HKPhO Pre-Training 2005
Cover Page
The materials in this note serve as a guideline for the mechanics topics of HKPhO in addition
to the S4-5 syllabus before 2003. Videos of the lecture and the tutorial will be available in
http://hkpho.phys.ust.hk/. While the key issues are addressed explicitly, details of secondary
importance are mostly left out. Students are encouraged to spend additional time to further
digest the contents. Small gaps are left for the students to work out the details, and learn
physics through the process. Additional references should be sought to further understand the
topics like vectors, elementary calculus, etc., most of which can be found in relevant F6-7
textbooks.
We hope that the pre-training will give a kick start for those who hope to do well in HKPhO
2006 and beyond. We also encourage the trainees to share the materials with their classmates
back in schools.
Z. Yang
HKPhO Committee Chairman
1
HKPhO Pre-Training 2005
Mechanics
1
Vector
1.1




Any vector A  Ax x0  Ay y0  Az z 0
(1.1)
Addition of two vectors:
  



C  A  B  ( Ax  Bx ) x0  ( Ay  B y ) y0  ( Az  Bz ) z 0
(1.2)
Dot product of two vectors
 
A  B  AB cos  Ax Bx  Ay B y  Az Bz
(1.3)


It is also referred to as the projection of A on B , or vise versa.
Amplitude of the vector

 
(1.4)
| A | Ax2  Ay2  Az2  A  A
1.2




Position vector of a particle: r  xx0  yy0  zz 0

B

A

C

A


B
z
(x,y,z)

r
(1.5).
o
If the particle is moving, then x, y, and z are function of time t.
Velocity:

 dr dx  dy  dz 



v

x0 
y0 
z0  v x x0  v y y0  v z z0 (1.6)
x
dt dt
dt
dt
Acceleration

 dv dv x  dv y  dv z 


d 2x  d 2 y  d 2z 

a

x0 
y0 
z 0  2 x0  2 y 0  2 z 0  a x x0  a y y 0  a z z 0
dt
dt
dt
dt
dt
dt
dt
y
(1.7)
1.3 Uniform circular motion
y
Take the circle in the X-Y plane, so z = 0,
x  R cos(t  ) , y  R sin( t   )
(1.8)
 is the angular speed.  is the initial phase. Both are constants.
x
Using the above definition of velocity (1.6),



v   R sin( t  ) x0  R cos(t  ) y0


v is always perpendicular to r .
Its amplitude is v  R
(1.10)
(1.9),
The acceleration is:




a   R 2 sin( t   ) x0  R 2 cos(t  ) y 0   2 r
Its amplitude is a  R 2 
v2
R
  t  
(1.12)
2
(1.11).
HKPhO Pre-Training 2005
2 Relative Motion

A reference frame is needed to describe any motion of an object.
O’
r'
Consider two such reference frames S and S’ with their origins at

O and O’, respectively. The X-Y-Z axes in S are parallel to the X’- R

Y’-Z’ axes in S’. One is moving relative the other.
r
  
Note: r  r ' R .
(2.1)
O



 dr dr ' dR  
So the velocity is:
(2.2)
v


 v'  u
dt dt dt
  
Similar for acceleration:
(2.3)
a  a' A

 
This is the classic theory of relativity. If u is constant, then a  a' , i. e., Newton’s Laws work
in all inertia reference frames.
Properly choosing a reference frame can sometimes greatly simplify the problems.
3 Forces
Pull through a rope, push, contact forces (elastic force and friction force), air resistance, fluid
viscosity, surface tension of liquid and elastic membrane, gravity, electric and magnetic,
strong interaction, weak interaction. Only the last four are fundamental. All the others are the
net effect of the electric and magnetic force.

3.1
Tension
T
Pulling force (tension) in a thin and light rope:
Two forces, one on each end, act along the rope direction. The
two forces are of equal amplitude and opposite signs because
the rope is massless. It is also true for massless sticks.
3.2
Elastics
Elastic contact forces are due to the deformation of solids.
Usually the deformation is so small that it is not noticed. The
contact force is always perpendicular to the contact surface. In


the example both F1 and F2 are pointing at the center of the
sphere.

