1 KE = mv 2 2 W = F x m kg = J s
2
C inelastic if _ m1 > m2 _ v f = vi C inelastic if _ m1 < m2 _ v f = C elastic m1v1 m2
1 1 1 1 2 2 m1v12,i + m2 v 2,i = m1v12, f + m2 v 2, f 2 2 2 2
h Mgh ( m + M ) g W = F cos x 2 W = KE pulleysystem v sameheight = 1 (m
3.1.1 The Stability of Solutions 185
implies that there is a positive number a with
Reljs—a, j=1,.,r,
because Re A,- < 0 for j = 1, ., r. Assuming t 2 1, as we can do, we ﬁnd that
|w,(t) g C,t"“e“", t 2 1,
where C, is some positive constant. However, it i
3.1 The Zeros of an Analytic Function 177
Since A > 1, Re h(iy) is always negative. Further, Im h(iy) decreases from a positive
value at y = R to a negative value at y = —R. That is, h(iy) moves from the second
to the third quadrant as y decreases from R
3.1.1 The Stability of Solutions 187
= —a—2b, B=2ab+b2+c2, C= —a(b2+c2).
Suppose ﬁrst that a and b are negative; it clearly follows from the formulas
above that A, B, and C are all positive. Furthermore,
C—AB=2b(a2+2ab+b2+c2)<0,
so AB > C. Conversely, sup
3.2 Maximum Modulus and Mean Value 193
THEOREM 2 Schwarz’s Lemma Suppose that f is analytic in the disc |z| < 1, that f(0) = 0,
and that I f (Z)| < 1, for all z in the disc. Then
|f(Z)| S lzl, |z| < 1.
Equality can h01d for some 2 ¢ 0 only if f (z) = 12,
3.1.1 The Stability of Solutions 189
M4 =
HNMi-‘O
2
1
4
1
OOH-i=-
Mv-OOO
A computation shows that the determinants of M1 and M 2 are positive, but those
of M 3 and M4 are zero. Hence, p is not stable. El
EXERCISES FOR SECTION 3.1.1
1. Use the Routh—Hurwit
10.
11.
3.2 Maximum Modulus and Mean Value 195
is analytic on the disc {2: |z| < 1} and |S(z)| = 1 if |z| = 1, 2 ye 1. However,
8(0) = e > 1. Why does this not contradict the maximum-modulus principle?
Let D be a bounded domain with boundary B. Suppose th
3.2 Maximum Modulus and Mean Value 191
(Note that yR is oriented negatively; this is done because it is traditional to
evaluate F (iy) as y increases.) Show that if FR winds around —c a net positive
number of times in the clockwise direction, then 1 + (1/
3.3 Linear Fractional Transformations 197
and, after some cancellation,
bcz2 + adz1 = adzz + bczl,
01'
(ad — bc)z1 = (ad — bc)zz.
Consequently, 21 = 22. Hence, the function T maps distinct points onto distinct
images. Note also that T has a pole of order
3.3 Linear Fractional Transformations 199
_ w—w1 w2 — W3
SW) _ (w — w3>(w2 — W1),
so that S(w1) = 0, S(w2) = 1, and S(w3) = 00. L is given by
L(z) = S“(T(z).
