Classical Mechanics - Rotational Motion


One ice skater of mass m moves with speed 2v, while another of the same mass m moves with speed v toward the left, as shown in Figure I. Their paths are separated by a distance b. At t = 0, when they are both at x = 0, they grasp a pole of length b and negligible mass. For t > 0, consider the system as a rigid body of two masses m separated by distance b, as shown in Figure II. Which of the following is the correct formula for the motion after t = 0 of the skater initially at y = b/2?

A. x = 2vt y = b/2
B. x = vt + 0.5b sin (3vt/b)    y = 0.5b cos (3vt/b)
C. x = 0.5vt + 0.5b sin (3vt/b)     y = 0.5b cos (3vt/b)
D.  x = vt + 0.5b sin (6vt/b y = 0.5b cos (6vt/b)
E.     x = 0.5vt + 0.5b sin (6vt/b) y = 0.5b cos (6vt/b)
(GR9277 #78)
Solution:
x = xtranslational + xrotational

To find  xtrans:
Conservation of linear momentum: mv₁ + mv₂ = (m₁ + m₂)v'
m(2v) + m(−v) = 2mv'
→ v' = 0.5v
xtransv't = 0.5vt
→ A, B and E are FALSE.

Check answer C. x = 0.5vt + 0.5b sin (3vt/b)
xrot = a sin ωt = 0.5b sin (3vt/b) ω = 3v/b

To check if ω = 3v/b:

At t = 0,
L = L₁ + L = rmv + rmv₂ = 1/2bm2v + 1/2bmv = 3/2bmv

At t > 0,  
L' =
Moment inertia of the system: I = ½mb²
→ L' = ½mb²ω

Conservation of angular momentum:  L = L' 
3/2bmv = 1/2mb²ω ω = 3v/b

Answer: C

Optics - Dispersion Curve


The dispersion curve shown above relates the angular frequency to the wave number k. For waves with wave numbers lying in the range k₁ < k< k₂ which of the following is true of the phase velocity and the group velocity?

A. They are in opposite directions.
B. They are in the same direction and the phase velocity is larger.
C. They are in the same direction and the group velocity is larger.
D. The phase velocity is infinite and the group velocity is finite.
E. They are the same in direction and magnitude.
(GR9277 #79)
Solution:

Phase velocity: vp = ω/k
Group velocity: vg = /dk

In the range k₁ < k< k₂:
vp → positive quantity
vg → negative quantity (negative slope)
vp and vg are in opposite directions.

Answer: A

Nuclear & Particle Physics - X-rays

A beam of electrons is accelerated through a potential difference of 25 kV in an X-ray tube. The continuous X-ray spectrum emitted by the target of the tube will have a short wavelength limit of most nearly

A. 0.1 Å
B. 0.5 Å
C. 2 Å
D. 25 Å
E. 50 Å
(GR9277 #80)
Solution:

Energy: E = pc
Wavelength: λ = h/p = hc/E

E = 25 kV
h = 4.14 × 10−15 eV second
c = 3 × 108 m/s
hc = 1.24 × 10−6 eVm

λ = 1.24 × 10−6/25 × 10 = ½ × 10−10 m = 0.5 Å

Answer: B

Electromagnetism - RLC Circuit


In the RLC circuit shown, the applied voltage is ε(t) = εm cos ωt For a constant εm, at what angular frequency ω does the current have its maximum steady-state amplitude after the transients have died out?

A.1/RC
B.2L/R
C. 1/√(LC)
D. √((1/LC) − (R/2L)²)
E.  √((1/RC)² − (L/R)²)  
(GR9277 #81)
Solution:

The current is maximized when Inductive and Capacitive Reactance are equal in magnitude
→  XL = XC
→ ωL = 1/ωC
ω = 1/√(LC)

Answer: C

Classical Mechanics - Torque


A thin plate of mass M, length L, and width 2d is mounted vertically on a frictionless axle along the z-axis. Initially the object is at rest. It is then tapped with a hammer to provide a torque τ, which produces an angular impulse H about the z-axis of magnitude H = ∫ τ dt. What is the angular speed ω of the plate about the z-axis after the tap? 

