Classical Mechanics - Kinematics

A 5 kilogram stone is dropped on a nail and drives the nail 0.025 meter into a piece of wood. If stone is moving at 10 meters per second when it hits the nail, the average force exerted on the nail by the stone while the nail is going into the wood is most nearly 

A. 10 N
B. 100 N
C. 1000 N
D. 10,000 N
E. 100,000 N
(GR8677 #8)
Solution:

Force: F = ma

The acceleration when the stone hits the nail:
2ay = v² + v0²
v0 = 0
→ a = v²/2y = 10²/2(0.025) = 2 × 10³ m/s²

→ F = 5 × 2 × 10³ = 10,000 N

Answer: D

Electromagnetism - Drift Velocity

A wire of diameter 0.02 meter contains 1028 free electrons per cubic meter. For an electric current of 100 amperes, the drift velocity for free electrons in the wire is most nearly

A. 0.6 × 10−29 m/s
B. 1 × 10−19 m/s
C. 5 × 10−10 m/s
D. 2 × 10−4 m/s
E. 8 × 103 m/s
(GR8677 #9)
Solution:

Drift velocity is the average velocity of a carrier that is moving under the influence of an electric field.

Velocity: v =  L/t

In a wire with length L and cross sectional area A, there are n electrons with charge qe per cubic meter.
→ total number of mobile electrons in the wire, Q = nqeLA

Current: I = Q/t = nqeLA/t = nqevA
→ v = I/nqeA

qe = charge of an electron = 1.6 × 10−19 C
n = 1028 electrons/m³
A = πr² =  0.5π × 10−4
I =100 Ampere

→ v = 102 / (1028× 1.6 × 10−19× 0.5π × 10−4) = 10228+19+4 / (1.6 × 0.5π) ≈ 10−4

Answer: D

Electromagnetism - Gauss' Law


An isolated sphere of radius R contains a uniform volume distribution of positive charge. Which of the curves on the graph above correctly illustrated the dependence of the magnitude of the electric field of the sphere as a function of the distance r from its center?

A. A
B. B
C. C
D. D
E. E
(GR8677 #10)
Solution:

The graph shows different lines for condition inside the solid sphere, r < R

Gauss’ Law:  


Inside the sphere:



E is linearly proportional to r.

Answer: C

Electromagnetism - Vector Identity

Which of the following equation is a consequence of the equation ?

A.
B.
C.
D.
E.
(GR8677 #11)
Solution:

Vector identity:



Answer: A

Sound and Wave - Doppler Effect

A source of 1-kilohertz sound is moving straight toward you at a speed 0.9 times the speed of sound. The frequency you receive is 

A. 0.1 kHz
B. 0.5 kHz
C. 1.1 kHz
D. 1.9 kHz
E. 10 kHz
(GR8677 #12)
Solution:

Moving toward observer
f increases
f  > 1 kHz
→ A and B are FALSE

Speed 0.9 times the speed of sound
f increases greatly
→ E. TRUE

Answer: E

Calculation:





+ sign = receding
− sign = approaching



Optics - Interference

Two coherent sources of visible monochromatic light from an interference pattern on a screen. If the relative phase of the source is varied from 0 to 2π at a frequency of 500 hertz, which of the following best describes the effect, if any, on the interference pattern?
  1. It is unaffected because the frequency of the phase change is very small compared to the frequency of visible light.
  2. It is unaffected because the frequency of the phase change is an integral multiple of π.
  3. It is destroyed except when the phase difference is 0 to π.
  4. It is destroyed for all phase differences because the monochromaticity of the sources is destroyed.
  5. It is not destroyed but simply shifts positions at a rate too rapid to be detected by the eye.
(GR8677 #13)
Solution:


Answer:

Thermal Physics - Specific Heat

For an ideal gas, the specific heat at constant pressure Cp is greater than the specific heat at constant volume Cv because the
  1. Gas does work on its environment when its pressure remains constant while its temperature is increased.
  2. Heat input per degree increase in temperature is the same in processes for which either the pressure or the volume is kept constant.
  3. Pressure of the gas remains constant when its temperature remains constant.
  4. Increase in the gas’s internal energy is greater when the pressure remains constant than when the volume remains constant
  5. Heat needed is greater when the volume remains constant than when the pressure remains constant.
(GR8677 #14)
Solution:

A. TRUE.
Heat Capacity: C = Q/dT

First law of Thermodynamics: the change in internal energy of a system dU is equal to the heat Q added and the work, W done on or by the system 
→ dU = Q ± W

W done on the system → +W
W done by the system → −W

Gas (the system) does work on its environment
→ W done by the system
→ dU = Q − W

At constant V:
→ Work, W = PdV = 0
→ Q = dU
→ Cv = dU/dT

At constant P:
→ Work, W = PdV ≠ 0
→ Q = dU + W
→ Cp = dU/dT + PdV/dT =  Cv + PdV/dT
→ Cp > Cv

B. FALSE.
This means Cp = Cv, but according to A, Cp > Cv

C. FALSE.
Ideal gas law: PV = NkT
→ If T = constant, P changes if V changes.

D. FALSE.
Heat Capacity, C = Q/dT, does not depend on the gas’s internal energy, U

E. FALSE.
See A. At constant V, Q = dU. At constant P, Q = dU + W.

Answer: A

Thermal Physics - Probability

A sample of N atoms of helium gas is confined in a 1.0 cubic meter volume. The probability that none of the helium atoms is in a 10−6 cubic meter volume of the container is

A. 0
B. (10−6)N
C. (1 − 10−6)N
D. 1 − (10−6)N
E. 1
(GR8677 #15)
Solution:

Total Probability: P = P1 + P2 = 1
P1 = The probability that one atom is in a 10−6 cubic meter volume of the container
P2 =  The probability that none of the helium atoms is in a 10−6 cubic meter volume of the container
→ P2 = 1 − P1

P1 = 1/n
n = number of 10−6 m3 cubes in the 1 m3 volume
10−6 × n = 1 → n = 1/10−6 = 106 cubes
→ P1 = 1/n = 1/106 = 10−6
→ P2 = 1 − P1 = 1 − 10−6
For N atoms: P2 = (1 − 10−6)N

Answer: C

Nuclear & Particle Physics - Muon

Except for mass, the properties of the muon most closely resemble the properties of the

A. electron
B. graviton
C. photon
D. pion
E. proton
(GR8677 #16)
Solution:

A. muon and electron are elementary particles → fermions (half-integer spin) → leptons (one unit charge)
B. graviton is gravity force carrier → boson (integer spin)
C. photon is EM force carrier → boson (integer spin)
D. pion is composite particle → meson (consist of quark-antiquark)
E. proton is composite particle → baryon (consist of 3 quarks)


click image to enlarge

Answer: A

Nuclear & Particle Physics - Radioactivity

Suppose that decays by natural radioactivity in two stages to . The two stages would most likely be which of the following?

First StageSecond Stage
A.      β emission with an antineutrino     α emission
B.β emissionα emission with a neutrino
C.β emissionγ emission
D.Emission of a deuteronEmission of two neutrons
E.α emissionγ emission
(GR9677 #30)
Solution:
→ First Stage → Second Stage →
by natural decay.

A. TRUE.
β decay:  emission of an electron and electron anti-neutrino,
α decay: emission of a Helium nucleus resulting in the loss of two protons and two neutrons,
Therefore,

B. FALSE.
β decay always emits an anti-neutrino
α decay does not emit neutrino

C. FALSE.
γ decay: photon emission with no change in mass number A or atomic number Z
Therefore,

D. FALSE.
Deuteron decay:
Deuteron (nucleus of deuterium) is a stable particle.
Deuteron decay is rare, not natural.

E. FALSE.

Answer: A