THEORETICAL COMPETITION Questions and Solutions
THEORETICAL COMPETITION Questions and Solutions
Question 1
1A. SPRING CYLINDER WITH MASSIVE PISTON (5 points)
Consider n=2 moles of ideal Helium gas at a pressure
P and temperature T = 300 K placed in a vertical SPRING
cylindrical container (see Figure 1.1). A moveable frictionless horizontal piston of mass M = 50 kg
2
2
(assume g = 9.8 m/s ) and cross section A = 100 cm
PISTON
compresses the gas leaving the upper section of the container void. There is a vertical spring attached to the piston and the upper wall of the container. Initially the system in equilibrium and the spring is GAS unstreched. Then the piston is pushed down until the gas volume halved, and released. Disregarding any gas leakage through their surface contact, and neglecting the specific thermal capacities of the
Figure 1.1 container, piston and spring, calculate4 gV
a. the gas volume when the piston velocity is (3 points)
5 A
b. the frequency f of small oscillation of the piston, when it is slightly displaced from equilibrium position. (2 points) Let the spring constant k = MgA/V (note V is the initial volume of gas). All the
- 1 -1
processes in gas are adiabatic. Gas constant R = 8.314 JK mol . For monoatomic gas (Helium) use Laplace constant = 5/3.
THEORETICAL COMPETITION Questions and Solutions
Solution:
a) Gas Volume
At the initial condition, the system is in equilibrium and the spring is unstreched; therefore
Mg P A Mg P
or (1)
A
The initial volume of gas
nRT nRT A
V
(2)
P Mg
The work done by the gas from ½ V to V
1
V V
PV PV
V 0 0 0 0 1
W PdV dV
V
gas
V
V V1
2 2 2
(3)
Eq. (3) can also be obtained by calculating the internal energy change (without integration)
W E nC T T '
(4)
gas
V
T ' where is the temperature when the gas volume is V /2.
The change of the gravitational potential energy
1 V
V
(5)
2
mg h mg
PE A
THEORETICAL COMPETITION Questions and Solutions
The change of the potential energy of the spring
1
1
2
2 kx kx
spring
2
2
2
2 V
V V V / 2
1 MgA 1 MgA
(6)
2 V A
2 V A
2
1 MgV
V
1 MgV
1
2 A
V
8 A
The kinetic energy
1
1 4 gV
2 MgV
2 KE M v M
(7)
2
2
5 A
5 A By conservation of energy, we have
W KE
(8)
gas PE spring
V
2 1
V
PV
V MgV MgV MgV
1
V
1
2
1 0 0
2 V Mg
1
1
2 A
2 A
V
8 A
5 A
(9)
V
2 1 1
V
MgV
V
1 MgV
V
11 MgV
2 Mg
1
1
A
1 V
2 A
2 A
V
40 A
(10)
Let s = V/V , so the above equation becomes
1
1
1
1
1
2
11
1 s s
1 s
1
2
2
2
40
(11)
THEORETICAL COMPETITION Questions and Solutions
(12) Solving equation (12) numerically, we get
V V Mg
nRT A
0.74 0.74 75.3litres
1
Therefore
s 1 = 0.74 and s 2 = 1.30
With = 5/3 we get
s s
2
1 2 40 2
1 11 3
2/ 3 2 2 / 3
or V 2 = 1.30V =132.3 litres. THEORETICAL COMPETITION Questions and Solutions
b) Small Oscillation (2 points)
The equation of motion when the piston is displaced by x from the equilibrium position is
Mx kx PA Mg
(13)
P is the gas pressure PV PV P 0 0 0 0
P
V V Ax
(14)
Ax
1
V
Ax
Ax
V P P
1
Since then we have , therefore
V
Ax
Mx kx P A
1 Mg
V
A
Mx k P A x
V
(15) MgA Mg A
Mx A x
V A
V
MgA Mx
1 x
V The frequency of the small oscillation is
2
1 gA
1 Mg
f
1
1 (16)
2 V 2 nRT Numerically f = 0.255 Hz.
THEORETICAL COMPETITION Questions and Solutions
1 = 75.3 litres or V 2 132.3 litres
V
V Ax PV PV P
0 0 0 0
0.3 Pressure
kx PA Mg
0.5 Force Equation mx
b (2.0)
0.2 V
0.2 Approximation
2 = 1.30 V
1 = 0.74 V or V
0.3 V
s s
1 Ax
1 2 40 2
f nRT
2 Mg
1
1
2
0.3
P P
V
mx x
1 MgA
0.5 Equation
V
2
1 11 3
[Marking Scheme] THEORETICAL Question 1A
gas PV
0.3 Gravitational Potential Energy
V
V W
2
2 PE
1
1 0 0
1
0.3 Work done by gas
0.2 Initial volume V = nRT A/Mg
0.2 Initial Pressure P = Mg/A
a. (3.0)
Spring Cylinder with Massive Piston
1
mg
2/ 3 2 2 / 3
V A
Equation
0.9(*)
gas PE spring W KE
0.3 Conservation of energy
V A
spring MgV MgV
V V A
8
2
1
1
1
2
0.3 Spring Potential Energy
0.2 f = 0.255 Hz (*) Propagation errors reduce marks halved.
