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Solutions of Exam# 1
Q1
(a) A particle of mass m moves in a Yukawa potential
k
V  r    e  r / a ; where k , a are consts.
r
Find the equation of a bound orbit of the particle (having an orbital angular
momentum ( ℓ ) about the center of the potential) to the first order of r/a.
(b) A particle of mass m moves in a central force field that has a constant
magnitude F0 but always points toward the origin.
(i) Find the angular velocity ω φ required for the particle to move in a
circular orbit of radius ro .
(ii) Find the frequency of ω r of small radial oscillations about the circular
orbit. Both answers should be in terms of m, r0, F0.
(a) The bound motion means that E 
m v2
k
 V  0 where V   e r a .
2
r
The orbit of particle moving in this central force potential is given by
 r  
 / r 2  dr
r

rmin
2


2   E V 
2 
2 r 

r

2

rmin
1
r2
dr
E
2
ke  r / a

r
2 r 2
In first order of (r/ a), that is (e-r/a ~ 1-r/a)
(r) 
2 
dr
r2 E 
2
k
k


2
r 2r a

2 
dr
2
k k

r2  E    

a r 2r2
k

Now effectively, this is the orbit of particle of total energy  E   moving in potential

a
k
 . It is well known that this orbit is given by (see Chapter 8 Eqs 8.38, 8.40, 8.41)
r

r
Where  
2
k
and
  1
22
k2
 1  cos
k

 E  
a
If 0    1, the orbit is ellipsoid; if   0 , the orbit is circular.
(b)
(i) In equilibrium, for a circular orbit of radius r0,
F0  m 2r0   
(ii)
F0
m r0
The angular momentum (which is conserved) of a particle in circular orbit is
L  m r02   mF0 r03
The force acting on a particle, which is placed a distance r (r is very close to equilibrium
position ro ) from the center of force is
F  m 2r F0 

L3
 F0
m r3
L3
3L2
3L2

r

r

F


 0 0
 r r0    k r r0 
m r03 m r02
m r04
where k  3L2 /m r04 . So the frequency of oscillation is
r 
3F0
k
3L2


2 4
m
m r0
m r0
Q2
For the elastic scattering of T0= 5.0 MeV α-particles ( 24 He ) from gold
( 197
79 Au )(initially at rest) at a scattering angle in the LAB of ψ = 90◦. Find the
following:
(i)
The LAB kinetic energies of the scattered α-particle and gold.
(ii)
The recoil scattered angle of gold ζ.
(iii)
The scattering angles of α-particle (θ1) and gold (θ2) in the CM system.
(iv)
The impact parameter b and the distance of closest approach between
the particles.
(v)
The differential cross section (LAB) of scattering at ψ = 90◦.
(vi)
The ratio of the probabilities of scattering at ψ = 90◦ to that at ψ = 30◦.
(i)
m 1  mass of  ( 24 He) particle,
m 2  mass of gold (
197
79
Au )
u1 ,u1 : velocity of  particle in LAB and CM before collision
v1 ,v1 : velocity of  particle in LAB and CM after collision
u2 ,u2 : velocity of
197
79
Au in LAB and CM before collision
v2 ,v2 : velocity of
197
79
Au in LAB and CM after collision
u2  0 ,
T0 
m1u 12
 5.0 MeV
2
  90 is angle through which  particle is deflected in LAB
1 and
2
are the angles through which  particle and
 is recoil angle of
197
79
197
79
Au are deflected in CM
Au in LAB
(i) Conservation of momentum in LAB:
m 1 u1  m 2 v2 cos 
  v1  u1 tan 
m 1 v1  m 2 v2 sin  
Conservation of energy in LAB:
(1)
m 1u12 m 1v12 m 2v22


2
2
2
 m  m1  2
v 12   2
 u1
 m 2  m1 
Thus,
and
(2)

 2
2m12
v 22  
 u 1 (3)
m
m

m


2
1
2


The kinetic energy of  particle after collision in LAB is
T1 
The kinetic energy of particle
197
79
m1v 12  m 2  m1 

T 0  4.80 MeV
2
 m1  m 2 
Au after collision in LAB is
T2 
m 2v 22
2m1

T 0  0.20 MeV
2
m1  m 2
Evidently, conservation of energy is satisfied.
(ii) From Eq (1) , (2) and (3) we obtain the recoil scattering angle of
197
79
Au
 m  m1 
tan    2
  0.9799    44.42
 m 2  m1 
(iii)
The velocity of CM of system is
vCM 
The velocity of
197
79
m 1u1
m1  m2
Au in CM after collision is v2'  v2  vCM . From the above figure we can
obtain the scattering angle of particle
tan  2 
197
79
Au in CM to be (use Eqs 1, 3)
v 2 sin 
m 22  m12

