Transcript Chapter 3 Impulse

Impulse and Momentum
Some more extremely useful relations are the concepts known as
impulse and momentum. Conservation of momentum underlies
several important physical laws in a variety of scientific
disciplines from quantum electrodynamics to fluid mechanics. A
derivation for this fundamental quantity is shown below:


r2






dr

t Fdt  t mr dt  t m dt dt  r m dr  mr2  mr1  mv2  mv1
1
1
1
1
t2
t2
t2



We now define a quantity known as momentum, as P  mr  mv , and the
change in momentum over a discrete time interval is defined as the impulse, I :
t2
 t2


  
I   Fdt   d mv   mv2  mv1  P2  P1
t1
t1
Note: In comparison with potential and kinetic energy, which are scalar quantities,
impulse and momentum are vector quantities and must be added vectorially.
11 - 1
Conservation of Linear Momentum
Starting from a generalized expression for momentum



 Fdt   d mv    dP
We can take the derivative of both sides of the equation:


Fdt  dP
And we arrive at another expression for Newton’s Second law.
The force is equal to the rate of change of momentum.

 d  
F  mr  P  P
dt



F  0  P  P  const.
Thus, when the net force on an object is zero, then the
momentum must be constant. This is the famous law of
conservation of momentum.
11 - 2
11 - 3
Conservation of Linear Momentum






mAv Ai  mB vBi  mC vCi  mAv Af  mB vBf  mC vCf
 


Linear Momentum prior to impact
Linear Momentum after impact
Initial conditions:


vAi  v0  v0 xˆ cos30  yˆ sin30
After impact:



m Av Af  mB vBf  mC vCf 
m Av Af
mB vBf
mC vCf
  
 
xˆ sin 49.3   yˆ cos49.3  
xˆ cos45   yˆ sin 45 
ˆx sin 7.4o  yˆ cos 7.4 o 
o
o
o
o
11 - 4
Given the following information:
v0  4.0 ms , vC  2.1 ms , mA  mB  mC
Leads to 2 equations and 2 unknowns.
 
yˆ : v cos30   v
 
cos7.4  v
 
 
cos49.3  2 sin45 
xˆ : v0 cos 30o  vAf sin 7.4o  vBf sin 49.3o  2 ms cos 45o
o
0
o
Af
Bf
These can be solved simultaneously,
leading to the following result:
o
m
s
o
vBf  2.27 ms
v Af  2.01 ms
11 - 5
Relation between Force and Impulse
As shown in previous slides, the impulse can be written as:



I   Fdt   dP
If the force is constant over a discrete time interval, then it can
be taken out of the integral and the impulse is re-written as:
 



I  F  dt  Favg t  P  mv
This equation is very useful for modeling collisions, because
the force of impact is not usually known (or measurable), but
the initial and final velocities are known and/or measurable.
11 - 6
11 - 7
N

 m g  400 s t 0s  t  2s
 F  m g  T (t )    m g  800N 2s  t  4s

  m g 4s  t  5s
N

0 s Fdt  0s   m g  400t s dt  2s  m g  800N dt  4s  m gdt
5s
2s
4s
 Fdt  mv  mv
5s
f
5s
 vi 
0s
5s
 Fdt  m g5s   200
0s
vf 
N
2s 2  800N 2s   2400N  s  5s m g
s
2400
m
m
N  s  5s g  6  2.1
200kg
s
s
11 - 8
Angular Momentum
Like linear momentum, there is also an expression for angular
momentum, which describes how much inertia an object carries in
rotating about a coordinate center. While its derivation will be
reserved for later in this course, the mathematical expression is
given here as:

v
Path of
particle

r



HO  r  mv
The cross product denotes that the angular
momentum points in an entirely different
 direction
from the instantaneous position vector, r , and the
instantaneous velocity vector . The direction can
be found according to the right-hand rule. The O
means that the angular momentum is taken with
respect to the origin.

v
11 - 9
Conservation of Angular Momentum
The rate of change of angular momentum can be derived as follows:



 
 
 
 
  
d 
HO 
r  mv  r  mv  r  mv  rm
r  r  mr  r  F

dt
0
(i.e. the cross product of a vector with itself is always
zero (i.e. they point in the same direction)


 
M O  HO  r  F

We define this new vector, M, which is the rate of change of angular
moment, and it is called the moment of force about a pivot. Note: If the net
force (or moment in this case) is zero, then it leads directly to the law of
conservation of angular momentum.