T

F1

F2
3.3
Friction
The friction force between two contact surfaces is caused by the relative motion or the
tendency of relative motion. Its amplitude is proportional to the elastic contact force, so a

friction coefficient  can be defined.
v
N
When there is relative motion, the friction force is given by f =
kN, where k is the kinetic friction coefficient. The direction
of the friction is always opposite to the direction of the
relative motion.
3

f
HKPhO Pre-Training 2005

N
When there is no relative motion but a tendency for such

motion, like a block is being pushed by a force F , the
amplitude of f is equal to F until it reaches the limiting value
of sN if F keeps increasing, where s is the static friction
coefficient.
Once the block starts to move, the friction becomes f = kN. Usually k < s.

F

f
3.4
Viscosity
Air resistance and fluid viscous forces are proportional to the relative motion speed and the
contact area. A coefficient called viscosity  is used in these cases.
3.5
Inertial force
Inertial force is a ‘fake’ force which is present in a reference frame (say S’) which itself is
  
accelerating. Recall that a  a' A . Assume frame-S is not accelerating, then according to


 
Newton’s Second Law, F  ma  m(a ' A) . So in the S’-frame, if one wants to correctly



apply Newton’s Law, she will get F  mA  ma' , i. e., there seems to be an additional force


(3.1)
Fint  mA

a
acting upon the object.
Example 3.1

A block is attached by a spring to the wall and placed on the

F  ma
smooth surface of a cart which is accelerating. According to

the ground frame, the force F acting on the block by the spring


Ground frame
is keeping the block accelerating with the cart, so F  ma . In




the reference frame on the cart, one sees the block at rest but
Fint  ma
F  ma
there is a force on the block by the spring. This force is


‘balanced’ by the inertial force Fint  ma
Cart frame
3.6
Gravity
Between two point masses M and m, the force is

GMm 
F  2 r
(3.2)
r

GM 
The gravitation field due to M is g   2 r
(3.3).
r
The field due to a sphere at any position outside the
sphere is equal to that as if all the mass is concentrated at
the sphere center. (Newton spent nearly 10 years trying to
proof it.)
 GM
Potential energy U 
(3.4).
r
4

r
m
M
=
M
M
HKPhO Pre-Training 2005
Application of superposition: Uniform density
larger sphere with a smaller spherical hole.

r1
Near Earth surface, because the large radius
of Earth, the gravitation field of Earth can be
taken as constant, and its amplitude is

r2
-
=

r1
O
g
GM E
= 9.8 m/s2
RE2

r2
O
O
(3.5).
Its direction is pointing towards the center of Earth, which in practice can be regarded as
‘downwards’ in most cases.
Example 3.2
y
The ‘weight’, or the force of the ground on a person on
different places on Earth. Take the radius of Earth as R, and
the rotation speed being  (= 2/86400 s-1).
On the Equator, we have mg - N = ma = m2R,
so N = mg - m2R = m(g - 2R)

N

a

 f
mg


mg
At the South Pole (or North Pole), we have N = mg
At latitude , by breaking down the forces along
the X-direction and Y-direction, we have

N
x

mg

N
(mg  N ) cos   f sin   ma , and
(mg  N ) sin   f cos   0 .
Solving the two equations, we get (mg  N )  ma cos  , f  ma sin  , where a   2 R cos .
The negative sign of f means that its direction is the opposite of what we have guessed.
One can also break down the forces along the tangential and radial directions to obtain the
same answers. One can also take Earth as reference frame and introduce the inertia force to
account for the rotational acceleration.
3.7
Buoyancy
In a fluid (liquid or gas) of mass density  at depth H,
consider a column of it with cross section area A, then the
total mass of the column is AH, and the gravity on it is
AHg. The gravity must be balanced by the supporting force
from below, so the force of the column on the rest of the
liquid is F = AHg and pointing downwards. The pressure
5
H

mg
HKPhO Pre-Training 2005
P = F/A = Hg
(3.6).
Now consider a very small cubic of fluid with all six side area of A at depth H. The force on
its upper surface is AHg and pointing down, the force on its lower surface is AHg but
pointing upwards so the cubic is at rest. However, for the cubic not to be deformed by the two
forces on its upper and lower surfaces, the forces on its side surfaces must be of the same
magnitude. This leads to the conclusion that the pressure on any surface at depth H is AHg,
and its direction is perpendicular to the surface. One can then easily prove that the net force of
the fluid (buoyancy) on a submerged body of volume V is equal Vg. (See the HKPhO 2003
paper.) The buoyancy force is acting on the center of mass of the submerged portion of the
object.
3.8
Torque
When two forces of equal amplitude and opposite directions
acting upon the two ends of a rod, the center of the rod remains
stationary but the rod will spin around the center. The torque (of
a force) is introduced to describe its effect on the rotational
motion of the object upon which the force is acting. First, an
origin (pivot) point O should be chosen. The amplitude of the