 49.24   2  88.84
v 2 cos   v cm
m12
In CM, clearly after collision, particle  moves in opposite direction of that of 197
79 Au
i.e. θ1 =180 – 88.84 = 91.16◦.
(iv) The impact parameter in CM is given by Eq (9.139).
b
where k 
T 0 
CM,
k
 
cot  1 

2T 0
2
q1q 2 z 1z 2e 2

  2  79 1.44 MeV .fm  227.52 MeV .fm and
4 0
4 0
 m 1m 2  2
1
m1u12  m 2u 22  
u1  4.90 MeV
2
 2  m1  m 2  


is the total energy of system in
b
so
k
 
cot  1   2.28  1014 m
2T '0
2
We note that b is the impact parameter of  particle with respect to CM, so the impact
 m1  m 2 b  2.32  1014 m.
parameter of  particle with respect to 197
79 Au is
m2
To find the closest distance from  particle to the center of mass:
In CM system, the orbit equation of particle  is

r
 1   cos
 rm in 

2
m 1k

1 

  1 2
and
 1
 
 rmin

1 


2
where   0
corresponds to r  rm in
is closest distance from  particle to the center of mass, and
 m1u1b 
2
m 1k

2T 0'b 2
 2.24  1014 m
k
 m u b 
E 2
 1  2T 0' 1 1
m 1k
m 1k
2
4T 0'2b 2
1
k
 1.12 1014 m
But the actual minimum distance between particles is
rmin 
(v)
m1  m 2
  1.14 1014 m.
rmin
m2
Using formula
 LAB     CM
where x 
m1
,   90 ,
m2
 CM 1  
x cos  


 
1
k2
 4T 0
2
1  x 2 sin 2 
1  x 2 sin 2 
1
 5.18  1028 m 2 ;
4  1 
sin  
2
2
 LAB     CM 1  1  x
2
  CM 1 
m 
1   1   0.9998  CM 1 
 m2 
where T0’ = 4.90 MeV; θ1= 91.16◦, k = 227.52 MeV.fm
We find this differential cross section in LAB at  = 90°:

2
 LAB   90  5.18  1028 m2
(vi)
Since the probability of scattering at an angle ψ =
that the ratio of probability is 
 LAB   sin 
 LAB  ' sin  
For ψ’=30◦ the corresponding θ’ is related by tan ' 
 LAB  '   CM
where x 
 x cos '
 '
1
m1
,  '  30 ,
m2
1  x 2 sin 2  '

dN
  LAB ( ) sin  d d we see
N
sin 1 '
 1 '  30.58
m
cos 1 ' 1
m2
2
1  x 2 sin 2  '
 CM  '1  
k2
 4T 0
2
1
 1028 m 2 ;
4   '1 
sin 

 2 
Q3 (a) Find the center of mass of a uniformly solid cone of base diameter (2a) and height
(h) and a solid hemisphere (of the same material as the cone)of radius (a) where the
two basis are touching..
(b) A 104 kg spherical probe of radius R = 20 cm is launched vertically upward from the
surface of Earth with an initial speed 6000 m/s. If the air resistance F = -(1/2) cwρAv2;
where the constant cw = 0.2, ρ = air density ( =1.3 kg/m3 ), A = πR2 and v is the speed of
the probe, determine the maximum height reached and the time taken to reach this
height.
Solve the problem taking g = constant first then solve it again taking g = G M /r2;
where M is the mass of Earth.
(a)
For the hemisphere:
Put the shell in the z > 0 region, with the base in the x-y plane. By symmetry, x  y  0 .
2
 2
  
z  
  
0
2
0
r2
 0 r r1
2 r2
 zr2 drsin  d d
 0 r r1
 r2 drsin  d d
Using z = r cos  and doing the integrals gives
z
For the cone:
By symmetry, x  y  0 .


8 r  r 
3 r24  r14
3
2
3
1
Use cylindrical coordinates , , z.
0  m ass density
h
 h
a
0
  0   0 z 0
h
2
a
  h
4
0
  0   0 z 0
2
z
a
  
 z  d d dz
  
  d d dz

h
4
The center of mass is on the axis
3
of the cone h from the vertex.
4
For the compound problem: (see Fig)
By symmetry, x  y  0 .
The center of mass of the cone is at z 
1
h.
4
3
The center of mass of the hemisphere is at z   a r2  a,r1  0
8
So the problem reduces to
z1 
1
1
h; m1  1  a 2 h
4
3
3
2
z2   a; m2  2  a3
8
3
z
m 1z1  m 2z2 1h2  32a2

