Classical Mechanics - Pendulum

Which of the following best illustrates the acceleration of a pendulum bob at points a through e?

(GR0177 #01)

Solution:

The acceleration of pendulum: a = a_centripetal + a_tangential
a_cent = v²/r = ω²r
a_tan = αr

At equilibrium (position B): ω = constant
α = dω/dt = 0
a = a_cent

At maximum amplitude (position A and C): v = 0
a = a_tan

At other positions:
a = a_cent + a_tan

Answer: C

Classical Mechanics - Circular Motion

The coefficient of static friction between a small coin and the surface of a turntable is 0.30. The turntable rotates at 33.3 revolutions per minute. What is the maximum distance from the center of the turntable at which the coin will not slide?

A. 0.024 m
B. 0.048 m
C. 0.121 m
D. 0.242 m
E. 0.484 m

(GR0177 #02)

Solution:

The coin will not slide if  F_centripetal = F_friction
mω²r = μ_smg
r = μ_sg/ω² =  μ_sg/(4π²f²)

Given:
μ_s = 0.30
f = 33.3 revolutions per minute = 33.3/60 = (100/3)(1/60) = 5/9
and take π² = (3.14)² ≈ 10

Radius:
r = (0.3)(10)/[4(10)(25/81)] = (3)(81)/[4(10)(25)] = 243/1000 = 0.243

Answer: D


Notes:


F_centripetal = mv²/r = mω²r
F_friction = μ_sN = μ_smg

Classical Mechanics - Uniform Circular Motion

A satellite of mass m orbits a planet of mass M in a circular orbit of radius R. The time required for one revolution is

A. independent of M
B. proportional to √m
C. linear in R
D. proportional to R^3/2
E. proportional to R²

(GR0177 #03)

Solution:

F_c = F_G
mω²R = GmM / R²
ω² = GM / R³
(2π/T)² = GM / R³
T = (4π²R³/ GM)^1/2
T ∝ R^3/2

Answer: D

Classical Mechanics - Inelastic Collision

In a non relativistic, one-dimensional collision, a particle of mass 2m collides with a particle of mass m at rest. If the particles stick together after the collision, what fraction of the initial kinetic energy is lost in the collision?

A. 0
B. 1/4
C. 1/3
D. 1/2
E. 2/3

(GR0177 #04)

Solution:

Conservation of Momentum (inelastic collision):

m_av_a + m_bv_b = (m_a + m_b)v'

2mv_a + m · 0 = (2m + m)v'

2v_a =  3v'  → v'  = 2v_a / 3

KE_i − KE_f  = ( ½ m_av_a² + ½ m_bv_b² ) − ½ ( m_a + m_b ) (v' )²

= mv_a² − ½ ( 2m + m ) (2v_a / 3)²

= mv_a² − ⅔ mv_a²

= ⅓ mv_a²

  
Answer: C

Thermal Physics - Equipartition Law

A three-dimensional harmonic oscillator is in thermal equilibrium with a temperature reservoir at temperature T. The average total energy of oscillator is

A. ½kT
B. kT
C. ³⁄₂kT
D. 3kT
E. 6kT

(GR0177 #05)

Solution:

Equipartition of Energy: E = ½ fkT

f = Degree of Freedom (DoF)

3D harmonic oscillator has 6 DoF = 3 components of momentum (kinetic energy) and 3 components of position (potential energy)
→ E = ⁶⁄₂ kT = 3kT

Answer: D

Thermal Physics - Isothermal vs Adiabatic

An ideal monatomic gas expands quasi-statically to twice its volume. If the process is isothermal, the work done by the gas is W_i. If the process is adiabatic, the work done by the gas is W_a. Which is the following is true?