torque of force F is
 = rF
(3.7),

where r is the distance between F and the origin O. The
direction of the torque (a vector as well) is point out of the paper
surface using the right hand rule. One can choose any point as
origin, so the torque of a force depends on the choice of origin.
However, for two forces of equal amplitude and opposite
directions, the total torque is independent of the origin.

F
O
r
O

r
The general form of torque is defined as



  r F

F
(3.8),
which involves the cross-product of two vectors.
4 Oscillations
4.1
Simple Harmonic Motion
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HKPhO Pre-Training 2005
Frequency f and Period T:
T
x(t )  xm cos(t   )
(4.1)
1
f
(a) Effects of different amplitudes
(b) Effects of different periods
(c) Effects of different phases
Since the motion returns to its initial value after
one period T,
xmcos(t  )  xmcos[(t T )  ],
t   2  (t T )  ,
T 2 .
Thus

2
 2f .
T
(4.2)
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HKPhO Pre-Training 2005
Velocity
dx d
v(t ) 
 [ x cos(t   )],
dt dt m
v(t )  xm sin( t   )].
(4.3a)
Velocity amplitude: vm   xm
(4.4).
Acceleration
dv d
a(t ) 
 [xm sin( t   )],
dt dt
a(t )   2 xm cos(t   )].
(4.5)
(4.3b)
Acceleration amplitude a m   2 xm (4.6).
This equation of motion will be very useful in identifying simple harmonic motion and its
frequency.
4.2
The Force Law for Simple Harmonic Motion
Consider the simple harmonic motion of a block of mass m subject to the elastic force of a
spring
F  kx (Hook’s Law)
(4.7).
Newton’s law:
F  kx  ma.
d 2x
m 2  kx  0.
dt
d 2x k
 x  0.
dt 2 m
(4.8)
Comparing with the equation of motion for simple harmonic motion,
k
(4.9)
2  .
m
Simple harmonic motion is the motion executed by an object of mass m subject to a force that
is proportional to the displacement of the object but opposite in sign.
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HKPhO Pre-Training 2005
Angular frequency:

k
m
(4.10)
Period:
T  2
m
k
(4.11)
Examples 4.1
A block whose mass m is 680 g is fastened to a spring whose spring constant k is 65 Nm-1.
The block is pulled a distance x = 11 cm from its equilibrium position at x = 0 on a frictionless
surface and released from rest at t = 0.
(a) What force does the spring exert on the block just before the block is released?
(b) What are the angular frequency, the frequency, and the period of the resulting oscillation?
(c) What is the amplitude of the oscillation?
(d) What is the maximum speed of the oscillating block?
(e) What is the magnitude of the maximum acceleration of the block?
(f) What is the phase constant  for the motion?
Answers:
(a)
F  kx  65  0.11  7.2 N
k
65
(b)


 9.78 rad s -1
m
0.68

f 
 1.56 Hz
2
1
 0.643 s
f
xm  11 cm
vm   xm  1.08 ms -1
T
(c)
(d)
(e)
(f)
am   2 xm  9.782  0.11  10.5 ms - 2
At t = 0,
x(0)  xm cos  0.11
v(0)   xm sin   0
(2)
(2): sin   0    0
(1)
Example 4.2
At t = 0, the displacement of x(0) of the block in a linear oscillator is – 8.50 cm. Its velocity
v(0) then is – 0.920 ms-1, and its acceleration a(0) is 47.0 ms-2.
(a) What are the angular frequency  and the frequency f of this system?
(b) What is the phase constant ?
(c) What is the amplitude xm of the motion?
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HKPhO Pre-Training 2005
(a)
At t = 0,
x(t )  xmcos(t   )  xmcos  0.085. (1)
v(t )   x sin(t  )   x sin  0.920. (2)
a(t )   2 xmcos(t   )   2 xmcos  47.0. (3)
m
m
(3)  (1): a(0)   2 .
(b)
x(0)
a(0)
47.0
 