m1  m2
4 1h  22a
for 1  2
z
h2  3a3
4 2a h
(b1)
When we add the expression for air resistance, the differential equation that describes
the projectile’s ascent is



dv
1
v
Fm
  m g  cW  A v2   m g 1  
dt
2

 vt 

2





(1)
where vt  2m g cW  A
25 km  s1 would be the terminal velocity if the object were
falling from a sufficient height (using 13 kg  m
differential equation gives
3


as the density of air). Solution of this

v0  gt

 
vt 
 vt 

v t  vt tan  tan 1 

This gives v = 0 at time    vt g tan1  v0 vt
(2)
300 s. The velocity can in turn be
integrated to give the y-coordinate of the projectile on the ascent. The height it reaches is
the y-coordinate at time :

h


vt2
v 
ln 1  0 
2g   vt 

2





(3)
which is  600 km.
(b2)
G M
Changing the acceleration due to gravity from –g to
e
 R e  y
2
  g  R e  R e  y  changes our differential equation for y to
2
  y  2  R  2
e
y   g     
 
  vt   R e  y 
(5)
Q4 A pojectile is fired due East from a point on the surface of Earth at a northern
latitude λ with an initial velocity V0 and at an angle α to the horizontal. Take South to
be x-axis and East to be y-axis and z-axis to be vertically upward from the surface of
Earth.
(a) Assume that the z motion to be unaffected by the Coriolis force, show that the
side deflection when the projectile strikes Earth is
4V03
x T  
sin  cos  sin 2 
2
g
(b) Now take the effect of the Coriolis force in the z-motion find the range of the
projectile (Range = y at z=0).
(a)
The coordinates x, y, z as in the diagram: South is x-axis and East is y-axis, z is vertically
upward.
Then, the velocity of the particle and the rotation frequency of the Earth are expressed as
v   0, y, z 


   x ,0,  z     cos  , 0,  sin   
(1)
so that the acceleration due to the Coriolis force is
a  2  r  2  yz , zx ,  yx 
(2)
Since the motion is in y-z plane, then we see that the lateral (side) deflection of the
projectile is in the x direction and that the acceleration is
ax  x  2 z y  2  sin   V0 cos  
(1)
Integrating this expression twice and using the initial conditions, x 0  0 and x 0  0 ,
we obtain
xt   V0t2 cos sin 
Now, we treat the z motion of the projectile as if it were undisturbed by the Coriolis
force. In this approximation, we have
(2)
zt  V0tsin  
1 2
gt
2
(3)
from which the time T of impact is obtained by setting z = 0:
T
2V0 sin 
g
(4)
Substituting this value for T into (2), we find the lateral deflection at impact to be
x T  
4V03
sin  cos sin2 
g2
(5)
(b) In the previous part we assumed the z motion to be unaffected by the Coriolis force.
Recall that the Coriolis acceleration is:
a  2  r  2  yz , zx ,  yx 
With
v   0, y, z    0,V0 cos  ,V0 sin   gt  

   x ,0,  z     cos  , 0,  sin   
The upward acceleration is given by 2 x y so that
z  2 V0 cos cos  g
(6)
from which the time of flight is obtained by integrating twice, using the initial
conditions, and then setting z = 0:
T 
2V0 sin 
g  2V0 cos cos
(7)
Now, the acceleration in the y direction is
a y  y  2x z
 2   cos   V0 sin   gt 
(8)
Integrating twice and using the initial conditions, y 0  V0 cos and y 0  0 , we have
yt 
1
 gt3 cos  V0t2 cos sin   V0tcos
3
(9)
Substituting (7) into (9), the range R is
R 
4V 03 sin3  cos
2V 02 cos cos
8 V 03 g sin3  cos


3  g  2V 0 cos cos  3  g  2V 0 cos cos  2 g  2V 0 cos cos
(10)
We can stop here.
However, if we want to compare R’ with the range for the case ω=0, we have the
following:
We now expand each of these three terms, retaining quantities up to order  but
neglecting all quantities proportional to  2 and higher powers of . In the first two
terms, this amounts to neglecting 2V0 cos cos compared to g in the denominator.
But in the third term we must use
2V 02 cos sin 
 2V 0

g 1
cos cos  
g



2V 02
cos sin 
g
 R 0 
 2V 0

cos cos  
1 
g


4V 03
sin  cos2  cos 
g2
(11)
where R0 is the range when Coriolis effects are neglected [see Example 2.7]:
R 0 
2V 02
cos sin 
g
(12)
The range difference, R   R   R0 , now becomes
R  
4V03
1


cos  sin  cos2   sin3  
2


g
3
(13)
Substituting for V 0 in terms of R0 from (12), we have, finally,
R  
2R 0
1
 cos  cot1 2   tan3 2  


g
3
(14)