A. W_i = W_a
B. 0 = W_i  W_a 
C. 0  W_i  W_a 
D. 0 = W_a  W_i
E. 0  W_a  W_i

(GR0177 #06)

Solution:

Isothermal and Adiabatic P-V Diagram

• Adiabatic connects high-T isotherm and low-T isotherm.
• Isothermal line is always higher than the adiabatic line and they both end at the same volume
• The area under the isothermal line is bigger than the adiabatic → W_a  W_i

Answer: E

Calculation:

Isothermal:

PV = constant
P_i V_i  = P_f V_f  = constant
Given V_f  = 2V_i

P_i V_i  = 2P_f V_i

P_f (iso) = ½ P_i

Adiabatic:

PV^γ = c
P_i V_i^γ = P_f V_f ^γ = constant
V_f  = 2V_i
P_iV_i^γ  = P_f (2V_i)^γ = 2^γP_f V_i^γ 
P_f (adi) = (1/2^γ)P_i 
  
Therefore,

P_{f (iso)} /P_{f (adi)} = 2^γ / 2
P_f (iso) = 2^γ−1P_{f (adi)}

→ P_adi  P_iso

W = ∫ PdV → W_adi  W_iso

Electromagnetism - Magnetic Field

Two long identical bar magnets are placed under a horizontal piece of paper, as shown in the figure above. The paper is covered with iron fillings. When the two north poles are a small distance apart and touching the paper, the iron filings move into pattern that shows the magnetic field lines. Which is the following best illustrates the pattern that results?

(GR0177 #07)

Solution:

(A) and (C) FALSE

(D) FALSE
It could be right if there is a current through the magnets.

(E) FALSE
No magnetic monopoles

Answer: B

Electromagnetism - Method of Image

A positive charge Q is located at a distance L above an infinite grounded conducting plane, as shown in the figure above. What is the total charge induced on the plane?

A. 2Q
B. Q
C. 0
D. –Q
E. –2Q

(GR0177 #08)

Solution:

Method of image:

• +Q attracts negative charges in the plate and repels positive ones. 
• As the positive charges want to “get away”, they succeed in doing so through the ground, leaving the negative charges behind. 
• The plate is left with a net negative charge.

Answer: D

Electromagnetism - Gauss' Law

Five positive charges of magnitude q are arranged symmetrically around the circumference of a circle of radius r. What is the magnitude of the electric field at the center of the circle? (k = 1/4πε₀)

A. 0
B. kq/r²
C. 5kq/r²
D. (kq/r²) cos (2π/5)
E. (5kq/r²) cos (2π/5)

(GR0177 #09)

Solution:

Gauss’ Law:  

There is no q_enc inside the Gaussian surface → E = 0  

Answer: A

Electromagnetism - Capacitor

A 3-microfarad capacitor is connected in series with 6-microfarad capacitor. When a 300-volt potential difference is applied across this combination, the total energy stored in the two capacitor is

A. 0.09 J
B. 0.18 J
C. 0.27 J
D. 0.41 J
E. 0.81 J

(GR0177 #10)

Solution:

C in Series:

¹/_Cnet  = ¹/_C1  + ¹/_C2 = ( ¹/₃  + ¹/₆ )  ( ¹/₁₀-6 ) = ¹/_2 ∙ 10-6 

C_net = 2 ∙ 10^-6 F 

U = ½ QV = ½ C_netV² = ½ (2 ∙ 10^-6) (300)² = 9 (10^-6) (10⁴) = 0.09 J

Answer: A

Optics - Converge Lens

An object is located 40 centimeters from the first of two thin converging lenses of focal lengths 20 cm and 10 cm, respectively, as shown in figure above. The lenses are separated by 30 cm. The final image formed by the two lenses is located 

A. 5.0 cm to the right of the second lens
B. 13.3 cm to the right of the second lens
C. infinitely to the right of the second lens
D. 13.3 cm to the left of the second lens
E. 100 cm to left right of the second lens

(GR0177 #11)

Solution:

¹/_S + ¹/_S' = ¹/_f

For the first lens:

¹/_S1 + ¹/_S1' = ¹/_f1

¹/₄₀ + ¹/_S1'  = ¹/₂₀

S₁' = 40 cm

The image is located 40 cm behind (to the right of) the first lens, which is 10 cm behind the second lens. Therefore, S₂ = −10.