 
 23.5 rad s - 1. (answer)
x(0)
 0.0850
sin 
(2)  (1): v(0)
 
  tan  .
x(0)
cos
tan   
v(0)
 0.920

 0.4603.
x(0)
(23.51)(0.085)
 = –24.7o or 180o – 24.7o = 155o.
(c)
One of these 2 answers will be chosen in (c).
(1):
x(0)
xm 
.
cos
 0.085
For  = –24.7o,
xm 
m  9.4 cm.
cos(24.7o )
For  = 155o, x   0.085 m  9.4 cm.
m
cos155o
Since xm is positive,  = 155o and xm = 9.4 cm.
(answer)
Example 4.3
A uniform bar with mass m lies symmetrically across two rapidly rotating, fixed rollers, A and
B, with distance L = 2.0 cm between the bar’s centre of mass and each roller. The rollers slip
against the bar with coefficient of kinetic friction k = 0.40. Suppose the bar is displaced
horizontally by a distance x, and then released. What is the angular frequency  of the
resulting horizontal simple harmonic (back and forth) motion of the bar?
Newton’s law:
 F y  F A  FB  mg  0.
 Fx  f kA  f kB  ma.
(1)
(2)
or  F   F  ma.
k A
k B
Considering torques about A,
 z  FA  0  FB  2 L  mg ( L  x)  f kA  0  f kB  0  0. (3)
or FB  2 L  mg ( L  x).
(3): F  mg ( L  x) .
B
2L
mg ( L  x)
.
(1): FA  mg  FB 
2L
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HKPhO Pre-Training 2005
(2):  k [ mg ( L  x )  mg ( L  x )]  ma.
2L
a
g k
L
x  0.
2
Comparing with d x   2 x  0 for simple harmonic motion,  2 
2
dt

k g
4.3
Energy in Simple Harmonic Motion
L

k g ,
L
(0.40)(9.8)
 14 rad s -1. (answer)
0.02
Potential energy:
Since x(t )  xmcos(t   ),
1
1
U(t)  kx 2  kxm2 cos2 (t   ). (4.12)
2
2
Kinetic energy:
Since v(t )   xm sin(t   ),
K (t )  mv 2  m 2 xm2sin 2 (t  ).
1
1
2
2
(4.13)
Since  2  k / m,
K(t)  kxm2 sin 2 (t   ).
1
2
Mechanical energy:
E  U  K  kxm2 cos2 (t   )  kxm2 sin 2 (t   )

1
2
1
1
2
2
kxm2 [cos2 (t   )  sin 2 (t   )].
Since cos2 (t  )  sin 2 (t  )  1,
E  U  K  kxm2 .
1
2
11
(4.14)
HKPhO Pre-Training 2005
(a) The potential energy U(t), kinetic energy K(t), and mechanical energy E as functions of
time, for a linear harmonic oscillator. Note that all energies are positive and that the potential
energy and kinetic energy peak twice during every period. (b) The potential energy U(t),
kinetic energy K(t), and mechanical energy E as functions of position, for a linear harmonic
oscillator with amplitude Xm. For x = 0 the energy is all kinetic, and for x = Xm it is all
potential.
The mechanical energy is conserved.
4.4
The Simple Pendulum
Consider the tangential motion acting on the mass.
Using Newton’s law of motion,
d 2
 mgsin   ma  mL 2 ,
dt
2
d g
(4.14)
 sin  0.
2
dt
L
When the pendulum swings through a small angle,
sin  . Therefore
d 2 g
   0.
dt 2 L
(4.15)
Comparing with the equation of motion for simple harmonic motion,  2 
T  2
L
.
g
5 The Centre of Mass
5.1
Definition
The centre of mass of a body or a system of
bodies is the point that moves as though all of
the mass were concentrated there and all
external forces were applied there.
For two particles,
xcm 
m1 x1  m2 x2 m1 x1  m2 x2