THEORETICAL COMPETITION Questions and Solutions
1B. THE PARAMETRIC SWING (5 points)
A child builds up the motion of a swing by standing and squatting. The trajectory followed by the center of mass of the child is illustrated in Fig. 1.2. Let r be the radial
u
distance from the swing pivot to the child’s center of mass when the child is standing, while r is the radial distance from the swing pivot to the child’s center of mass when
d 1/10
the child is squatting. Let the ratio of r d to r u be 2 = 1.072, that is the child moves its center of mass by roughly 7% compared to its average radial distance from the swing pivot. To keep the analysis simple it is assumed that the swing be massless, the swing amplitude is sufficiently small and that the mass of the child resides at its center of mass. It is also assumed that the transitions from squatting to standing (the A to B and the E to F transitions) are fast compared to the swing cycle and can be taken to be instantaneous. It is similarly assumed that the squatting transitions (the C to D and the G to H transitions) can also be regarded as occurring instantaneously.
Figure 1.2 How many cycles of this maneuver does it take for the child to build up the amplitude (or the maximum angular velocity) of the swing by a factor of two? (Note: you may use the following approximation but with this approximation you will lose 1 point, assume that the work done against gravity at the A to B transition is equal to the work recovered from gravity at the C to D transition, that is motion at THEORETICAL COMPETITION Questions and Solutions these transitions is assumed to be essentially vertical. Similar assumptions also apply to the E to F and G to H transitions).
THEORETICAL COMPETITION Questions and Solutions
Solution (5 points)
(1) The conservation of angular momentum (CAM)
2
(M= mass of the child which dominated the mass of the swing
L I Mr
= the and r = distance of the child’s center of mass to the swing’s pivot P, swing’s angular velocity with respect to P) during the passage of the swing through its equilibrium/vertical position when r changes from r to r in either
d u st
are the angular velocity before and after the (very brief 1
d u
direction. If and type) transition, then according to CAM,
2
2
2
L Mr L Mr so that ( / ) r r d d d u u u u d u d
(1) hence each time the swing repeat moving upward(twice in each cycle) its angular speed increases by factor of
(2) The conservation of Mechanical Energy of Rotation (CMER)
2 2
2
1
1
E E
V Mr Mgr Mgr kin
2
2
(2) E is constant during the motion between the equilibrium position (where
st
V Mgr , ) after 1 transition type and maximum angular position (where nd
is zero before the 2 type of transition causing r changes r to higher r ) or
u d
nd
vice versa. During the 2 type transition with fixed, V increases from the
2
2
1
value to higher one and with this potential energy, the swing
1 Mgr Mgr
2 u 2 d
moves downward to give a higher value of the new angular velocity at
d st equilibrium position before the 1 type of transition starts again. Application eq.
nd st
(2) to the end of the 2 type and the start of the 1 leads to
2 2 2
2
2
2
1
1
1
1
Mgr Mr ; Mgr Mr
' '
2 u u 2 u u
2 d
2 d d d(3)
2
2
'
r r r
u u u d u
'
In eq.(3) , so that
u
d u
r r r '
d d u d d
(4)
st
This expression becomes the input of the next transition of 1 type giving an
2 r r u d
output for half a cycle ' '
u u d r r d u
(5)
THEORETICAL COMPETITION Questions and Solutions
For n complete cycles, the growth of angular velocity amplitude as well as the
2 n r
d
angular amplitude increases by a factor of
A A n , r u
1 1 n 2 n r
10
d 10 52
2 For then with one gets 2 2 2 n
5 A n ,
r u SOLUTION (approximation 4 points).