For the second lens:

¹/_S2 + ¹/_S2' = ¹/_f2

¹/ ₍₋₁₀₎ + ¹/_S2' = ¹/₁₀

S₂' = 5 cm

The final image is located 5 cm to the right of the second lens.

Answer: A

Optics - Concave Mirror

A spherical concave mirror is shown in the figure above. The focal point F and the location of the object O are indicated. At what point will the image be located?

A. I
B. II
C. III
D. IV
E. V

(GR0177 #12)

Solution:

The mirror equation:



The picture shows:   is negative.
The image is VIRTUAL (inside the mirror) i.e. at point V

Answer: E

Note:
For concave mirror:

• If an object is located BEFORE the focal point then image is VIRTUAL (inside the mirror).
• If an object is located AFTER the focal point then image is REAL (outside the mirror). 

Optics - Telescope

Two stars are separated by an angle of 3 × 10⁻⁵ radians. What is the diameter of the smallest telescope can resolve the two stars using visible light (λ = 600 nanometers)? (ignore any effect due to Earth’s atmosphere)

A. 1 mm
B. 2.5 cm
C. 10 cm
D. 2.5 m
E. 10 m

(GR0177 #13)

Solution:

Rayleigh telescope resolution limit: sin θ = 1.22 λ/D

Since θ ≪, assume sin θ ~ θ

D = ^1.22λ/_θ 

λ = 600 nanometers = 600 × 10⁻⁹ m

D = ^{1.22(600×10−9)}/_{(3×10⁻⁵)} = 2.44 × 10^-2 m ≈  2.5 cm

Answer: B

Nuclear & Particle Physics - Gamma Rays Detector

An 8-cm diameter by 8-cm long NAI(TI) detector detects gamma rays of a specific energy from a point source of radioactivity. When the source is placed just next to the detector at the center of the circular face, 50 percents of all emitted gamma rays at that energy are detected. If the detector is moved to 1 meter away, the fraction of detected gamma rays drops to

A. 10⁻⁴
B. 2 ×10⁻⁴
C. 4 ×10⁻⁴
D. 8 ×10⁻⁴
E. 16 ×10⁻⁴

(GR0177 #14)

Solution:

The net power radiated by the source: P = I A
I = intensity
A = area of the surface

No power loss mentioned: P₁ = P₂
I₁ A₁ = I₂ A₂

A₁ = area of circle with diameter 8 cm
r₁  = 4 × 10⁻² m

A₂ = area of sphere with radius 1 m

The problem asks for the fraction, not how much the intensity is detected.
Thus, the fraction is just the ratio of the areas:



Answer: C

Lab Methods - Precision

Five classes of students measure the height of a building. Each classes uses a different method and each measures the height many different times. The data for each class are plotted below. Which class made the most precise measurement?

(GR0177 #15)

Solution:

The accuracy is how close the peak is to the reference value.

The precision is how narrow the peak is.

So we look for the graph with the narrowest peak.

Answer: A

Lab Methods - Uncertainty

A student makes 10 one-second measurement of the disintegration of a sample of a long lived radioactive isotope and obtains a following values: 3, 0, 2, 1, 2, 4, 0, 1, 2, 5. How long should the student count to establish the rate to an uncertainty of 1 percent?

A. 80 s
B. 160 s
C. 2000 s
D. 5000 s
E. 6400 s

(GR0177 #16)

Solution:

Radioactive decay can be described by Poisson Distribution.