.
m1  m2
M
12
g
and
L
HKPhO Pre-Training 2005
For n particles,
xcm 
m1 x1  mn xn
.
M
In general and in vector form,
1 n 

rcm 
m r .
M i 1 i i
(5.1)
If the particles are in a uniform gravity field, then the
total torque relative to the center of mass is zero. The
same applies for the inertia force when the particles
are in an accelerating reference frame.
Proof:
n
 n



 
  n


   mi (ri  rcm )  g  (  mi ri  rcm  mi )  g  ( Mrcm  Mrcm )  g  0
i 1
5.2
i 1
i 1
Rigid Bodies
1
1
 xdm 
 x  dV ,
M
M
1
1
ycm  
 y dm   y  dV ,
M
M
1
1
z cm 
 z dm 
 z  dV
M
M
xcm 
(5.2)
where  is the mass density.
If the object has uniform density,

dm

dV
M
.
V
Rewriting dm   dV and m  V , we obtain
13
(5.3)
HKPhO Pre-Training 2005
1
 x dV ,
V
1
ycm   y dV ,
V
1
zcm   z dV .
V
xcm 
(5.4)
Similar to a system of particles, if the rigid body is in a uniform gravity field, then the total
torque relative to its center of mass is zero. This is true even when the density of the object is
non-uniform. The same applies to the inertia force. The proof is very much the same as in the
case for particles. One only needs to replace the summation by integration operations.
5.3
Newton’s Second Law for a System of Particles
In terms of X-Y-Z components,
 Fext , x  Macm , x ,
 Fext , y  Ma cm , y ,
(5.5)
 Fext , z  Ma cm , z .
5.4
Linear Momentum
For a single particle, the linear momentum is


p  mv.
(5.6)
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HKPhO Pre-Training 2005
Newton’s Law:



dv d

 dp
 (mv )  .
(5.7)
 F  ma  m
dt dt
dt
This is the most general form of Newton’s Second Law. It accounts for the change of mass as
well.



I    F dt  p f  pi .
(5.8)
The change of momentum is equal to the time integration of the force, or impulse.
For a system of particles, the total linear momentum is
 



P  p1    pn  m1v1    mn vn .
(5.9)
Differentiating the position of the centre of mass,



Mvcm  m1v1    mn vn .


P  Mvcm .
(5.10)
The linear momentum of a system of particles is equal to the product of the total mass M of
the system and the velocity of the centre of mass.
Apply Newton’s Law’s to the particle system,



mi ai (t )  Fi ext   Fij ,
j i
 1

X
 mi xi , ( M   mi  total mass)
i
M i
 1

V 
 mi vi
M i
 1
 

1   ext
 A
 mi ai    Fi   Fij 
j i
M i
M i 



According to the Third Law, Fij   F ji . So

 Fij  0
(5.11)
j i
and

 1   ext
Ftotext

A
 Fi  (0) 
 M
M  i
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(5.12)
HKPhO Pre-Training 2005
Newton’s law:


dvcm
dP

M
 Ma .
cm
dt
dt
(5.13)
Hence


dP
 Fext  .
dt
(5.14)



If  Fext  0 , then MV   mi vi  const
(5.15)
i
The total momentum of a system is conserved if the total external force is zero.
5.5
Rigid body at rest
The necessary and sufficient conditions for the balance of a rigid body is that net external
force = 0, and the net torque due to these external forces = 0, relative to any origin (pivot).
Choosing an appropriate origin can sometimes greatly simplify the problems. A common trick
is choosing the origin at the point where an unknown external force is acting upon.
Example 5.1
A uniform rod of length 2l and mass m is fixed on one end by
a thin and horizontal rope, and on the wall on the other end.
Find the tension in the rope and the force of wall acting upon
the lower end of the rod.
Answer:
The force diagram is shown. Choose the lower end as the

origin, so the torque of the unknown force F is zero. By
balance of the torque due to gravity and the tension, we get
mglsin - Tlcos = 0, or T = mg tan

T


F

mg

Breaking F along the X-Y (horizontal-vertical) directions, we get Fx = T, and Fy = mg.
It is interesting to explore further. Let us choose another point of origin for the consideration
of torque balance. One can easily verify that with the above answers the total torque is
balanced relative to any point of origin, like the center of the rod, or the upper end of the rod.
Can you prove the following?
If a rigid body is at rest, the total torque relative to any pivot point is zero.
5.6
Conservation of Linear Momentum
If the system of particles is isolated (i.e. there are no external forces) and closed (i.e. no
particles leave or enter the system), then
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HKPhO Pre-Training 2005

P  constant .
(5.16)
Law of conservation of linear momentum:
 