A. H/2Md²
B. H/Md²
C. 2H/Md²
D. 3H/Md²
E. 4H/Md²
(GR9277 #82)
Solution:

Torque: τ = H = ∫ τ dt = t
Angular acceleration: α = ω/ω = αt H = Iω
Moment inertia for the plate about the z-axis: I =  ⅓ Md²
ω = H/I = 3H/Md²

Answer: D

Electromagnetism - Coulomb's Law


Two pith balls of equal mass M and equal charge q are suspended from the same point on long massless threads of length L as shown in the figure above. If k is the Coulomb's law constant, then for small values of θ, the distance d between the charged pith balls at equilibrium is

A. (2kq²L/Mg)1/3
B. (kq²L/Mg)1/3
C. (2kq²L/Mg)1/2 
D. (kq²L/Mg)1/2
E. L/4
(GR9277 #83)
Solution:

In Equilibrium:  
T cos θ = Mg  
T sin θ = Fcoulomb = kq²/d²

For θ ≪, cos (θ → 0) → 1 → T = Mg  
sin θ½d/L

Mgd/2L = kq²/d²
Mgd³ = 2kq²L
d = (2kq²L/Mg)1/3

Answer: A

Special Relativity - Relativistic Energy

A free electron (rest mass me = 0.5 MeV/c²) has a total energy of 1.5 MeV. Its momentum p in units of MeV/c is about

A. 0.86
B. 1.0
C. 1.4
D. 1.5
E. 2.0
(GR9277 #85)
Solution:

E² = p²c² + m₀²c
1.5² = p²c² + 0.5²
p²c² = 1.5² − 0.5² = 2
p = √2/c = 1.41/c

Answer: C

Classical Mechanics - Circular Motion

A particle of mass M moves in a circular orbit of radius r around a fixed point under the influence of an attractive force F = Kr³, where K is a constant. If the potential energy of the particle is zero at an infinite distance from the force center, the total energy of the particle in the circular orbit is

A. − Kr² 
B. − K2r²
C. 0
D. K2r²
E. Kr²
(GR9277 #87)
Solution:

Attractive force = Centripetal Force
→  Kr³ = mv²r
mv² = Kr²

Kinetic energy:  T = ½mv² = K2r²
Potential energy: V(r) = − ∫ F dr = −K1r³ dr = K2r²
Attractive force → negative potential energy → V(r) = − K2r²
Total energy:  T + V = K2r²  −  K2r² = 0

Answer: C

Electromagnetism - Capacitor

A parallel-plate capacitor is connected to a battery. V₀ is the potential difference between the plates, Q₀ is the charge on the positive plate, E₀ the magnitude of the electric field, and D₀ the magnitude of the displacement vector. The original vacuum between the plates is filled with a dielectric and then the battery is disconnected. If the corresponding electrical parameters for the final state of the capacitor are denoted by a subscript f, which of the following is true?

A. Vf >V₀ 
B. Vf <V
C. Q = Q
D. Ef > E
E. Df > D
(GR9277 #88)
Solution:

A. FALSE
The battery is disconnected after the plates is filled with a dielectric
V= V= constant

B. FALSE
See A.

C. FALSE
Charge: Q = CV
V= V₀ and when dielectric is inserted: Cf   = κC
Q = Cf V = κCV₀ = κQ

D. FALSE
Potential: V = Ed
Since V= V₀ and d is not changed
E= E

E. TRUE
Displacement vector in empty space: D₀ = εE
with dielectric constant: Df   = κεE
Df > D

Answer: E

Quantum Mechanics - Harmonic Oscillator

The energy levels for the one-dimensional harmonic oscillator are (n + ½), n = 0,1,2,⋯ How will the energy levels for the potential shown in the graph above differ from those for the harmonic oscillator?

A. The term ¹⁄₂ will be changed to ³⁄₂
B. The energy of each level will be doubled.
C. The energy of each level will be halved.
D. Only those for even values of n will be present.
E. Only those for odd values of n will be present.
(GR9277 #89)
Solution:

See GR9677 #98

Answer: E