Download Q No - Air University

Air University
Mid Term Examination
(Spring-2011)
Course: Math-446, MMSC
Date 04-06-2011
Max. Marks: 45
Time Allowed: 2 hrs
Note: Attempt all 5 questions. Please encircle your final answers. You may use basic scientific
non-programmable calculator. DO NOT write in pencil or red ink on answer sheet. DO NOT write
anything on the question paper.
1. a) A particle of mass m , positive charge q and speed v moving along x -axis enters a uniform electric
P
field of intensity E that is pointing in the y -direction as shown in Fig 1. The field is spread over a
distance d along x-axis. Ignore gravity.
Fig 1
E
v
d
a) What is the magnitude of its acceleration and the duration of time it is in the electric field?
1
b) Assume that the particle enters the region of field at the origin at time t  0 Find expressions for the
distances traveled by the particle in time t in the x - and y -directions.
2
c) Find the equation of the trajectory of the particle, sketch y as a function of x and identify this sketch
with some familiar curve.
3
Solution: Magnitude of the force acting on the particle F  qE
Magnitude of the acceleration of the particle a 
F qE

m m
Time spent by the particle in the region of the electric field t 
d
v
b) Distance traveled in the x -direction after time t , xt   vt
Distance traveled in the y -direction after time t , y t  
y
1 2 qE 2
at 
t
2
2m
c) Eliminating time t from the above two equations we get,
y
qE x 2
qE 2

x
2
2m v
2mv 2
x
Figure shows a sketch of y as a function of x . The particle trajectory will be a parabola.


2. a) A 1MeV (10 6 eV  1.6  10 13 J ) proton m  1.67  10 27 kg, q  1.60  10 19 C initially moving
in the x  z plane enters a region of uniform magnetic field of strength B  0.01T pointing in the z direction making an angle of   45 0 with the direction of the field.
a) Calculate its speed and find its first order relativistic correction
1
b) Calculate the radius and the pitch of its spiral trajectory in the magnetic field.
3
c) Assuming that the particle enters the region of field at the origin at time t  0 , find expressions for the
distances traveled by the particle in time t in the x - y - and z -directions.
2
Solution: a) Using the non-relativistic expression for kinetic energy we get,
© Air University, 2010
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v
2K

m
2  1.6  10 13
 1.4  10 7 m / s
 27
1.67  10
The relativistic expression for kinetic energy is
p2
mv 2
mv 2

K


1  v 2 / c 2 ,
2
2
2m 21  v / c 
2
Since c  3  10 8 m / s
v 2 1.4

 10  2  2  10 3 , or about 0.2%
9
c2
2
The 1st order correction term is 
b) As the proton enters the region of uniform magnetic field the component of its velocity
perpendicular to the field makes it move in a circular trajectory of radius r given by,
mv 2 sin 2 
 evB sin  , or
r
r
mv sin  1.67  10 27  1.4  10 7

 10.4m
eB
1.6  10 19  10 2  2
Time period T 
2r
2  3.14  10.4

2  6.5s
v sin 
1.4  10 7
the proton drift along the field due to the component of its velocity along the field. Therefore the
distance it covers along the field while moving once around the field, or its pitch p is given by,
p  cos   T 
1.4  10 7  6.5  10 6
2
 65m
c) On entering the region of the uniform magnetic field the motion of the proton perpendicular
to the field in converted into circular motion of radius r with angular velocity w given by,
w
2
2

 966rad / s
T
6.5  10 3
Therefore, we have,
xt   10.4 cos966t m , yt   10.4c sin 966t m
7
At the same time it drifts along the z -axis with constant speed v  10 m / s giving,
z t   10 7 t m
B
Fig 2
3. An 1KeV electron ( m  9.1  10 31 kg) enters a cyclotron perpendicular to its axis. A magnetic
E
field of intensity B  0.1T and an accelerating electric potential 10V is applied along the axis
and across the gaps of the chamber as shown in Fig 2.
a) Find the initial velocity of the electron and the corresponding switching frequency of the
potential.
2
b) What is the change in the energy of the electron in one cycle?
1
© Air University, 2010
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c) Find the change in the energy of the electron after one second. and correct it for relativistic
effects.
3
Solution: a) From the relativistic expression for energy-momentum relation,
2
m0 c 4
, the expression fro kinetic energy is,
p c  m0 c 
v2
1 2
c
2
2
2
4
2
m0 c 4
v2
2 4