Pi  Pf .
(5.17)
Example 5.2
Imagine a spaceship and cargo module, of total mass M, traveling in deep space with velocity
vi = 2100 km/h relative to the Sun. With a small explosion, the ship ejects the cargo module,
of mass 0.20M. The ship then travels 500 km/h faster than the module; that is, the relative
speed vrel between the module and the ship is 500 km/h. What then is the velocity vf of the
ship relative to the Sun?
Using conservation of linear momentum,
Pi  Pf
MVi  0.2M (v f  vrel )  0.8Mv f
vi  v f  0.2vrel
v f  vi  0.2vrel
= 2100 + (0.2)(500)
= 2200 km/h (answer)
Example 5.3
Two blocks are connected by an ideal spring and are free to slide on a frictionless horizontal
surface. Block 1 has mass m1 and block 2 has mass m2. The blocks are pulled in opposite
directions (stretching the spring) and then released from rest.
(a) What is the ratio v1/v2 of the velocity of block 1 to the velocity of block 2 as the
separation between the blocks decreases?
(b) What is the ratio K1/K2 of the kinetic energies of the blocks as their separation decreases?
Answer
(a) Using conservation of linear momentum,
Pi  Pf
0  m1v1  m2v2
v1
m
 2
v2
m1
1
2
2
2
K1 2 m1v1 m1  v1 
m1  m2  m2
  

 


(b)
K 2 1 m v 2 m2  v2  m2  m1 
m1
2 2
2
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HKPhO Pre-Training 2005
Example 5.4
A firecracker placed inside a coconut of mass M, initially at rest on a frictionless floor, blows
the fruit into three pieces and sends them sliding across the floor. An overhead view is shown
in the figure. Piece C, with mass 0.30M, has final speed vfc=5.0ms-1.
(a) What is the speed of piece B, with mass 0.20M?
(b) What is the speed of piece A?
Answer:
(a)
Using conservation of linear momentum,
Pix  Pfx
(b)
Piy  Pfy
mC v fC cos 80o  mBv fB cos 50o  mAv fA  0
(1)
mC v fC sin 80  mBv fB sin 50  0
mA = 0.5M, mB = 0.2M, mC = 0.3M.
(2): 0.3Mv fC sin 80o  0.2Mv fB sin 50o  0
(2)
o
o
(0.3)(5) sin 80o
 9.64 ms -1  9.6 ms -1 (answer)
0.2 sin 50o
(b) (1): 0.3Mv fC cos 80o  0.2Mv fB cos 50o  0.5Mv fA
v fB 
(0.3)(5) cos 80o  (0.2)(9.64) cos 50o
 3.0 ms -1
0.5
(answer)
v fA 
5.7
Elastic Collisions in One Dimension
In an elastic collision, the kinetic energy of each colliding body can change, but the total
kinetic energy of the system does not change.
In a closed, isolated system, the linear momentum of each colliding body can change, but the
net linear momentum cannot change, regardless of whether the collision is elastic.
In the case of stationary target, conservation of
linear momentum:
m1v1i  m1v
1f
 m2v2 f .
Conservation of kinetic energy:
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HKPhO Pre-Training 2005
1
1
1
m1v12i  m1v12f  m2 v22 f .
2
2
2
Rewriting these equations as
m1 (v1i  v1 f )  m2 v2 f ,
m1 (v 2  v12f )  m2 v22 f .
1i
Dividing,
v1i  v1 f  v2 f .
We have two linear equations for v1f and v2f. Solution:
v1 f 
m1  m2
v ,
m1  m2 1i
v2 f 
2m1
v .
m1  m2 1i
Motion of the centre of mass:
vcm 
m1
P

v .
m1  m2 m1  m2 1i
Example 5.5
In a nuclear reactor, newly produced fast neutrons must be slowed down before they can
participate effectively in the chain-reaction process. This is done by allowing them to collide
with the nuclei of atoms in a moderator.
(a) By what fraction is the kinetic energy of a neutron (of mass m1) reduced in a head-on
elastic collision with a nucleus of mass m2, initially at rest?
(b) Evaluate the fraction for lead, carbon, and hydrogen. The ratios of the mass of a nucleus
to the mass of a neutron (= m2/m1) for these nuclei are 206 for lead, 12 for carbon and
about 1 for hydrogen.
Answer
(a)
Conservation of momentum
m1v1i  m1vif  m2v2 f
For elastic collisions,
1
1
1
m1v12i  m1v12f  m2v22 f
2
2
2
m1 (v1i  v1 f )  m2v2 f
(1)
m1 (v12i  v12f )  m2v22 f
(2)
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HKPhO Pre-Training 2005
Dividing (1) over (2), v1i  v1 f  v2 f
(1): m1 (v1i  v1 f )  m2 (v1i  v1 f )
v1 f 
m1  m2
v1i
m1  m2
Fraction of kinetic energy reduction

