Classical Mechanics - 1D Vertical Motion

A rock is thrown vertically upward with initial speed v₀. Assume a friction force proportional to –v, where v is the velocity of the rock, and neglect the buoyant force exerted by air. Which of the following is correct?

1. The acceleration of the rock is always equal to g.
2. The acceleration of the rock is equal to g only at the top of the flight.
3. The acceleration of the rock is always less than g.
4. The speed of the rock upon return to its starting point is v₀.
5. The rock can attain a terminal speed greater than v₀ before it returns to its starting point.

(GR8677 #01)

Solution:

There is a friction force:

• acceleration is not constant → (A) and (C) FALSE
• energy is not conserved so it's initial and final speed is not the same → (D) FALSE
• frictional force slows down the object, so its speed at time t has to be less than its initial speed  → (E) FALSE

Answer: B


Math analysis:

Friction force: F_f  = − kv
The equation of motion with friction force: ma = − mg − kv.

a = − g − (kv/m)
(A) FALSE

At the top of the flight, v = 0
a = − g − (k∙0/m) = − g
(B) TRUE

Moving up: v positive
a = − g − (kv/m) = − g − c
a − g
Moving Down: v negative
a = − g − [k(−v)/m] = − g + c
a − g
(C) FALSE

Classical Mechanics - Satellite

A Satellite orbits the Earth in a circular orbit. An Astronaut on board perturbs the orbit slightly by briefly firing a control jet aimed toward the Earth’s center. Afterward, which of the following is true of the satellite’s path?

A. It is an ellipse
B. It is a hyperbola
C. It is a circle with larger radius
D. It is a spiral with increasing radius
E. It exhibits many radial oscillations per revolution.

(GR8677 #02)

Solution:

Initially, the object orbits the Earth in a circular orbit.

Perturbs the orbit slightly means giving some extra momentum, so the orbit won't be circular any longer, and will be elliptic, not enough to be Parabolic or Hyperbolic.

Also, logically the astronaut will not want the satellite to have v = v_escape (parabolic) or v v_escape (hyperbolic).

Answer: A


Notes:

Types of Orbits	Eccentricity	Energy	Velocity
Circular	e = 0	E = Vmin
Elliptic	0 e 1	Vmin E 0	v vescape
Parabolic	e = 1	E = 0	v = vescape It will escape the gravitational pull of the planet. If v is increased it will become a hyperbolic orbit.
Hyperbolic	e 1	E 1	v vescape It escapes the gravitational pull of the planet and continues to travel infinitely until it is acted upon by another body with sufficient gravitational force.

The orbital eccentricity, e is the amount by which its orbit deviates from a perfect circle.

Special Relativity - Velocity of Light

For blue light, a transparent material has a relative permittivity (dielectric constant) of 2.1 and a relative permeability of 1.0. If the speed of light in vacuum is c, the phase velocity of blue light in an unbounded medium of this material is

A. √3.1 c
B. √2.1 c
C. c/√1.1
D. c/√2.1
E. c/√3.1

(GR8677 #03)

Solution:

Permittivity, ϵ = 2.1 ϵ₀
Permeability, μ = μ₀
Speed of light in vacuum:
The velocity of light in medium:  

Answer: D

Classical Mechanics - Wave equation

The equation where A, T, and λ are positive constants, represents a wave whose

A. Amplitude is 2A
B. Velocity is in the negative x–direction
C. Period is T/λ
D. Speed is x/t
E. Speed is λ/T

(GR8677 #04)

Solution:

(A) FALSE.
Amplitude  = maximum displacement = y_max = A.

(B) FALSE.
For a wave traveling to the right (positive x-direction): y = ƒ(x − vt)
(x − vt) = constant.
As t increases, x must increases to keep (x − vt) = constant

For a wave traveling to the left (negative x-direction): y = ƒ(x + vt)
As t increases, x decreases to keep (x + vt) = constant.

The problem gives:  y = ƒ(vt − x)
As t increases, x must increases to keep (x − vt) = constant.
The waves is traveling to the right.

(C) FALSE.
The unit of T/λ (second/meter) does not match with the unit is period (second).

(D) FALSE.
The speed of  the wave is λ/T.

(E) TRUE.
See D.