m
c
0
2
2
p2
c2
K
 1 v / c 2

m0 c 2 , giving,
2
2
2m0
2m0 c
2 1 v / c

2 K / m0

1  2 K / m0 c 2
v

2 Kc 2
2  1.6  10 16  9  1016

 1.9  10 7 m / s
m0 c 2  2 K
9.1  10 31  9  1016  2  1.6  10 16
This shows that the initial velocity of the electron is non-relativistic. On entering the chamber the
electron moves in a circle of radius r given by,
r
mv 9.1  10 31  1.9  10 7

 1.1mm
eB
1.6  10 19  10 1
The time it initially takes to complete one revolution is given by,
T
2r 6.3  1.1  10 3

 3.6  10 10 s
7
v
1.9  10
Therefore the initial switching frequency f of the accelerating voltage is given by,
f 
2
6.3

 1.81010 Hz
10
T
3.6  10
b) In each cycle the change in kinetic energy K of the electron is given by,
K  2eV  2  10  20eV  32  10 19 J
c) Since there are 1 / f  5.7  10 9 cycles in one second, the change in the energy of the
electron K in one second would be,
K  K / f  20  5.7  109  114  109  114GeV
The above value is, however grossly exaggerated as after a few cycles the electron would begin
to move with relativistic velocities. This means that we need to use the relativistic mass for
calculating the frequency.
f 
2 K / m0 c 2
2 v eB
eB
eB
1
 
1 v2 / c2 
1

, therefore we may
2
T
r m0
m
m0 1  2 K / m0 c 2
1  2 K / m0 c
write,
© Air University, 2010
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dK
e 2VB
 2eVf 
dt
m0
m0 c 2
m0 c 2
1
2.56  10 38  10  0.1
8
,
2

5
.
6

10
2K
9.1  10 31
m0 c 2  2 K
m0 c 2  2 K
1
m0 c 2
Ignoring the relatively small initial kinetic energy of the electron, its kinetic energy after one
second is given by,

12
m0 c 2  2 K
23

15 10 
K
3/ 2
14 2 / 3
 5.6  10 8 m0 c 2  5.6  10 8 8  10 13  5.0  10 14 , or
 8  10 14
2
 1.4GeV , which is about one hundredth the non-relativistic value.
4. a) Assume that the moon’s orbit around the earth is circular. Derive expression for the time
period of revolution T of the moon in terms of its distance r from the center of the earth, the
gravitational acceleration g  9.8ms 2 on earth’s surface, and the radius of the
earth R  6400km .
3
b) Calculate the value of r given that the value of T  27.3 days.
2
c) Estimate the diameter of the moon given that its angular diameter from earth is about 0.5 0 1
Solution: a) If r is the distance of the center of the moon from the center of the earth then it
centripetal acceleration around the earth is given by,
v 2 gR 2
 2 , giving,
r
r
2
v
4 2 gR 2
2

w

 3
r2
T2
r
a
So that the time period of the moon T around the earth is given by,
4 2 r 3
gR 2
T
b) For T  27.3days  27.3  24  3600  2.4  10 6 s
 gR 2T 2 
r
2 
 4 
1/ 3

 

 9.8  6.4  10 6 2  2.4  10 6 2 


39


1/ 3

 59  10 24

1/ 3
 3.9  10 8 m  390  10 3 km
d


rad ,
r  R 360
 r  R  390  6.4  3.14

 10 3  2500km =
Therefore, diameter of moon d 
380
360
c) Angular radius of moon as observed from earth  
5. At time t  0 an object of mass m is moving with a velocity
vector
F
r
0
v perpendicular to its position
0
in the inverse square central force field of the earth,
mgR 2 
 2 r
r
a) Write down the equations describing the motion of the object at any subsequent time t .
b) Sow that the following are solutions of these equations of motion
© Air University, 2010
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2
mrv  L  mr0 v0