Poisson Distribution (PD):
Probability distribution of discrete events over an interval (time. distance, etc)

In PD, Standard Deviation, σ = √μ  (see problem GR8677 #40)
μ = λT = expected value
λ = average rate
T = time interval

% Uncertainty = (σ/μ) × 100%

σ/μ = 0.01
√μ/μ = 10⁻²
μ/μ² = 10⁻⁴
1/μ = 1/10⁴
μ = λT = 10⁴

λ = (3 + 0 + 2 + 1 + 2 + 4 + 0 + 1 + 2 + 5)/10 = 2
T = 10⁴/2 = 5000

Answer: D

Nuclear & Particle Physics - Electron Configuration

The ground state electron configuration for phosphorus, which has 15 electrons, is 

A. 1s² 2s² 2p⁶ 3s¹ 3p⁴
B. 1s² 2s² 2p⁶ 3s² 3p³
C. 1s² 2s² 2p⁶ 3s² 3d³
D. 1s² 2s² 2p⁶ 3s¹ 3d⁴
E. 1s² 2s² 2p⁶ 3p²3d³

(GR0177 #17)

Solution:

Phosphorus (15 electrons):  1s² 2s² 2p⁶ 3s² 3p³

Answer: B



Note: 

Electron configuration

	l = 0	l = 1	l = 2	l = 3	# of electrons
n = 1	1s2				2
n = 2	2s2	2p6			8
n = 3	3s2	3p6	3d10		18
n = 4	4s2	4p6	4d10	4f14	32

Nuclear & Particle Physics - Helium

The energy required to remove both electrons from the Helium atom in its ground state is 79.0 eV. How much energy is required to ionize Helium (i.e. to remove one electron)?

A. 24.6 eV
B. 39.5 eV
C. 51.8 eV
D. 54.4 eV
E. 65.4 eV

(GR0177 #18)

Solution:

Helium: 2 Protons, 2 Neutrons, 2 Electrons.

The energy required to remove one electron from He in its ground state, leaving behind He⁺ (a Hydrogen like atom)

E_n = 13.6 Z²/n² eV

Z = Helium atomic number = 2
n = 1 (ground state)

E₁ = 13.6(4) eV = 54.4 eV

The energy required to remove both electrons from He in its ground state leaving behind He⁺⁺ ion = 79 eV.

Thus, the energy required to remove one electron: 79 − 54.4 = 24.6 eV

Answer: A

Nuclear & Particle Physics - The Sun’s Energy

The primary source of the sun’s energy is a series of thermonuclear reactions in which the energy produces is c² times the mass difference between

A. two hydrogen atoms and one helium atom
B. four hydrogen atoms and one helium atom
C. six hydrogen atoms and two helium atoms
D. three helium atoms and one carbon atom
E. two hydrogen atoms plus two helium atoms and one carbon atom

(GR0177 #19)

Solution:

In the Sun’s core, the protons of the hydrogen atoms (which largely make up the Sun) collide into each other and stick together or “fuse” to create helium nuclei.

4H → He + energy

It takes 4H nuclei to create a He nucleus due to the conservation of atomic mass.

Hydrogen atomic mass = 1 proton = 1.
Helium atomic mass = 2 protons + 2 neutrons = 4.

Answer: B

Nuclear & Particle Physics - X-rays

In the production of X-rays, the term “bremsstrahlung” refers to which of the following?

1. The cut-off wavelength, λ_min, of the X-ray tube
2. The discrete X-ray lines emitted when an electron in an outer orbit fills a vacancy in an inner orbit of the atoms in the target metal of the X-ray tube
3. The discrete X-ray lines absorbed when an electron in an inner the X-ray tube
4. The smooth, continuous X-ray spectra produced by high-energy blackbody radiation from the X-ray tube
5. The smooth, continuous X-ray spectra produced by rapidly decelerating electron in the target metal of the X-ray tube

(GR0177 #20)

Solution:

Bremsstrahlung is German for ‘braking radiation’.

This is the radiation that is released as the electron slows down or is “braked” and results in a continuous spectra.

Answer: E