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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 ( s ) – it is the distance moved in a constant direction
iii)
average speed =
total distance moved
time taken


s
iv)
average velocity = v av 
=
total displacement moved
t
v)
vi)
vii)
time taken



lim
s
dv

instantaneous velocity = v 
st  0 t
dt


v
velocity c h a n g e
average acceleration = a av. 
=
t
time
taken



lim v dv

instantaneous acceleration = a 
st  0 t dt
** 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  vt

s
t2
t1
vdt
s = total displacement moved in time
interval t1
- Hence, displacement is equal to the area under the graph.
axis is positive while area below time axis is negative.
t2
Area above time
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
t
v
0
u
 adt   dv
at  v  u
v  u  at
(ii)
v
ds
dt
 ds   vdt
 ds   (u  at )dt
s
t
0
0
1
s  ut  at 2
2
(iii)
 s = average velocity x time taken
s 
(iv)
uv
t
2
dv dv ds


dt ds dt
dv
a
v
ds
ads  vdv
a
s
a  ds 
0
v
 vdv
u
1 2
(v  u 2 )
2
2
v  u 2  2as
as 
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 from coiling to floor and
b) the distance it has fallen relative to the elevator shaft
P.3
g=10ms-2)
(Take
 -s1 + s2 = 3
1
1
 (2t  (10) t2)+2t+ (1)t 2 = 3
2
2
t = 0.74s
** ALTERNATIVE 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
1
s  ut  at 2
2
t = 0.74s
b) u = 2ms-1,
a = -10ms-2,
t = 0.74s
1
s1 = ut + at2 = -1.26m
2
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
u y  u sin 
horizontal motion
a  g
consider OA,
a0
0  u 2 sin 2   sgH
R  u xt
(v 2  u 2  2as)
from(2)
H
u x  u cos
u sin 
.................(1)
2g
2
2
R  u cos 
2u sin 
g
u 2 sin 2
...............(3)
g
Hence R is maximum at θ = 45˚.
0  u sin   gt1
R
(v = u + at)
u sin 
t' 
g
time of flight = 2t’
2u sin 
t
..............(2)
g
c) Consider the velocity v at point B,
v x  u cos ............(4)
v y  u y  at
2u sin 
g
v y  u sin  ............(5)
 u sin   g
velocity at B,
v  v x2  v y2
v  u
tan  
(it can also easily be obtained by using
conservation of energy)
vy
vx
  tan 
  
(-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.
 x  u cos  t
x
t 
...............(1)
u cos
y  tu sin  
1 s
gt ..............(2)
2
( s  ut 
1 2
at )
2
sub. (1) into (2),
x
gx 2
u sin   2
u cos
2u cos 2 
g
y  x tan   2
x2
2u cos 2 
y
This is of the form y = ax + bx2 which is the equation of a parabola.
practice air resistance causes slight deviation from a parabolic path.
In
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 motion 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 when 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
d ( mv)
dt
F = force acting at the
instant t
FK
d (mv)
dt
m = mass at time t
dv
= accel. At time t
dt
In S.I. unit, k=1
F=
d (mv)
dv
dm
 m v
dt
dt
dt
F= m
dv
 ma
dt
v = vel. At time t
dm
= rate of change of
dt
mass
(if mass of the system is unchanged)
P.6
** From this, 1 newton is defined 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 100cm2. If the
water strikes a wall normally, calculate force on the call in newtons, assuming
the velocity of the water normal to the wall is zero after collision. (density of
water = 1gcm-3)
Solution:
Volume of water striking = 100×20 = 2000cm3s-1 the wall per second
m
= mass of striking the = 2000gs-1 = 2kgs-1
t
v – u = velocity change of water = 0 – 20 = -20cms-1
= -0.2ms-1
F = force acting on the water
force acting on the wall

m
(v  u )
t
= -0.4N
= -F (by Newton’s third law action a reaction)
= 0.4N
e.g. Sand drops vertically 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)
dm
 2kgs 1
dt
F m
dv
dm
v
dt
dt
(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 
1 dm 2
v
2 dt
1
 2  0.12
2
 0.01W

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 extra
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 increase its velocity by △v after a time △t. Find the force acting on the
rocket.
Solution:
Fr = force acting on the rocket
Fg
=
force acting on the ejected gases
 Fr   Fg (Newton’s third law)

uM  vM
t
 Fr  (u  v)
dm
dt
(Newton’s second law)
(
dm
M

whent    0, M  ve, dm  ve)
dt
t
acceleration of the rocket can be found to be
a  Fr / M
If
M=
mass of a rocket at certain instant = 10000kg
dm
=
dt
exhausted rate of the gases = -5kgs-1
(u-v)=
velocity if the ejected gases relative to the rocket = -2000ms-1
then,
dm
dt
 (2000)  (5)
 10000 N
Fr  (u  v)
acc. at this instant = Fr / M = 1ms-2
d) Weight
- the weight W of a body is the force of gravity acting on it towards the centre of
the earth.
W  mg
(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.
(discussed 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 quantity 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.