  
HO  r  F  r  0  0

H O  const.
11 - 10
Momentum according to Isaac Asimov
Isaac Asimov wrote in "Understanding Physics": “This
tendency for motion (or for rest) to maintain itself steadily
unless made to do otherwise by some interfering force can be
viewed as a kind of "laziness," a kind of unwillingness to make
a change. And indeed, Newton's first law of motion is referred
to as the principle of inertia, from a Latin word meaning
"idleness" or "laziness."
He added a footnote: "In Aristotle's time the earth was
considered a motionless body fixed at the center of the
universe; the notion of 'rest' therefore had a literal meaning.
What we ordinarily consider 'rest' nowadays is a state of being
motionless with respect to the surface of the earth. But we
know (and Newton did, too) that the earth itself is in motion
about the sun and about its own axis. A body resting on the
surface of the earth is therefore not really in a state of rest at
all."
11 - 11
Central Forces
In many forces, such as gravitational, electrical forces, etc., the magnitude of
the force only depends on the distance away from the source (i.e. r 2 ), and
the direction is always towards or away from the source rˆ .





F r,   ma  Fr rˆ  Fˆ  m r  r2 rˆ  m r  2r ˆ

 F  0  m r  2r

Using these assumptions, we can derive an expression for the rotation
frequency:
 
1 d 2
r   r  2r  0
r dt
m r2  const.  H
H

 2
mr
11 - 12
Using this result, we can arrive at an expression for radial force
written purely in terms of the radial distance
H

 2
mr
2
2




H
H


2



Fr  m r  r  m r  r  2   m r  2 3 


m
r
mr 







Now the question becomes, what function f(r) satisfies the
solution to this differential form?

H2 
f r   m r  2 3 
mr 

Or alternatively, what is the solution to this equation:
2
H
mr 
 f r   0
3
mr
11 - 13
The trick is to write it in differential form:
H2
mrr 
r  rf r   0
3
mr
For example, consider the dot product of the velocity vector with itself.
   
2
v  v  v  r  r  rrˆ  rˆ  rrˆ  rˆ



2
2
H
H


v 2  r 2  r 2 2  r 2  r 2  2   r 2  2 2
mr
 mr 
What is the time derivative of this quantity?
d 2 d  
d  2 H2 
v  v  v    r  2 2 
dt
dt
dt 
mr 
d 2
H2
v  2rr  2 2 3 r
dt
mr
11 - 14
If you now multiply this function by ½ m, you will find that:
d 1 2
H2
r
 mv   mrr 
3
dt  2
r
 
 m
H2
mrr 
r  rf r   0
3
r

 m
The result is that the equation for radial motion is now reduced to:
d 1 2
 m v   rf r   0
dt  2

The next step is to write f(r) in a form that allows for solution of
this differential equation.
11 - 15
Consider for the moment that the function f(r) is actually
derivative of another function.
dU r 
dU r   dr dt 
dU r  1
f r   



dr
dr  dt dr 
dt r
After some manipulation, this formula becomes:
dU r 
  rf r 
dt
So, now we can re-write the equation on the previous slide as:
d 1 2
d  1 2  dU r  d  1 2






m
v

r
f
r

m
v


m
v

U
r





0
dt  2
dt  2
dt
dt  2



11 - 16
d 1 2

 m v  U r   0
dt  2

This equation is easily solvable. It is simply a constant.
1 2
mv  U r   Const .
2
But we also know that the expression for the square of velocity.
2
H
2
2
v  r  2 2
mr
This equation then becomes upon multiplication by ½ m…
1 2 1 H2
mr 
 U r   Const.
2
2
2mr
11 - 17
1 2 1 H2
r   Const .
mr 
U
2

2  
2m
r


This is the
kinetic energy
stored in the
linear velocity
This is the kinetic
energy stored in
the angular
momentum
This is the
potential
energy term
When we make the substitution…


P  mv
P m v
2
2 2
2
2
P
1
H
The above equation becomes:

 U r   Const .
2
2m 
2m
r


Linear momentum
Energy term
Angular momentum
Energy term
11 - 18
Problem 1
D
10 mm
20o
A
B
C
15o
A 25-g steel-jacket bullet is fired
horizontally with a velocity of
600 m/s and ricochets off a steel
plate along the path CD with a
velocity of 400 m/s. Knowing that
the bullet leaves a 10-mm scratch
on the plate and assuming that its
average speed is 500 m/s while it
is in contact with the plate,
determine the magnitude and
direction of the average impulsive
force exerted by the bullet on the
plate.
11 - 19
Problem 1
D
10 mm
20o
A 25-g steel-jacket bullet is fired
A
B
horizontally with a velocity of
o
15
600 m/s and ricochets off a steel
plate along the path CD with a
velocity of 400 m/s. Knowing that the bullet leaves a 10-mm
scratch on the plate and assuming that its average speed is
500 m/s while it is in contact with the plate, determine the
magnitude and direction of the average impulsive force exerted
by the bullet on the plate.
C
1. Draw a momentum impulse diagram: The diagram
shows the particle, its momentum at t1 and at t2, and the impulses
of the forces exerted on the particle during the time interval t1 to t2.
11 - 20
Problem 1
D
10 mm
20o
A 25-g steel-jacket bullet is fired
A
B
horizontally with a velocity of
o
15
600 m/s and ricochets off a steel
plate along the path CD with a
velocity of 400 m/s. Knowing that the bullet leaves a 10-mm
scratch on the plate and assuming that its average speed is
500 m/s while it is in contact with the plate, determine the
magnitude and direction of the average impulsive force exerted
by the bullet on the plate.
2. Apply the principle of impulse and momentum: The final
momentum mv2 of the particle is obtained by adding its initial
momentum mv1 and the impulse of the forces F acting on the
particle during the time interval considered.
C
mv1 +S F t = mv2
S F is sum of the impulsive forces (the forces that are large
enough to produce a definite change in momentum).
11 - 21
Problem 1 Solution
D
10 mm
20o
A
B
Draw a momentum impulse
diagram.
C
15o
y
y m v1
y
o
15
m v2
x
x
+
Fx  t
=
x
20o
Fy  t
Since the bullet leaves a 10-mm scratch and its average speed
is 500 m/s, the time of contact  t is:
 t = (0.010 m) / (500 m/s) = 2x10-5 s
11 - 22
Apply the principle of impulse
and momentum.
y
y m v1
x
o
15
+
Problem 1 Solution
y
m v2
x
Fx  t
=
x
20o
Fy  t
mv1 +S F t = mv2
+
x components:
(0.025 kg)(600 m/s)cos15o+Fx2x10-5s = (0.025 kg)(400 m/s)cos20o
Fx = - 254.6 kN
+
y components:
-(0.025 kg)(600 m/s)sin15o+Fy2x10-5s=(0.025 kg)(400 m/s) sin20o
Fy = 365.1 kN
11 - 23
Problem 1 Solution
y
y m v1
y
o
15
m v2
x
x
+
Fx  t
=
x
20o
Fy  t
Fx = - 254.6 kN,
F=
Fy = 365.1 kN
( -254.6 kN )2 + ( 365.12 kN )2 = 445 kN
F = 445 kN
40.1o
11 - 24
Problem 2
650 kg
1.2 m
140 kg
The 650-kg hammer of a drop-hammer
pile driver falls from a height of 1.2 m
onto the top of a 140-kg pile, driving
it 110 mm into the ground. Assuming
perfectly plastic impact (e = 0 ),
determine the average resistance of
the ground to penetration.
11 - 25
Problem 2
650 kg
1.2 m
140 kg
The 650-kg hammer of a drop-hammer
pile driver falls from a height of 1.2 m
onto the top of a 140-kg pile, driving
it 110 mm into the ground. Assuming
perfectly plastic impact (e = 0 ),
determine the average resistance of
the ground to penetration.
1. Apply conservation of energy principle : When a particle
moves under the action of a conservative force, the sum of the
kinetic and potential energies of the particle remains constant.
T1 + V1 = T2 + V2
where 1 and 2 are two positions of the particle.
1a. Kinetic energy: The kinetic energy at each point on the path
is given by:
1
T = 2 m v2
11 - 26
Problem 2
650 kg
1.2 m
140 kg
The 650-kg hammer of a drop-hammer
pile driver falls from a height of 1.2 m
onto the top of a 140-kg pile, driving
it 110 mm into the ground. Assuming
perfectly plastic impact (e = 0 ),