We can carry out a similar sort of computation if one of 21, 22, and 23 or W}, w,
and W3 is 0
3.3 Linear Fractional Transformations 203
Figure 3.8
A dipole located at the origin;
+ Z the circulation is given by
. . e
Circulation of 7-;(2) = :+ 1 1(2) : z: I
with fixed points i c
_ e % % %e
(a) (b)
Figure 3.9
the family of circles centered o
3.1 The Zeros of an Analytic Function 181
20. Suppose that f is analytic on a domain containing {2: |z| s 1} and that
I f (ew)| < 1, 0 s 6 s 21:. Show that f has exactly one ﬁxed point in the disc
|z| < 1; that is, the equation f (z) = 2 has precisely one
3.1.1
3.1.1 The Stability of Solutions 183
differential equation
6F
0—hH4WE—c—0F=Q
(b) Use (a) and Exercise 35 to show that the Legendre polynomials satisfy
(n + 1)P,+1(z) — (2n + 1)zP,(z) + nP,_1(z) = 0,
and, using (8),
(2n + 1)P.(Z) = P.1+1(Z) — P.2-r(z
10^9 G 10^6 M 10^3 k 10^-2 c 10^-3 m 10^-6 10^-9 n 10^-12 p
=
m v d
V lim
0
V lim
x dx = 0 t dt
d C =0 dx d x =1 dx dn x = nx n 1 dx d ax n = a nx n 1 dx d 1 d 1 1 = x = 1x 11 = 2 dx x dx x d sin ( 3 ) = 3 cos( 3 ) dx d cos(2 ) = 2 sin(+ ) Vi 2 V f dx V
PHY 110
Fall 2013
Quiz 10
Solution
Show all your work neatly! A 30 kg weight hangs
from the end of a light beam that is 2 m long. The
beam is attached to the wall at the left by a frictionless
pin. A frictionless pulley hangs from the ceiling, and a
cable
PHY 110
Fall 2013
1. (5 pts) The quantity
Quiz 1
Solution
G
is well known to string theorists. In that expression, has units
c3
kg m 2
m3
of
, G has units of
, and c is the speed of light. What is
s
kg s 2
a) a length
b) a time
c) a velocity
kg m 2 m 3
s
PHY 110
Fall 2013
Quiz 3
Solution
Circle the best answer from among the choices given for each question.
1. (2 pts) An object is at a position in meters given by (x,y) coordinates (1.000, 2.000). One millisecond later,
it is at the position (1.001, 1.997)
PHY 110
Fall 2013
Take Home Quiz 2
Solution
1. Remember that froghopper from the homework? Suppose he uses his special jumping legs to
accelerate upward at 4 km/s2 for a distance of 2 mm. Ignore gravity for this acceleration phase.
What is his upward velo
PHY 110
Fall 2013
Quiz 4
Solution
1. (3 pts) A 9 kg object is moved from rest by a constant vector force F . In 6 seconds, the
object undergoes a vector displacement 4 i + 3 m. What is F ?
j
a) 2i + 1.5 N
j
b) 4 i + 3 N
j
c) 8i + 6 N
j
d) 16i + 9 N
j
1
r
PHY 110
Fall 2013
Quiz 6
Solution
1. (3 pts) A marble (mass = 15 g) is released from rest in a jar of liquid. After it has fallen 10 cm
through the shampoo, its velocity is 75 cm/s. What is the work done on the marble by the drag force of
the liquid?
Sinc
PHY 110
Fall 2013
The REAL Quiz 10
Solution
Consider a particle that undergoes simple harmonic motion in the x-direction according to the
following expression (x is in meters, t is in seconds):
x = 0.8 sin ( 5 t 0.7 )
1. What is the maximum acceleration t
PHY 110
Fall 2013
Quiz 9
Solution
The wheel shown rotates so that the top of the wheel moves in
the direction described as into the page and the bottom of
the wheel moves out of the page. The wheel has large
angular momentum and rotates freely on the thin
PHY 110
Fall 2013
Quiz 7
Solution
1. (3 pts) A ball is thrown from the top of a building. Ignoring the effects of air resistance, which of
the following statement(s) is (are) true?
a)
b)
c)
d)
e)
The linear momentum of the ball is conserved.
The kinetic e
PHY 110
Fall 2013
Quiz 5
Solution
1. (3 pts) A block slides 1 m down a rough incline, slowing down as it moves. Which of the following
statements is (are) true? You may circle more than one answer.
a) The work done by gravity is negative.
b) The work done
3.3 Linear Fractional Transformations 201
Hz) = 1
Z
A
Figure 3.7
by inversion to a circle (which goes through the origin when and only when a line
was inverted) (Fig. 3.7).
There is an interesting connection between certain linear fractional transfor-
mat