Fa  Fr
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 (rB, mBg); (RA, mAg); (RM, mMg); (FA, FB); (FM,FA)
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 mechanics 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 mass appears to change under high velocity.
P.9
h) Other comments on Newton’s Law
(i) 1st law is a special case of 2nd law.
(ii)2nd law provides a method to measure the inertial mass of a body.
 ma  m0 a 0
 m  m0
a0
a
(same force)
The inertial mass m can be found by comparing its acceleration with that 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 of 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 if the
normal reaction is constant.
3.a) When the surface are at rest, the limiting frictional force F is directly
proportional to the normal force N.
i.e. F  N
F  N
(μ = coefficient of static friction)
b) When motion occurs, the dynamic frictional force F1 is directly
proportional to the normal force N.
ie. F '  N
F '  'N
(μ’ = coefficient of dynamic / sliding
friction)
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 contact 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 be 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,
FN
accounting for law 3
d) (OPTIONAL) Rolling friction
- Ideally there should be on rolling friction, because there should be no relative
motion between 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 deformed 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 horizontal
force 20N along an inclined plane which is 30º to the horizontal. Find the
coefficient of sliding friction μ between the object and the plane.
Solution:
20 cos 30  f  20 sin 30.............(1)
N  20 sin 30  20 cos 30............(2)
f  N ........................................(3)
(f = friction between the plane and the object
N = normal reaction between the plane and the object)
From (1), (2) and (3),
  0.27
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 no external force acting on them.
Proof: Consider the following case,
By Newton’s second law
F1 = (m1v1 – m1u1) / t --------------------(1)
F2 = (m2v2 – m2u2) / t --------------------(2)
By Newton’s third law
F1 = -F2
→
m1u1 + m2u2 = m1v1 +m2v2
initial total momentum = final total momentum
2.7 Work, energy and power
a)
Work
(i) Definition–
application along
actually a process of
 
W= F S
W = FS cos θ
work is done when a force moves its point of
the direction of its line of action. Work is
transferring energy.
(θ is the angle between F and S)
e.g.
e.g
W = 5×3×cos90º
W = 3×4×cos60º
= 0J
- work done = positive (support the motion)
= 6J
- work done = negative (against the motion)
e.g.
e.g.
W = 3×(-2)
(ii)
W = 3×2
= -6J
Force – displacement graph
= 6J
w  Fxs
w  SFds  FSds  Fs(ifFiscons tan t )
**(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.
P
=
dw
dt
F .ds
 F .v
dt
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. =
1
mv 2
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.
v 2  u 2  2as
a  v 2 / 2s....................(1)
from (1),
F  ma
F  mv 2 / 2s.................(2)
K.E. gained = work done on the body by the force
 Fs
mv 2

xs
2s
1
 mv 2
2
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, equal 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 which 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 internal energy.
P.13
e.g. A body of mass m accelerates uniformly from rest to a speed v1 in time t1.
a) show that the work done on the body as a function of time t, in terms of v1 and
1 v2
t1 is m( 12 )t 2
2 t1
b) As a function of time t, what is the instantaneous 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 acceleration
At time = t1v1  u  at1
 at1
a  v1 / t1 .........(1)
At time = tv  u  at
From (1),  (v1 / t1 )t.........(2)
Work done = K.E. gained
from (2),

1 2
mv
2

1
m(v1 / t1 ) 2 t 2
2
b) instantaneous power =

c)
dw
dt
1
m(v1 / t1 ) 2 t
2
m  1000kg, v1  30ms 1 , t1  10s, t  10s
(i) instantaneous power = m(v1 / t1 ) 2 t  90000W
(ii) average power
= 90kW
= total work done / time taken
= total K.E. gained / time taken
1 2
mv1 / t1
2
 45000W
 45kW

2.9 Collisions
a) Coefficient of resitution (e)
Before collision (u1>u2)
after collision (v2<v1)
Relative velocity of separation = v2 – v1
Relative velocity of approaching = u1 – u2
velocityofseparation
e
 (v2  v1 ) /(u1  u 2 )
velocityofapproachin g
P.14
b) Different types of collisions
- momentum is always 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 velocity 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 molecules
(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 m1 strikes on another body with mass m2 which is
initially at rest. They stick together after the collision with a common velocity v.
Before collision
after collision
K .E.  K .E.i  K .E. f
1
1
m1u 2  (m1  m2 )v 2
2
2
since there is no net external force, momentum is conserved
m1u  (m1  m2 _ v