Ki  K f
Ki
1
1
m1v12i  m1v12f v 2  v 2
v2
2
2
 1i 2 1 f  1  12f
1
v1i
v1i
m1v12i
2
2
 m  m2 
4m1m2
 
 1   1
(answer)
2
 m1  m2  (m1  m2 )
(b) For lead, m2 = 206m1,
4m1 (206m1 )
4(206)

 1.9% (answer)
Fraction 
2
(m1  206m1 )
207 2
For carbon, m2 = 12m1,
4m1 (12m1 )
4(12)

 28% (answer)
Fraction 
2
(m1  12m1 )
132
For hydrogen, m2 = m1,
4m1 (m1 )
 100% (answer)
(m1  m1 ) 2
In practice, water is preferred.
Fraction 
5.8
Inelastic Collisions in One Dimension
In an inelastic collision, the kinetic energy of the
system of colliding bodies is not conserved.
In a completely inelastic collision, the colliding
bodies stick together after the collision.
However, the conservation of linear momentum still holds.
m1v  (m1  m2 )V , or V 
m1
v.
m1  m2
Examples 5.6
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(3)
HKPhO Pre-Training 2005
The ballistic pendulum was used to measure the speeds of bullets before electronic timing
devices were developed. Here it consists of a large block of wood of mass M = 5.4 kg,
hanging from two long cords. A bullet of mass m = 9.5 g is fired into the block, coming
quickly to rest. The block + bullet then swing upward, their centre of mass rising a vertical
distance h = 6.3 cm before the pendulum comes momentarily to rest at the end of its arc.
(a) What was the speed v of the bullet just prior to the collision?
(b) What is the initial kinetic energy of the bullet? How much of this energy remains as
mechanical energy of the swinging pendulum?
Answer
(a) Using conservation of momentum during collision,
mv  ( M  m)V
Using conservation of
energy after collision,
1
( M  m)V 2  ( M  m) gh
2
V  2 gh
M m
v
V
m
M m
5.4  0.0095

2 gh 
2(9.8)(0.063)  630 ms -1
m
0.0095
(b) Initial kinetic energy
1
1
K  mv2  (0.0095)6302  1900 J
2
2
Final mechanical energy
E  ( M  m) gh  (5.4  0.0095)(9.8)(0.063)  3.3 J
(only 0.2%) (answer)
Example 5.7
(The Physics of Karate) A karate expert strikes downward with his fist (of mass m1 = 0.70 kg),
breaking a 0.14 kg wooden board. He then does the same to a 3.2 kg concrete block. The
spring constants k for bending are 4.1  104 Nm-1 for the board and 2.6  106 Nm-1 for the
block. Breaking occurs at a deflection d of 16 mm for the board and 1.1 mm for the block.
(a) Just before the board and block break, what is the energy stored in each?
(b) What fist speed v is required to break the board and the block? Assume that mechanical
energy is conserved during the bending, that the fist and struck object stop just before the
break, and that the fist-object collision at the onset of bending is completely inelastic.
Answer
1 2 1
kd  (4.1 104 )0.0162  5.248 J  5.2 J (answer)
2
2
1 2 1
For the block, U  kd  (2.6  106 )0.00112  1.573 J  1.6 J (answer)
2
2
(a) For the board, U 
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HKPhO Pre-Training 2005
(b)
For the board, first the fist and the board undergoes
an inelastic collision. Conservation of momentum:
m1v  (m1  m2 )V
(1)
Then the kinetic energy of the fist and the board is
converted to the bending energy of the wooden board.
Conservation of energy:
(2): V 
(1): v 
2(5.248)
2U

 3.534 ms -1
m1  m2
0.7  0.14
m1  m2
 0.7  0.14 
V 
3.534  4.2 ms-1 (answer)
m1
0.7


