Solutions for University Physics with Modern Physics, 15th Edition by Young (All Chapters included)

University Physics with Modern
Physics, 15th Edition by
Hugh D. Young



Complete Chapter Solutions Manual
are included (Ch 1 to 44)




** Immediate Download
** Swift Response
** All Chapters included

,Table of Contents are given below

1.Units, Physical Quantities, and Vectors
2.Motion Along a Straight Line
3.Motion in Two or Three Dimensions
4.Newton's Laws of Motion
5.Applying Newton's Laws
6.Work and Kinetic Energy
7.Potential Energy and Energy Conservation
8.Momentum, Impulse, and Collisions
9.Rotation of Rigid Bodies
10.Dynamics of Rotational Motion
11.Equilibrium and Elasticity
12.Fluid Mechanics
13.Gravitation
14.Periodic Motion
15.Mechanical Waves
16.Sound and Hearing
17.Temperature and Heat
18.Thermal Properties of Matter
19.The First Law of Thermodynamics
20.The Second Law of Thermodynamics
21.Electric Charge and Electric Field
22.Gauss's Law
23.Electric Potential
24.Capacitance and Dielectrics
25.Current, Resistance, and Electromotive Force
26.Direct-Current Circuits
27.Magnetic Field and Magnetic Forces
28.Sources of Magnetic Field
29.Electromagnetic Induction
30.Inductance
31.Alternating Current
32.Electromagnetic Waves
33.The Nature and Propagation of Light
34.Geometric Optics
35.Interference
36.Diffraction
37.Relativity
38.Photons: Light Waves Behaving as Particles
39.Particles Behaving as Waves
40.Quantum Mechanics I: Wave Functions
41.Quantum Mechanics II: Atomic Structure
42.Quantum Mechanics II: Atomic Structure
43.Molecules and Condensed Matter
44.Nuclear Physics
45.Particle Physics and Cosmology

,Solutions Manual organized in reverse order, with the last chapter displayed first, to ensure that
all chapters are included in this document. (Complete Chapters included Ch44-1)




PARTICLE PHYSICS AND COSMOLOGY
44
VP44.1.1. IDENTIFY: This problem is about a cyclotron accelerating protons.
SET UP and EXECUTE: (a) We want the frequency. Use Eq. (44.7).
eB e(0.600 T)
f    9.15 MHz.
2 m 2 mp
(b) We want the maximum kinetic energy of the proton. Use Eq. (44.8) with B = 0.600 T and
R = 0.800 m.
e2 B 2 R 2
Kmax   11.0 MeV.
2mp
(c) We want the proton’s speed. The rest energy of the proton is 938 MeV which is much greater than
the kinetic energy of 11 MeV, so we do not have to use the relativistic equation. Solving K  12 mv 2 for
v gives
v  2 K /mp  4.60  107 m/s  0.153c.

EVALUATE: Since v ≈ 15% of c, it is reasonable not to use relativity. If we had used relativity, our
result would be
K  mc2 (  1)    1.012

v  c 1  1/ 2  0.152c.
The percent difference between the two answers is 0.001/0.152 = 0.66%, which is extremely small.
VP44.1.2. IDENTIFY: This problem is about a cyclotron accelerating hydrogen ions (protons).
SET UP and EXECUTE: (a) We want the radius R. R = mpv/eB. Using B = 0.460 T and the given speed,
we get R = 0.136 m.
(b) We want the angular frequency. Use Eq. (44.7) with q = e and B = 0.460 T, which gives  = eB/m
= 4.41  107 rad/s.
EVALUATE: This angular frequency is rather small compared to many modern cyclotrons.
VP44.1.3. IDENTIFY: This problem deals with the operation of a cyclotron.
SET UP and EXECUTE: (a) We want the magnetic field. The rest energy of the proton is 938 MeV and
its kinetic energy here is 80.0 keV, which is much less than its rest energy. Therefore we do not need to
use special relativity. We use Eq. (44.8), with R = (11.4 cm)/2 and Kmax = 80.0 keV, and solve for B,
which gives
2mp K max
B  0.717 T.
eR
(b) We want the frequency. Use Eq. (44.7) with B = 0.717 T and q = e.




44-1

, 44-2 Chapter 44


eB e(0.717 T)
f    10.9 MHz.
2 m 2 mp
EVALUATE: Modern cyclotrons produce much greater kinetic energy, but after all, this was the first
one.
VP44.1.4. IDENTIFY: This problem deals with the operation of a cyclotron.
SET UP and EXECUTE: (a) We want the magnetic field and know that the frequency is 5.80 MHz. Use
Eq. (44.7) and solve for B using q = 2e and the given mass of an alpha particle.
2 m f
B   
2e
(b) We want the kinetic energy when R = 0.650 m. Using Eq. (44.8) with B = 0.756 T gives
(2e) 2 B 2 R 2
K max   11.6 MeV.
2m
EVALUATE: A magnetic field of 0.756 T is certainly physically reasonable in a physics laboratory.
VP44.3.1. IDENTIFY: This problem deals with the available energy when particles collide.
SET UP and EXECUTE: We follow the procedure of Example 44.2.
(a) We want the available energy Ea. The total available energy must be the total rest energy in the
center-of-momentum frame.
Ea  2mpc 2  mnc 2  2(938 MeV)  958 MeV  2834 MeV.
(b) We want the kinetic energy K of the incoming proton. First use Eq. (44.10) to find Em (the total
energy of the incoming proton). Then use Em = K + mpc2 to find K.
Ea2
Em  2
 mpc2  K  mpc 2
2mpc
Ea2
K 2
 2mpc 2
2mpc
Using Ea = 2834 MeV and mpc2 = 938 MeV gives us K = 2410 MeV = 2.41 GeV.
EVALUATE: Colliding beams would require far less kinetic energy because nearly all the incoming
kinetic energy is available energy.
VP44.3.2. IDENTIFY: This problem deals with the available energy during collisions.
SET UP and EXECUTE: (a) We want Ea. The minimum available energy is the rest energy of the
particles after the collision. Ea = mpc2 + mc2 = 938 MeV + 1232 MeV = 2170 MeV = 2.17 GeV.
(b) We want the kinetic energy. Use the same method as in problem VP44.3.1.
Ea2
K  2mpc 2
2mpc 2
Using Ea = 2170 MeV and mpc2 = 938 MeV gives K = 634 MeV.
EVALUATE: The incoming proton could have more than 634 MeV of kinetic energy. In that case, the
products would each have kinetic energy.
VP44.3.3. IDENTIFY: This problem deals with the available energy during collisions.
SET UP and EXECUTE: (a) We want Ea. The minimum available energy is the rest energy of the
product (the 0), which is 1232 MeV.
(b) We want the minimum kinetic energy K of the pion. The target and incident particle have different
masses, so use Eq. (44.9) to find the available energy. Then use this to find Em, and finally use Em = K
+ mc2 to find K.
The quantities in Eq. (44.9) are:
M = target = proton, so Mc2 = 938 MeV
m = incident particle = π–, so mc2 = 140 MeV