 K .E. 
K .E. 
m12 u 2
1
1
m1u 2  (m1  m2 )
2
2
(m1  m2 ) 2
1 m1 m2
(
)u 2  0
2 m1  m2
For a given u, when m1=m2
K .E. 
1
mu 2  halftheinitialK .E.
4
EXERCISE Calculate the total time of continuous partly elastic collisions by a rubber
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 m1 with velocity u1 strikes on another mass m2 which is
initially at rest. The velocities of m1 and m2 after the collision are v1 and v2
respectively.
(before collision)
(after collision)
Since the collision is elastic, K.E. is conserved
1
1
1
m1u12  m1v12  m2 v 22
2
2
2
2
2
2
m1u1  m1v1  m2 v 2
m1 (u1  v1 )(u1  v1 )  m2 v 22 .....................(1)
By conservation of momentum,
m1u1  m1v1  m2 v2
m1 (u1  v1 )  m2 v2 ..................................(2)
sub. (2) into (1), we get
u1  v1  v2
v1  v2  u1 . . . . . . . .3(.) .
sub. (3) into (2),
m1 (u1  v2  u1 )  m2 v2
2m1u1
v2 
..................(4)
m1  m2
from (3), (4)
we have
v1 
(m1  m2 )
u1 .............(5)
(m1  m2 )
v1 = 0
v1 > 0
v1 < 0
EXERCISE
if m1 = m2
if m1 > m2
if m1 < m2
Consider the following case, prove that the coefficient of resitution e
is equal to l if elastic collision occurs.
And show that m1, m2 exchange their velocity if m1 = m2
Elastic collision
(before collision)
(after collision)
P.17
SOLUTION:
by conservation of momentum:
m1u1  m2 u 2  m1v1  m2 v2
m1 (u1  v1 )  m2 (v2  u 2 )...............(1)
by conservation of K.E. (elastic collision)
1
1
1
1
m1u12  m2 u 22  m1v12  m2 v22
2
2
2
2
m1 (u1  v1 )(u1  v1 )  m2 (v2  u 2 )(v2  u 2 )...(2)
( 2)
(1)
u1  v1  v2  u2  u1  u2  v2  v1
e
v 2  v1
1
u1  u 2
v1  v 2  u 2  u1 ................(1)
m1 (u1  v2  u2  u1 )  m2 (v2  u2 )
2m1u1  m1v 2  m1u 2  m2 v 2  m2 u 2
2m1u1  m1u 2  m2 u 2  v 2 (m1  m2 )
2m1u1  m1u 2  m2 u 2
v2 
m1  m2
(3) into (1)
2.12 Further applications
a) Principle of measuring inertial mass
By principle of conservation of momentum,
m1u1  m2 u 2  m1v1  m2 v 2
m1u1  m1v1  m2 v 2  m2 u 2
 m1 (v1  u1 )  m2 (v 2  u 2 )
m1 v 2  u 2
v

 2
m2 v1  u1
v1
m1
v
 2
m2
v1
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.
v y
mx

my
v x

(0.96)
6
0.16
Hence, the mass of X is 6 times the mass of Y.
b) Recoil of rifles
By conservation 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
m1v1 = momentum of the gases after the explosion
MV = recoil momentum of the gun.
For a typical rifle,
M  3.6kg,V  3.8ms 1 , m  10 g  0.01kg, v  900ms 1
Find the momentum of the gases.
Solution:
by conservation of momentum,
MV  mv  m1v1
m1v1  3.6  3.8  0.01  900
 4.7 kgms1
** It contributes about 34% of the recoil momentum of the rifle.
c) Oblique elastic collision of 2 identical particles with one of them initially at rest
By cos of momentum

mu  mv1  mv 2

u  v1  v 2
By conservation of kinetic energy (since it is elastic),
1
1
1
mu12  mv12  mv22
2
2
2
2
2
2
u1  v1  v2 .......................(1)
Equation (1) satisfy Pythagoras’ theorem. The only way that we can draw the
vector diagram for u1, v1 and v2 so that this equation apples is for u1 to be the
hypotenuse of a right-angled triangle.
Hence,
    90
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.
P.19
d) Chadwick’s estimation of neutron mass
Consider the linear elastic collision between a fast moving neutron and a
stationary atom.
before collision
m1
m2
u1
v1
v2
after collision
(i)
mass of a neutron in a.m.u.
mass of the atom is a.m.u.
initial velocity of the neutron
final velocity of the neutron
final velocity of the atom
findv2 int ermsofu1 , m1 andm2
from2.12equation(4), wehave
v2 
(ii)
=
=
=
=
=
2m1
u1
m1  m2
Take experimental observations on the collision of neutrons to hydrogen
and nitrogen respectively
Data: velocity of hydrogen atom after collision = 3.3×107ms-1
Velocity of nitrogen atom after collision = 4.7×106ms-1
Calculations:
v2 = 3.3×107ms-1
v2 = 4.7×106ms-1
m2 = 1 unit
m2 = 14 units
Since
since
v2  (
2m1
)u1
m1  m2
3.3  10 7 
v2 
2m1
u1 ..........(1)
m1  1
2m1
u1
m1  m2
4.7  10 6 
2m1
u1 ......(2)
m  14
(1) 3.3  10 7 m1  14


(2) 4.7  10 6
m1  1
m1 = 1.15 unit
Within the experimental error, mass of a neutron (m1) may be taken as 1.
e) Apparent non-conservation of momentum in β decay.
Observation - Electrons emitted from radioactive nuclei are known as βparticles. 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 process 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:
n  p  e
A
A

Z X  Z 1Y  e
(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:
n  p  e   v~
A
Z
X  Z A1Y  e   v~
- 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 that 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 decreased 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)