The moment of inertia with respect to the swing pivot
2 I = Mr (1)
Since the A to B transition is fast one has by conservation of angular momentum,
I I
(2)
A A B B
The energy at point A is
1
2 E
I (3)
A A A
2 The energy at point B is
1
I Mgh
2 E
(4)
B B B
2 where h = r d – r u is the vertical distance the child’s center of mass moves. The energy at point C (conservation of energy)
1
2 E E
I Mgh (5)
C B B B
2 As the child squats at the C to D transition, the swing losses energy of the amount Mgh so
1
2 E
I (6)
D B B
2 Energy at point E is equal to energy at point D (conservation energy)
1
2 E E
I (7)
E D B B
2 But we have also THEORETICAL COMPETITION Questions and Solutions
1
2 E (8)
I
E E E
2 From Eq. (7) and (8) we have,
I
2 B
2
E B (9)
I
EUsing Eq (2) this equation yields,
I
2
2 A
E A (10)
I
B Where we have used I E = I A .Using Eq. (1) one obtains from Eq. (10)
2
r
2
2
d
(11)
E A
2
r
u
From this one obtains,
E r
d
(12)
r
A u
This ratio gives the fractional increase in the amplitude for one half cycle of the swing motion. The fractional increase in the amplitude after n cycles is thus,
2 n
r
E n d
(13)
r
A u
Where is the initial amplitude and is the amplitude after n
A E n
cycles. Substitute the values,
2 n
(14)
10
2 2 or,
n = 5 (15)
Thus it takes only 5 swing cycles for the amplitude to build up by a factor of two.
THEORETICAL COMPETITION Questions and Solutions
0.25 Equation
0.25 Equation
E d A u
r
r
Equation
0.5
1.5 Equation
2 n E n d A u r r
Equation
0.25
2
Conservation of energy
10
2 2
n
Equation
6
10
2 2
n
0.25
0.25
n = 5 n = 5/3
0.25
E E I
[Marking Scheme] THEORETICAL Question 1B The Parametric Swing
0.25 Conservation of angular momentum C to D
(5.0)
0.25 Moment of inertia I = Mr
2
0.25 Conservation of angular momentum A to B
A A B B
I I
0.25 Correct expression of energy at point A
0.25 Correct expression of energy at point B
0.25 Correct expression of energy at point C
0.25 Correct expression of energy at point D
0.25 Correct expression of energy at point E
0.25
2 E D B B
0.50 Conservation of energy
2
1
2 C B B B
E E
I Mgh
Conservation of energy ………
0.25
0.50 Conservation of energy
2
1
Note: Propagation errors will not be considered here.
THEORETICAL COMPETITION Questions and Solutions
MAGNETIC FOCUSING Question 2
There exist many devices that utilize fine beams of charged particles. The cathode ray tube in oscilloscopes, in television receivers or in electron microscopes. In these devices the particle beam is focused and deflected in much the same manner as a light beam is in an optical instrument.
Beams of particles can be focused by electric fields or by magnetic fields. In problem 2A and 2B we are going to see how the beam can be focused by a magnetic field.
2A. MAGNETIC FOCUSING SOLENOID (4 points)
Figure 2.1 shows an electron gun situated inside (near the middle) a long solenoid. The electrons emerging from the hole on the anode have a small transverse velocitycomponent. The electron will follow a helical path. After one complete turn, the electron will return to the axis. By adjusting the magnetic field B inside the solenoid correctly, all the electrons will converge at the same point F after one complete turn, and form an image of the hole on the anode. Use the following data:
The voltage difference that accelerates the electrons V = 10 kilovolt The distance between the anode and the focus point F, L = 0.5 meter
- 31
kg The mass of an electron m = 9.11x 10
- 19
C The charge of an electron e = 1.6 x 10 Treat the problem non-relativistically
a) Calculate B! (3 points)
b) Find the current in the solenoid if the latter has 500 turns per meter. (1 point) Anode
F L Figure 2.1 THEORETICAL COMPETITION Questions and Solutions
SOLUTION a) In magnetic field, the particle will be deflected and follow a helical path.
Lorentz Force in a magnetic field B,
2 mv ev B
(1)
R Where v is the transverse velocity of the electron, R is the radius of the path.
2
R
Since v ( is the particle angular velocity and T is the period), then,
T
2
m eB
(2)
T L v is the parallel
To be focused, the period of electron T must be equal to , where
// v
// component of the velocity.
We also know,
1
1
2
2
2 eV m v v mv (3)
// //
2
2 All the information above leads to 1 2 3 mV e 2 (4)
B
2
L
Numerically
B = 4.24 mT
b) The magnetic field of the Solenoid:
in
(5)
B= THEORETICAL COMPETITION Questions and Solutions
B i n
1
2 // //
1
1
2
2
eV m v v mv
1.0 Formula 3 2
2
2
2
mV e B L
0.5 Numerical value B = 4.23 mT
b. (1.0)
in
0.3 B
i n
0.2 i = 6.75 A. Note: Propagation errors will not be considered.
2
(6)
R
Numerically i = 6.75 A.