determine the average resistance of
the ground to penetration.
1b. Potential energy: The potential energy of a weight W close
to the surface of the earth at a height y above a given datum is
given by:
Vg = W y
11 - 27
Problem 2
650 kg
1.2 m
140 kg
The 650-kg hammer of a drop-hammer
pile driver falls from a height of 1.2 m
onto the top of a 140-kg pile, driving
it 110 mm into the ground. Assuming
perfectly plastic impact (e = 0 ),
determine the average resistance of
the ground to penetration.
2. Apply conservation of momentum principle: During an impact
of two bodies A and B, the total momentum of A and B is
conserved if no impulsive external force is applied.
mA vA + mB vB = mA v’A + mB v’B
where vA and vB denote the velocities of the bodies before the
impact and v’A and v’B denote their velocities after the impact.
For perfectly plastic impact (e = 0), v’A = v’B = v’ and
mA vA + mB vB = (mA + mB) v’
11 - 28
Problem 2
650 kg
The 650-kg hammer of a drop-hammer
pile driver falls from a height of 1.2 m
onto the top of a 140-kg pile, driving
it 110 mm into the ground. Assuming
perfectly plastic impact (e = 0 ),
determine the average resistance of
the ground to penetration.
1.2 m
140 kg
3. Apply principle of work and energy: When a particle moves
from position 1 to position 2 under the action of a force F, the
work of the force F is equal to to the change in the kinetic energy
of the particle.
T1 + U1 2 = T2
where
T1 = 1 m v12 ,
2
T2 = 1 m v22 and
2
U1
2
= F . dr
11 - 29
Apply conservation of energy principle. Problem 2 Solution
Motion of the hammer during the drop just before impact.
650 kg
v1 = 0
650 kg
y
Position 1
1.2 m
Position 2
v2
1.2 m
T1 + V1 = T2 + V2
0 + mg (1.2 m) = 1 m v22 + 0
2
v22 = 2 ( 9.81 m/s2 )(1.2 m)
v2 = 4.85 m/s
11 - 30
Apply conservation of
momentum principle.
Problem 2 Solution
Impact process:
Before
impact:
mH vH
mP vP = 0
After
impact:
mH v’
mP v’
mH vH + mP vP = ( mH + mP ) v’
(650 kg)( 4.85 m/s) + 0 = (650 kg + 140 kg) v’
v’ = 3.99 m/s
11 - 31
Apply principle of work and energy.
Problem 2 Solution
Hammer and pile move against ground resistance.
v’
W
110 mm
v=0
R
Position 1
Work
Position 2
T1 + U1 2 = T2
1 ( m + m ) v’2 + (W + W - R) y = 0
H
P
H
P
2
1 (650 + 140) (3.99 m/s)2 + [(650 + 140)(9.81) - R](0.110) = 0
2
R = 65,000 N
11 - 32
Problem 3
O
B
45o
a
C
A
A small sphere B of mass m is attached
to an inextensible cord of length 2a,
D
which passes around the fixed peg A
and is attached to a fixed support at O.
The sphere is held close to the support
C’’ at O and released with no initial velocity.
It drops freely to point C, where the cord
becomes taut, and swings in a vertical
C’
plane, first about A and then about O.
Determine the vertical distance from
line OD to the highest point C’’ that the
sphere will reach.
11 - 33
O
B
D
Problem 3
45o
A small sphere B of mass m is attached
C’’ to an inextensible cord of length 2a,
which passes around the fixed peg A
and is attached to a fixed support at O.
C
C’
The sphere is held close to the support
at O and released with no initial velocity. It drops freely to point C,
where the cord becomes taut, and swings in a vertical plane, first
about A and then about O. Determine the vertical distance from
line OD to the highest point C’’ that the sphere will reach.
a
A
1. Apply conservation of energy principle : When a particle
moves under the action of a conservative force, the sum of the
kinetic and potential energies of the particle remains constant.
T1 + V1 = T2 + V2
where 1 and 2 are two positions of the particle.
11 - 34
O
B
D
Problem 3
45o
A small sphere B of mass m is attached
C’’ to an inextensible cord of length 2a,
which passes around the fixed peg A
and is attached to a fixed support at O.
C
C’
The sphere is held close to the support
at O and released with no initial velocity. It drops freely to point C,
