HKPhO Pre-Training Workshop
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
 Dot product of two vectors
A  B  AB cos   Ax Bx  Ay By  Az Bz

B

A

C
(1.2)

A
(1.3)


It is also referred to as the projection of A on B , and vise versa.
 Amplitude of the vector
| A | Ax2  Ay2  Az2  A  A


B
(1.4)
z
1.2




Position vector of a particle: r  xx0  yy0  zz 0
(x,y,z)

r
(1.5).
o
If the particle is moving, then x, y, and z are functions of time t.
Velocity:

x
 dr dx  dy  dz 



v

x0 
y0 
z0  v x x0  v y y0  v z z0 (1.6)
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.
  t  
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 cos(t  ) x0  R 2 sin( t  ) y0   2 r
1
(1.11).
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Its amplitude is a  R 2 
v2
R
(1.12)
2 Relative Motions

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’. S is moving relative to S’.
r
  
Note: r  r ' R .
(2.1)
O



 dr dr ' dR  
So the velocity is:
(2.2)
v


 v'  u
dt dt dt
dv dv ' du
a


 a ' A
Similar for acceleration:
(2.3)
dt
dt dt

 
This is the classic theory of relativity. If u is constant, then A  0 , and 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
There are apparently many kinds of forces. Pull through a rope, push, contact forces (elastic
force and frictional force), air resistance, fluid viscosity, surface tension of liquid and elastic
membrane, gravity, electric and magnetic, strong interaction, weak interaction, just to name a
few. Only the last four are fundamental. All the others are the net effect of the electric and
magnetic force due to electrons and nuclei.
3.1
Tension


T
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 in opposite directions 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.
3.3
Friction
The frictional 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 N,
so a friction coefficient  can be defined.
2

v

F1

F2

N

f
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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.

N
When there is no relative motion but a tendency for such


motion, like a block is being pushed by a force F , the
F
amplitude of f is equal to F until it reaches the limit value

of s N when F keeps increasing, where s is the static
f
friction coefficient. Once the block starts to move, the
friction becomes f = kN. Usually k is smaller than s.
3.4
Viscosity
Air resistance and fluid viscous forces are proportional to the speed of relative motion 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 in Eq. (2.3) 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
acting upon the object.

a
Example 3.1
A block is attached by a spring to the wall and placed on the
smooth surface of a cart which is accelerating. According to

the ground (inertial) frame, the force F acting on the block by
the spring 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 there is a force on the block by the spring.


This force is ‘balanced’ by the inertial force Fint  ma


F  ma
Ground frame




Fint  ma
F  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
r
3

r
M
(3.3).
m
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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
The Gravitational Potential is U 
(3.4).
r
g
GM E
= 9.8 m/s2
2
RE
M
M
Application of superposition: Uniform density
larger sphere with a smaller spherical hole.

r1
Near the Earth surface, because the large
radius of the Earth, the gravitation field of the
Earth can be taken as constant, and its
amplitude is
=

r2
-
=

r1
O

r2
O
O
(3.5).
Its direction is pointing towards the center of the Earth, which in practice can be regarded as
‘downwards’ in most cases.
Example 3.2
y
The ‘weight’ is defined as the force of the ground on a person
on different places on the 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)
At the South/North Poles, we have N = mg.
At latitude , by breaking down the forces along
the X-direction and Y-direction, we have
(mg  N ) cos   f sin   ma , and
(mg  N ) sin   f cos   0 .

N

a

 f
mg


mg

N
x

mg

N
Here f is the friction force, without which the object cannot be balanced. 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 of the circle to
obtain the same answers. One can also take the Earth as the reference frame and introduce the
inertia force to account for the rotational acceleration. All these approaches will lead to the
same answer as above.
3.7
Buoyancy
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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 acting 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
P = F/A = Hg
H

mg
(3.6).
Now consider a very small cube 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 cube is at rest. However, for the cube 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 magnitude and opposite directions acting
upon the two ends of a rod, the center of the rod remains stationary but
the rod will spin around its 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 magnitude of the

torque of force F is defined as
 = 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
magnitude and in opposite directions, the total torque is independent of the
origin.
The general form of torque is defined as
 
  r F


F
O
r
O

r
(3.8),
which involves the cross-product of two vectors.
4 Oscillations
5

F
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4.1
Simple Harmonic Motion
Frequency f and Period T:
Equation of motion:
T
1
f
x(t )  xm cos(t   )
(4.1)
(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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Velocity
dx d
v(t ) 
 [ x cos(t   )],
dt dt m
v(t )  xm sin( t   )].
(4.3a)
Velocity amplitude: vm   xm
(4.4).
(4.3b)
Acceleration
dv d
a(t ) 
 [xm sin( t   )],
dt dt
a(t )   2 xm cos(t  )].
(4.5)
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 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
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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.
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
f 
0.68

 1.56 Hz
2
(c)
1
 0.643 s
f
xm  11 cm
(d)
vm   xm  1.08 ms -1
T
(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?
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HKPhO Pre-Training Workshop
(b) What is the phase constant ?
(c) What is the amplitude xm of the motion?
(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 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).
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(3): F  mg ( L  x) .
B
2L
mg ( L  x)
.
(1): FA  mg  FB 
2L
(2):  k [ mg ( L  x )  mg ( L  x )]  ma.
2L
d 2 x g k

x  0.
dt 2
L
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 )  x cos(t  ), x
m
1
1
U (t )  kx2  kx2 cos2 (t  ).
2
2 m
(4.12)
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,
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(4.14)
E  U  K  kxm2 .
1
2
(a) The potential energy U(t), the
kinetic energy K(t), and the total
mechanical energy E as functions of
time for a 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
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 forces acting on the mass.
Using Newton’s law of motion,
mg sin  ma  mL
d
,
dt
2
2
2
Or d   g sin  0.
dt
2
(4.14)
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
11
g
and
L
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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.
For two particles,
m1 x1  m2 x2 m1 x1  m2 x2