For the concrete block,
V
v
2(1.573)
2U

 0.8981 ms -1 .
m1  m2
0.7  3.2
m1  m2
 0.7  3.2 
V 
0.8981  5.0 ms-1 (answer)
m1
 0.7 
The energy to break the concrete block is 1/3 of that for the wooden board, but the fist speed
required to break the concrete block is 20% faster! This is because the larger mass of the
block makes the transfer of energy to the block more difficult.
5.9
Collisions in Two Dimensions
Conservation of linear momentum:
x component: m1v1i  m1v1 f cos1  m2 v2 f cos 2 ,
y component: 0  m1v1 f sin 1  m2 v2 f sin  2.
Conservation of kinetic energy:
1
1
1
1
m1v12i  m2 v22i  m1v12f  m2 v22f .
2
2
2
2
Typically, we know m1 , m 2 , v1i and 1. Then we can solve for v1 f , v 2 f and 2.
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HKPhO Pre-Training 2005
Examples 5.8
Two particles of equal masses have an elastic collision, the target particle being initially at
rest. Show that (unless the collision is head-on) the two particles will always move off
perpendicular to each other after the collision.
Using conservation of momentum,



mv1i  mv1 f  mv2 f



v1i  v1 f  v2 f
The three vector form a triangle.
In this triangle, cosine law:
v12i  v12f  v22 f  2v1 f v2 f cos 
(1)
Using conservation of energy:
1 2 1 2 1 2
mv1i  mv1 f  mv2 f
2
2
2
2
2
2
v1i  v1 f  v2 f
(2)
(1) – (2): 2v1 f v2 f cos   0
  90o (answer)
Example 5.9
Two skaters collide and embrace, in a completely inelastic collision. That is, they stick
together after impact. Alfred, whose mass mA is 83 kg, is originally moving east with speed vA
= 6.2 km/h. Barbara, whose mass mB is 55 kg, is originally moving north with speed vB = 7.8
km/h.

(a) What is the velocity V of the couple after impact?
(b) What is the velocity of the centre of mass of the two skaters before and after the
collision?
(c) What is the fractional change in the kinetic
energy of the skaters because of the
collision?
(a) Conservation of
momentum:
mAvA  (mA  mB )V cos 
mBvB  (mA  mB )V sin 
(2)  (1):
(1)
(2)
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HKPhO Pre-Training 2005
tan  
mB vB (55)(7.8)

 0.834 , so  = 39.8o  40o (answer)
mAv A (83)(6.2)
m Av A
(83)(6.2)

= 4.86 km/h  4.9 km/h (answer)
(mA  mB ) cos  (83  55) cos 39.8o
(b) Velocity of the centre of mass is not changed by the collision. Therefore V = 4.9 km/h and
 = 40o both before and after the collision. (answer)
1
1
(c) Initial kinetic energy Ki  mAv A2  mB vB2 = 3270 kg km2/h2.
2
2
1
Final kinetic energy K f  (mA  mB )V 2 = 1630 kg km2/h2.
2
K  K f 1630  3270

 50% (answer)
Fraction  i
Ki
3270
(1): V 
6 General equations of motion in the X-Y plane with constant forces
6.1 General formulae
F 
v x (t )  v x o   x t
m
 Fy 
v y (t )  v yo   t
m
1 F
x(t )  xo  v x o t   x
2 m
(6.1)
2
t

(6.2)
1  Fy  2
y (t )  yo  v yo t   t
2 m 
These general formulae can be applied to any situations, once the initial conditions (x0, y0, vx0,
vy0) are given.
A word of caution: Friction forces are not always ‘constant’. Pay attention to their directions
because they change with the direction of the velocity.
6.2
Projectile motion near Earth surface
The only force on the object is the gravity which is along the –Y direction. Accordingly, we
have
F 
v x (t )  v x o   x t  v x o
m
(6.3)
 Fy 
v y (t )  v yo   t  v yo  gt
m
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HKPhO Pre-Training 2005
for the velocity and
1F 
x(t )  xo  v x o t   x t 2  xo  v x o t
2 m 
1  Fy 
1
y (t )  y o  v yo t   t 2  y o  v yo t  gt 2
2 m 
2
(6.4)
for the position.
These general formulae can be applied to any situations, once the initial conditions (x0, y0, vx0,
vy0) are given.
.end.
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