where the cord becomes taut, and swings in a vertical plane, first
about A and then about O. Determine the vertical distance from
line OD to the highest point C’’ that the sphere will reach.
a
A
1a. Kinetic energy: The kinetic energy at each end of the path
is given by:
1
T = 2 m v2
11 - 35
O
B
D
Problem 3
45o
A small sphere B of mass m is attached
C’’ to an inextensible cord of length 2a,
which passes around the fixed peg A
and is attached to a fixed support at O.
C
C’
The sphere is held close to the support
at O and released with no initial velocity. It drops freely to point C,
where the cord becomes taut, and swings in a vertical plane, first
about A and then about O. Determine the vertical distance from
line OD to the highest point C’’ that the sphere will reach.
a
A
1b. Potential energy: The potential energy of a weight W close
to the surface of the earth at a height y above a given datum is
given by:
Vg = W y
11 - 36
O
B
Problem 3
D
45o
A small sphere B of mass m is attached
A
C’’ to an inextensible cord of length 2a,
which passes around the fixed peg A
and is attached to a fixed support at O.
C
C’
The sphere is held close to the support
at O and released with no initial velocity. It drops freely to point C,
where the cord becomes taut, and swings in a vertical plane, first
about A and then about O. Determine the vertical distance from
line OD to the highest point C’’ that the sphere will reach.
2. Apply the principle of impulse and momentum: The final
momentum mv2 of the particle is obtained by adding its initial
momentum mv1 and the impulse of the forces F acting on the
particle during the time interval considered.
a
mv1 +S F t = mv2
S F is sum of the impulsive forces (the forces that are large
enough to produce a definite change in momentum).
11 - 37
Apply conservation of energy principle. Problem 3 Solution
Motion of the sphere from point B to point C (just before the
cord is taut).
y
O
O
D
D
B
B
45o
45o
vB = 0 a
a
A
A Position 1
Position 2
C
vC
T1 + V1 = T2 + V2
0 + 0 = 1 m vC2 - m g (2 a sin 45o )
2
vC = 1.682 g a
11 - 38
Apply impulse and momentum principle. Problem 3 Solution
Consider the sphere at point C as the cord becomes taut and the
velocity of the sphere changes to be in direction normal to the cord.
O
B
D
45o
a
C
vC
o
45
t
D
45o
a
A
C
A
v’C
t
m vC cos 45o = m v’C
v’C = vC cos 45o = 1.682
v’C = 1.1892
O
B
ga
g a cos 45o
Momentum is conserved
in the tangential direction
since the external impulse
(the cord on the sphere)
11 - 39
is in the normal direction.
Apply conservation of energy principle. Problem 3 Solution
Motion of the sphere from point C to point C’’.
O
D
B
2 a sin45o
C
a
45o
A
Position 2
v’C = 1.1892 g a
y
O
B
D
a
45o
A
Position 3
vC’’= 0 d
C’’
T2 + V2 = T3 + V3
1 m (v’ )2 - m g (2 a sin 45o ) = 0 - m g d
C
2
1 m (1.1892)2 g a - m g (2 a sin 45o ) = 0 - m g d
2
d = 0.707 a
11 - 40
Problem 4
C
lA
D
A
B
A
B
lB
A small sphere A attached to a cord
AC is released from rest in the
position shown and hits an identical
sphere B hanging from a vertical
cord BD. If the maximum angle B
formed by cord BD with the vertical
in the subsequent motion of sphere
B is to be equal to the angle A ,
determine the required value of the
ratio lB / lA of the lengths of the two
cords in terms of the coefficient of
restitution e between the two
spheres.
11 - 41
Problem 4
lA
C
A small sphere A attached to a cord
AC is released from rest in the
A
position shown and hits an identical
A
B
sphere B hanging from a vertical
B
cord BD. If the maximum angle B
formed by cord BD with the vertical in the subsequent motion of
sphere B is to be equal to the angle A , determine the required
value of the ratio lB / lA of the lengths of the two cords in terms of
the coefficient of restitution e between the two spheres.
D
lB
1. Apply principle of conservation of energy: When a particle