.
m1  m2
M
M  m1  m2 is the total mass of the system.
For n particles,
xcm 
m1 x1  mn xn
.
M
M  m1  m2  ...mn is the total mass of the system.
xcm 
In general and in vector form,
1 n 

rcm 
m r .
M i 1 i i
(5.1)
If the system is in a uniform gravity field, then the
total torque of gravity relative to the center of mass is
zero. The same applies to the inertia force when the
system is in a linear accelerating reference frame.
Proof:
n
 n



 
  n


   mi (ri  rcm )  g  (  mi ri  rcm  mi )  g  ( Mrcm  Mrcm )  g  0
i 1
i 1
i 1
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5.2
Rigid Bodies
1
1
 x dm 
 x  dV ,
M
M
1 
1
 y dm 
ycm 
 y  dV ,

M
M
1
1
zcm 
 z dm 
 z  dV
M
M
xcm 
(5.2)
where  is the mass density.
If the object has uniform density,

dm

dV
M
.
(5.3)
V
Rewriting dm   dV and m  V , we obtain
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 gravitational 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 ,
 Fext , z  Ma cm , z .
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(5.5)
HKPhO Pre-Training Workshop
5.4
Linear Momentum
For a single particle, the linear momentum is


p  mv.
(5.6)
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    Fdt  p f  pi .
(5.8)
The change of momentum is equal to the time integration of the total force  F , 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 (Eq. 5.1),



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.
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HKPhO Pre-Training Workshop
Apply Newton’s Laws 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
(5.12)
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 a rigid body at rest is that the 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 to
choose the origin at the point where an unknown external force is acting upon.
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HKPhO Pre-Training Workshop
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 at 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,
T

so the torque of the unknown force F is zero. By balance of the
torque due to gravity and the tension, we get
1

mglsin – 2Tlcos = 0, or T  mg tan 

2

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 a system of particles is isolated (i.e. there are no external forces) and closed (i.e. no particles
leave or enter the system), then

P  constant .
(5.16)
Law of conservation of linear momentum:
 
Pi  Pf .
(5.17)
Example 5.2
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 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
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HKPhO Pre-Training Workshop
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
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 =5.0 ms-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 and Piy  Pfy
mC v fC cos 80o  mBv fB cos 50o  mAv fA  0
(1)
(2)
mC v fC sin 80o  mBv fB sin 50o  0
mA = 0.5M, mB = 0.2M, mC = 0.3M.
o
o
(2): 0.3Mv fC sin 80  0.2Mv fB sin 50  0
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HKPhO Pre-Training Workshop
(0.3)(5) sin 80o
 9.64 ms -1  9.6 ms -1 (answer)
o
0.2 sin 50
o
o
(b) (1): 0.3Mv fC cos 80  0.2Mv fB cos 50  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:
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
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HKPhO Pre-Training Workshop
vcm 
Motion of the centre of mass:
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,
Dividing (1) over (2),
1
1
1
m1v12i  m1v12f  m2v22 f
2
2
2
m1 (v1i  v1 f )  m2v2 f
(1)
m1 (v12i  v12f )  m2v22 f
(2)
v1i  v1 f  v2 f
(3)
(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,
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HKPhO Pre-Training Workshop
4m1 (12m1 )
4(12)

 28% (answer)
2
(m1  12m1 )
132
For hydrogen, m2 = m1,
Fraction 
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
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
v
M m
V
m
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HKPhO Pre-Training Workshop
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
1
For the block, U  kd 2  (2.6  106 )0.00112  1.573 J  1.6 J (answer)
2
2
(b)
For the board, first the fist and the board undergoes
an inelastic collision. Conservation of momentum:
(a) For the board, U 
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.
By 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,
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HKPhO Pre-Training Workshop
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.
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 vectors form a triangle.
In this triangle, cosine law:
v12i  v12f  v22 f  2v1 f v2 f cos 
(1)
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HKPhO Pre-Training Workshop
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:
(1)
mAvA  (mA  mB )V cos 
(2)
mBvB  (mA  mB )V sin 
(2)  (1):
mv
(55)(7.8)
tan   B B 
 0.834 , so  = 39.8o  40o
mAv A (83)(6.2)
(answer)
m Av A
(83)(6.2)

(mA  mB ) cos  (83  55) cos 39.8o
= 4.86 km/h  4.9 km/h (answer)
(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 
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HKPhO Pre-Training Workshop
6 General equations of motion in the X-Y plane with constant forces
6.1 General formulae
F 
vx (t )  vx 0   x  t
m
 Fy 
v y (t )  v y 0    t
m
1 F 
x(t )  x0  vx 0t   x  t 2
2 m 
(6.1)
(6.2)
1  Fy  2
y (t )  y0  v y 0t    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: Frictional 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 
vx (t )  vx 0   x  t  vx 0
m
(6.3)
 Fy 
v y (t )  v y 0    t  v y 0  gt
m
for the velocity and
1 F 
x(t )  x0  vx 0t   x  t 2  x0  vx 0t
2 m 
(6.4)
1  Fy  2
1 2
y (t )  y0  v y 0t    t  y0  v y 0t  gt
2 m 
2
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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