity of nitrogen atom after collision = 4.7
×
10
6
ms

1
Calculations:
v
2
= 3.3
×
10
7
ms

1
v
2
= 4.7
×
10
6
ms

1
m
2
= 1 unit
m
2
= 14 units
Since
since
m
1
= 1.15 unit
Within the experimental error, mass of a neutron (m
1
) may be taken as 1.
e)
Apparent non

conservation of momentum in
β
decay.
Observation

Electrons emitted from radioactive nuclei are known as
β

particles. Unlike
α

particles which are emitted from the parent
nucleus with a definite speed, the
β

particles are emitted with a
continuous range of speeds from zero
up to a maximum.
Problems:
(
i
)
there is no electrons in the nuclei.
(ii)
emitted electrons, all of which result from the same process,
should have identical energies. But, as we have observed,
this is not so. It appears that bets decay is a p
rocess in
which momentum and energy may not be conserved.
P.
20
Explanation:
(
i)
a neutron can itself decay into a proton and an electron
which becomes the emitted
β
particle.
Simple equation:
(ii)
To solve this problem, W. Pauli suggested that there must be
another uncharged particle involved. It is called neutrino (v)
afterwards and its mass is very
little in order to fit the
observation.
Modified equation:

When a fast moving
β

particle is emitted, the neutrino moves slower.

When a slow moving
β

particle is emitted, the neutrino moves faster.

Thus the neutrino ensures t
hat sufficient energy and momentum are carried away
for the conservation laws (conservation of energy & conservation of
momentum) to apply.
f)
Terminal velocity

If an object is falling from above, its velocity will be accelerated but not
throughout the whole
process.

It is
because
there exist air resistance (viscosity). Its magnitude is directly
proportional to the magnitude of the velocity of the falling object and depends
on its size, shape and surface.

Hence the acceleration of the body will gradually dec
reased to zero in this
case.
At such moment, the object reaches its maximum velocity (or speed) and we
call
this
the terminal velocity. (e.g. parachutist)
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