moves under the action of a conservative force, the sum of the
kinetic and potential energies of the particle remains constant.
T1 + V1 = T2 + V2
where 1 and 2 are two positions of the particle.
11 - 42
Problem 4
lA
C
A small sphere A attached to a cord
D
AC is released from rest in the
A
position shown and hits an identical
A
B
sphere B hanging from a vertical
B
cord BD. If the maximum angle B
formed by cord BD with the vertical in the subsequent motion of
sphere B is to be equal to the angle A , determine the required
value of the ratio lB / lA of the lengths of the two cords in terms of
the coefficient of restitution e between the two spheres.
1a. Kinetic energy: The kinetic energy at each point on the path
is given by:
1
T = 2 m v2
1b. Potential energy: The potential energy of a weight W close
to the surface of the earth at a height y above a given datum is
given by:
11 - 43
Vg = W y
lB
Problem 4
lA
C
A small sphere A attached to a cord
AC is released from rest in the
A
position shown and hits an identical
A
B
sphere B hanging from a vertical
B
cord BD. If the maximum angle B
formed by cord BD with the vertical in the subsequent motion of
sphere B is to be equal to the angle A , determine the required
value of the ratio lB / lA of the lengths of the two cords in terms of
the coefficient of restitution e between the two spheres.
2. Apply conservation of momentum principle: During an impact
of two bodies A and B, the total momentum of A and B is
conserved if no impulsive external force is applied.
mA vA + mB vB = mA v’A + mB v’B
where vA and vB denote the velocities of the bodies before the
impact and v’A and v’B denote their velocities after the impact.
11 - 44
D
lB
Problem 4
lA
C
A small sphere A attached to a cord
AC is released from rest in the
A
position shown and hits an identical
A
B
sphere B hanging from a vertical
B
cord BD. If the maximum angle B
formed by cord BD with the vertical in the subsequent motion of
sphere B is to be equal to the angle A , determine the required
value of the ratio lB / lA of the lengths of the two cords in terms of
the coefficient of restitution e between the two spheres.
D
lB
3. Apply the relationship for the coefficient of restitution: For
impact of two particles A and B:
v’B - v’A = e (vA - vB)
where v’B - v’A and vA - vB are the relative velocities, normal to the
impact plane, after and before the impact, respectively, and e is
the coefficient of restitution.
11 - 45
Problem 4 Solution
C
lA
D
A
Motion of sphere A from its
release until it hits sphere B.
lB
B
A
B
C
lA
A
Position 1
A
vA1= 0
y
lA( 1 - cos A )
vA2
Position 2
11 - 46
Problem 4 Solution
C
lA
Apply principle of conservation
of energy.
y
A
Position 1
A
vA1= 0
lA( 1 - cos A )
vA2
Position 2
Motion of sphere A from its
release until it hits sphere B.
T1 + V 1 = T2 + V2
0 + m g lA( 1 - cos A ) = 1 m (vA2)2 + 0
2
vA2 =
2 g lA( 1 - cos A )
11 - 47
Problem 4 Solution
C
lA
lB
D
A
Collision of balls A and B.
B
A
B
C
C
D
vA2
B
vB2 = 0
Before impact
D
v’A2
B
v’B2
After impact
11 - 48
C
C
D
vA2
Problem 4 Solution
Apply conservation of
momentum principle.
D
B
vB2 = 0
Before impact
v’A2
B
v’B2
Apply the relationship
for the coefficient of
restitution.
After impact
m vA2 = m v’A2 + m v’B2
vA2 = v’A2 + v’B2
(1)
( v’B2 - v’A2 ) = e ( vA2 )
(2)
Eliminating v’A2 from equations (1) and (2) gives:
v’B2 =
vA2
2
( 1+ e )
11 - 49
Problem 4 Solution
C
lA
D
A
Motion of sphere B following
the collision.
lB
B
A
B
D
lB( 1 - cos B )
lB
B
B
Position 2
vB3
Position 3
v’B2
11 - 50
Problem 4 Solution
lB( 1 - cos B )
D
y
B
lB
B
vB3
Apply principle of
conservation of energy.
Position 3
Motion of sphere B
following the collision.
v’B2
Position 2
T2 + V2 = T3 + V3
1 m ( v’ )2 + 0 = 0 + m g l ( 1 - cos  )
B2
B
B
2
Substituting v’B2 =
vA2
2
( 1+ e ) and vA2 =
2 g lA( 1 - cos A )
and B = A gives:
lB
=
lA
(
1+e
2
)
2
11 - 51