[Marking Scheme] THEORETICAL Question 2A Magnetic Focusing Solenoid
a. (3.0)
0.3 Lorentz Force
2 mv ev B
R
0.1 Transverse velocity v
0.5 Conservation energy
0.1
2 T
0.3 Equation
2
m eB
T
0.2 Equation T =
// L v
0.5 B=
THEORETICAL COMPETITION Questions and Solutions
2B. MAGNETIC FOCUSING (FRINGING FIELD) (6 points)
Two pole magnets positioned on horizontal planes are separated by a certain distance such that the magnetic field between them be B in vertical direction (see Figure 2.2). The poles faces are rectangular with length l and width w. Consider the fringe field near the edges of the poles (fringe field is field particularly associated to the edge effects). Suppose the extent of the fringe field is b (see Fig. 2.3). The fringe field has two components B x i and B z k. For simplicity assume that |B x |= B|z|/b or explicitly:
- when the particle enters the fringe field B = +B z /b,
x
- when the particle enters the fringe field after travelling through the magnet, B = Bz b /
x
z x=b yB z z x x x
B x
Figure 2.3. Fringe fieldw l v
Figure 2.2. Overall viewA parallel beam of particles of mass m and charge q enters the magnet (near the center) with a high velocity v parallel to the horizontal plane. The vertical size of the beam is comparable to the distance between the magnets. The beam enters the magnet at an angle from the center line of the magnet and leaves the magnet at an angle - (see
Figure 2.4. Assume is very small). You may treat this problem relativistically or non- relativistically.l y y v w w x
x v
Figure 2.4. Top viewTHEORETICAL COMPETITION Questions and Solutions
The beam will be focused due to the fringe field. Calculate the approximate focal length if we define the focal length as illustrated in Figure 2.5 (assume b<<l and assume that the z-component of the deflection in the magnetic field B is very small).
xf
Figure 2.5. Side viewSolution:
The magnetic force due to the fringe field on charge q with velocity v is
(1)
F qv B
The z-component of the force obtained from the cross product is
qv sin Bz
F q v B v B qv B
(2)
z x y y x y x b
The vertical momentum gained by the particle after entering the fringe field
qvBz sin qvBz sin b
P F dt t qBz tan
(3)
z z
b b v cos
The particle undergoes a circular motion in the constant magnetic field B region
2 v
m qvB
(4)
R THEORETICAL COMPETITION Questions and Solutions
qBR qBl v
(5)
m 2 sin m
Therefore,
qBl
sin
(6) 2 mv After the particle exits the fringe field at the other end, it will gain the same momentum.
The total vertical momentum gained by the particle is
2
2 qBl q B zl P
2 P 2 qBz tan 2 qBz (7)
z z total
2 mv mv Note that for small , we can approximate tan sin
Meanwhile, the momentum along the horizontal plane (xy-plane) is
p = mv (8)
From the geometry in figure 4, we can get the focal length by the following relation,
P Z
z
(9)
p f 2 2 m v f
2
2 q B l
(10) THEORETICAL COMPETITION Questions and Solutions
[Marking Scheme] THEORETICAL Question 2B
(6.0)
Magnetic Focusing (Fringing Field)
0.25 Lorentz Force F qv B
0.25
2 z total q B zl P mv
0.5
2
z z
total P P (factor of 2)
0.25
2
qBl
mv
0.5 Horizontal momentum p = mv
1.0 Equation
z total P Z p f
1.0
2 2
2
2
m v f q B l Note: No propagation error will be considered here.
2
z-component z x y y x
z-component gained momentum z z
F q v B v B
0.25
z-component
sin
z y x qv Bz F qv B b
0.5
P F dt
0.25 sin
0.75 tan
z P qBz
0.5 Equation
2 v m qvB
R THEORETICAL COMPETITION Questions and Solutions
RELATIVISTIC MIRROR Question 3 Reflection of light by a relativistically moving mirror is not theoretically new.
Einstein discussed the possibility or worked out the process using the Lorentz transformation to get the reflection formula due to a mirror moving with a velocity
v . This formula, however, could also be derived by using a relatively simpler
method. Consider the reflection process as shown in Fig. 3.1, where a plane mirror
v v e ˆ e ˆ
x x
M moves with a velocity (where is a unit vector in the x-direction)
observed from the lab frame F. The mirror form forming an angle with respect to
n
90 the velocity (note that ). The plane of the mirror has as its normal. The light beam has an incident angle and reflection angle which are the angles
n
between and the incident beam 1 and reflection beam 1' , respectively. It can be shown that,
v sin sin sin sin ( )
(1)
c
Figure 3.1. Reflection of light by a relativistically moving mirrorTHEORETICAL COMPETITION Questions and Solutions
3A. Einstein’s Mirror (2.5 points)
About a century ago Einstein derived the law of reflection of an electromagnetic
v v e ˆ
x
wave by a mirror moving with a constant velocity (see Fig. 3.2). By applying the Lorentz transformation to the result obtained in the rest frame of the mirror, Einstein found that : 2
v v
(1 ) cos
2 2
c c
cos 2 (2)
v v
1 2 cos
2
c c
Derive this formula using Equation (1)!
y ? x
? ?
