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 = xtrans + xrot

To find  xtrans:
Conservation of linear momentum: mv₁ + mv₂ = (m₁ + m₂)v'
m2v + m(−v) = 2mv'
v' = 0.5v
xtrans = 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

Electromagnetism - EM Radiation

An electron oscillates back and forth along the + and − x axes, consequently emitting electromagnetic radiation. Which of the following statements concerning the radiation is NOT true?
  1. The total rate of radiation of energy into all directions is proportional to the square of the electron's acceleration.
  2. The total rate of radiation of energy into all directions is proportional to the square of the electron's charge.
  3. Far from the electron, the rate at which radiated energy crosses a perpendicular unit area decreases as the inverse square of the distance from the electron
  4. Far from the electron, the rate at which radiated energy crosses a perpendicular unit area is a maximum when the unit area is located on the + or - x-axes
  5. Far from the electron, the radiated energy is carried equally by the transverse electric and the transverse magnetic fields.
(GR9277 #84)
Solution:



Answer:

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

Electromagnetism - Capacitor


The circuit shown above is used to measure the size of the capacitance C. The y-coordinate of the spot on the oscilloscope screen is proportional to the potential difference across R, and the x-coordinate of the spot is swept at a constant speed s. The switch is closed and then opened. One can then calculate C from the shape and the size of the curve on the screen plus a knowledge of which of the following?

A. V₀ and R
B. s and R
C. s and V
D. R and R'
E. The sensitivity of the oscilloscope
(GR9277 #86)
Solution:



Answer:

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