L2
AL2
, with  
and
,
2
e

1  e cos(  0 )
gR 2
gR 2
c) Take the x -axis along the initial direction of the position vector r 0 to get 0  0 and find
the value of the unknown constant A from the given initial conditions.
2
r
Solution: a) For describing the motion of an object in central force field we use polar
coordinates r,  .If is the position vector of the object with the earth’s center as the origin,
r


with r and  unit vectors along the r -and  -directions, then,
dr
dt
d2 r
dt 2

dr  d 
r r
 , and
dt
dt
d 2r 
dr d 
d 2   d  
gR 2 
 2 r 2
  r 2   r  r  F  2 r
dt dt
dt
dt
r
 dt 
2


Equating the coefficients of r and  on the two sides of the above equation we get,
d 2r
gR 2
 d 

r

, and


dt 2
r2
 dt 
dr d
d 2
2
r 2 0
dt dt
dt
2
(5a)
(5b)
b) EQ (5b) can be rewritten as,
dr d
d 2 d  2 d 
2
 r 2  r
  0 , or,
dt dt
dt  dt 
dt
d
r2
 cons tan t , but,
dt
d
 d  mrv L
r2
 r r
 ,

dt
m
m
 dt 
Since the mass of the object isn’t changing this means that its angular
momentum L  mr 2
d
 cons tan t  mr0 v0 is a solution of the 2nd of EQ (5b)
dt
Also, we have,



1  e cos(  0 ) 1  e cos 
dr
e sin 
d e 2 d
eL

 r
sin  
sin  , giving
2
dt 1  e cos( ) dt 
dt
m
r
d 2 r eL
d
eL2
L2  
L2
L2


cos


cos



1


,


2 3
dt m 2 r 2
dt 2 m
m 2 r 2  r
m 2 r 2
 m r
So that substituting these expressions in EQ (5a) we get,
eL2
L2
L2
L2
gR 2
 d 
, giving,
cos


r







m 2 r 2
m 2 r 2 m 2 r 2 m 2 r 3
r2
 dt 
2
Therefore,
r 

1  e cos 
L2
 2 2
m gR
is a solution of EQ (5a) provided,
© Air University, 2010
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c) If we take the x -axis along the initial position vector of the object then at t  0 we
have   0 , giving,

r0 
, or
1 e
AL 
1 L2
, giving,
e


1

ro gm 2 R 2
gR 2 r0
A
L
gR 2 v0
gR 2



L
m mr0 v0
m 2 r0
6. A computer simulation yields the following n results for a particular quantity q ;
q1 , q2 ,....qn

a) Write expressions for the average value of q and its standard deviation q
1

b) Find the values of q and q for the following sequence of numbers.
1
1,-1,1,-1,1,-1,0,0,-1,-1,0,-1,1,1,-1,0,0,0,1,0,1,1,0,-1,0,-1,1,-1,0,0,0,1,1,1,-1,1,0,1,1,1,1,0,-1,1,1,0,-1,-1,1,-1,-1,-1,1,-1,0
c) Make at least two statements with some justification that are applicable to this sequence. 3

Solution: a) Average value of value of q , q 
1 n
 qi
n i 1




2
1 n 
1 n 2 q n

2
Standard deviation q  
q

q


q

q

q


q

q
  i 
 n
i
n i 1 
n i 1
i 1
b) For the given sequence of q value
2
2
1,-1,1,-1,1,-1,0,0,-1,-1,0,-1,1,1,-1,0,0,0,1,0,1,1,0,-1,0,-1,1,-1,0,0,0,1,1,1,-1,1,0,1,1,1,1,0,-1,1,1,0,-1,-1,1,-1,-1,-1,1,-1,0
We have n  n1  n0  n1  55 , with,
n1  20 , n0  16 and n1  20 , so that,

1
q   1  20  0  15  1  20)  0 , and
55
1
20  20   8  0.85
q  
55
11
c) i. It is a random integer sequence, since it shows no regularity in the occurrence of.
the three integer values.
ii. It seems to be evenly distributed between  1 and 1 , since the number of each of the three
given integer values is about the same and the standard deviation is also of the order of the
range of the interval of the random numbers.
7) Chose one of the following take-home assignments that should be submitted by 1500 hours
on the following day. No two students should choose the same assignment.
10
© Air University, 2010
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