Figure 3.2. Einstein mirror moving to the left with a velocity v.3B. Frequency Shift (2 points)
If the incident light is a monochromatic beam hitting M with a frequency , find the new frequency ' after it is reflected from the surface of the moving mirror. If 30 and 0.6 c in percentage of . , find frequency shift
3C. Relativistically moving Mirror Equation (5.5 Points)
Derive Equation (1)!
THEORETICAL COMPETITION Questions and Solutions
Solution:
a) EINSTEIN’S MIRROR
/ 2 v v
By taking and replacing with in Equation (1) we obtain
v sin sin sin ( )
(3)
c
This equation can also be written in the form of
v v
1 cos sin1 cos sin
(4)
c c
cos The square of this equation can be written in terms of a squared equation of , as follows, 2 2
v v v v v
2 2 21 2 cos cos
2 1 cos cos 2 cos 1 cos 2 2
c c c c c
(5) which has two solutions,
2
v v
2
2 cos 1 cos
2
c c
(6) cos
1
2 v v
1 2 cos
2 c c
and
2 v v
2 1 cos
2
c c
(7) cos
2
2 v v
1 2 cos
2 c c cos cos v
However, if the mirror is at rest ( ) then ; therefore the proper solution is 2
v v
2 1 cos 2 c c
(8)
cos 2 2 v v
1 2 cos 2 c c THEORETICAL COMPETITION Questions and Solutions
b) FREQUENCY SHIFT The reflection phenomenon can be considered as a collision of the mirror with a beam of photons each carrying an incident and reflected momentum of magnitude
p h / c p ' h '/ c
and , (9)
f f
The conservation of linear momentum during its reflection from the mirror for the component parallel to the mirror appears as 2
v
(1 )sin 2
c
p sin p 'sin or 'sin ' sin
2 (10)
f f v v
(1 ) 2 cos 2
c c
Thus 2
v v
(1 ) 2 cos 2
c
c
'
2 (11)
v
(1 ) 2
c v . 6 c
For 30 and , 2 2
1
v v
cos 3, 1 0.64, 1
1.36
(12) 2 2
c c
2 so that ' 1.36 0.6 3
(13)
0.5
0.64
Thus, there is a decrease of frequency by 50% due to reflection by the moving mirror.
THEORETICAL COMPETITION Questions and Solutions
c) RELATIVISTICALLY MOVING MIRROR EQUATION
t t
Figure 3.3 shows the positions of the mirror at time and . Since the observer is moving to the left, system is moving relatively to the right. Light beam 1 falls on pointt t
a at and is reflected as beam 1' . Light beam 2 falls on point d at and is reflected as
tbeam . Therefore, ab is the wavefront of the incoming light at time . The atoms
2 '
at point a is disturbed by the incident wavefront ab and begin to radiate a wavelet. The
t
disturbance due to the wavefront ab stops at time when the wavefront strikes point
d. As a consequence ac bd c ( t t ).
(14)
ed ag
From this figure we also have , and
dg bd ac af sin sin
, . (15)
ag ag ef
Figure 3.4 displays the beam path 1 in more detail. From this figure it is easy to show thatao v ( t t ) sin dg ae
(16)
cos cos
and
ao v ( t t ) sin
af
(17)
cos cos
THEORETICAL COMPETITION Questions and Solutions
From the triangles aeo and afo we have
tan ao eo
and
tan ao of
. Since
of eo ef
, then Figure 3.3.
( )sin tan tan ef v t t
(18) By substituting Equations (14), (16), (17), and (18) into Equation (15) we obtain
cos
sin
sin t t ag v c
(19) THEORETICAL COMPETITION Questions and Solutions
Figure 3.4.
and sin
c v
cos sin
(20)
ag
v sin (tan tan )
t t
ag /( t t )
Eliminating from the two Equations above leads to
1
1
1
1
v sin (tan tan ) c v sin
(21) sin sin sin cos sin cos
v sin
By collecting the terms containing we obtain
v cos cos sin sin
sin
(22)
c sin sin sin sin
or