Answer: E

Classical Mechanics - Pendulum

Two small spheres of putty, A and B of mass M and 3M, respectively, hang from the ceiling on strings of equal length l. Sphere A is drawn aside so that it is raised to a height h₀ as shown above and then released. Sphere A collides with sphere B; they stick together and swing to a maximum height h equal to

A. (1/16) h₀
B. (1/8) h₀
C. (1/4) h₀
D. (1/3) h₀
E. (1/2) h₀

(GR8677 #5)

Solution:

Conservation of energy of A before and when it hits B:


Conservation of momentum when and after collision:


Conservation of energy of A and B at h = 0 and h_max:


Answer: A

Classical Mechanics - Kinematics

A particle is initially at rest at the top of a curved frictionless track. The x- and y-coordinate of the track are related in dimensionless units by y = x²/4, where the positive y-axis is in the vertical downward direction. As the particle slides down the track, what its tangential acceleration?

A. 0
B. g
C. gx/2
D. gx/√(x²+4)
E. (gx²)/√(x²+16)

(GR8677 #06)

Solution:

A. FALSE.
The particle slides down the curved track → the tangential acceleration is not zero.

B. FALSE.
The particle slides down the curved track, not in the vertical downward direction (the direction of gravity acceleration) → the tangential acceleration is not equal to g.

C. FALSE.
The unit of gx/2 is not the unit of acceleration.

D. TRUE.
The unit of gx/√(x²+4) = g → the unit of acceleration.

E. FALSE.
The unit of (gx²)/√(x²+16) is not the unit of acceleration.

Answer: D

Calculation:

Classical Mechanics - Static Equilibrium

A 2-kilogram box hangs by a massless rope from a ceiling. A force slowly pulls the box horizontally to the side until the horizontal force is 10 Newtons. The box is then in equilibrium as shown above. The angle that the rope makes with the vertical is closest to

A. arctan 0.5
B. arcsin 0.5
C. arctan 2.0
D. arcsin 2.0
E. 45^o

(GR8677 #07)

Solution:

F_x = F − T sin θ = 0
T sin θ = F 

F_y = T cos θ − mg = 0
T cos θ = mg

T sin θ / T cos θ = tan θ = F/mg =  10/(2 × 10) = 0.5
θ = arctan 0.5

Answer: A

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 #08)

Solution:

Force: F = ma

The acceleration when the stone hits the nail: 2ay = v² + v₀²

v₀ = 0 → 2ay = v²

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 #09)

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 q_e per cubic meter.

Total number of mobile electrons in the wire, Q = nq_eLA

Current: I = Q/t = nq_eLA/t = nq_evA
v = I/nq_eA

q_e = 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)
= 10^2−28+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: ∇ · (∇ × A) = 0



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 form 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:

Interference pattern is observed only when the sources are coherent (both sources have identical λ, f, and phase relationship).

The problem states that the light sources are coherent, thus the interference patterns are observed, not destroyed → (C) and (D) are FALSE.

Phase change will affect the constructive and destructive patterns of interference → (A) and (B) are FALSE.

Note: 500 Hz or 500/second is too rapid for human eye which only can perceive 60 Hz to 80 Hz flickering light.

Answer: E

Thermal Physics - Specific Heat

For an ideal gas, the specific heat at constant pressure C_p is greater than the specific heat at constant volume C_v 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
C_v = dU/dT

At constant P:
Work, W = PdV ≠ 0
Q = dU + W
C_p = dU/dT + PdV/dT = C_v + PdV/dT
C_p  C_v

(B) FALSE.
This means C_p = C_v, but according to A, C_p  C_v

(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’ 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) TRUE
Muon and electron are elementary particles → fermions (half-integer spin) → leptons (one unit charge)

(B) FALSE
Graviton is gravity force carrier → boson (integer spin)

(C) FALSE
Photon is EM force carrier → boson (integer spin)

(D) FALSE
Pion is composite particle → meson (consist of quark-antiquark)

(E) FALSE
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 Stage	Second Stage
A.	β− emission with an antineutrino	α emission
B.	β− emission	α emission with a neutrino
C.	β− emission	γ emission
D.	Emission of a deuteron	Emission of two neutrons
E.	α emission	γ 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