Classical Mechanics - Rotational Kinetic Energy

Three equal masses m are rigidly connected to each other by massless rods of length l forming an equilateral triangle. The assembly is to be given an angular velocity ω about an axis perpendicular to the triangle. For fixed ω, the ratio of the kinetic energy of the assembly for an axis through B compared with that for an axis through A is equal to

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

(GR9677 #32)

Solution:

KE = ½Iω²

For fix ω, the ratio KE_B/KE_A = I_B/I_A

I = ∑_i m_ir_i²

From A:
m₁= m₂ = m₃ = m
x₁= x₂ = x₃ = l/√3 (see notes)

I_A = m₁x₁² + m₂x₂² + m₃x₃²
= 3m( l/√3)² = ml²

From B:
I_B = m₁x₁² + m₂x₂² 
= ml² + ml² = 2ml²

KE_A/KE_B = 2ml²/ml² = 2

Answer: B

Notes:

Cos 30^o = ½l / x₁ 
Cos 30^o = ½√3
½l / x₁ = ½√3
x₁ = x₂ = x₃ = l/√3 

Electromagnetism - Resistor

In the circuit shown above, the resistances are given in ohms and the battery is assumed ideal with emf equal to 3.0 volts The resistor that dissipated the most power is

A. R₁
B. R₂
C. R₃
D. R₄
E. R₅

(GR9277 #32)

Solution:

Power, P = VI = I²R → P ∝ R

R₃ and R₄ are in parallel → R₃₄
R₃₄ and R₅  are in series → R₃₄₅
R₂ and R₃₄₅  are in parallel → R₂₃₄₅





R₁ is the highest resistance → dissipated the most power.

Answer: A  

Nuclear & Particle Physics - Photoelectric

Questions 31-33 refer to the apparatus used to study the photoelectric effect (see GR8677 #31).

The photoelectric equation is derived under the assumption that

1. Electrons are restricted to orbits of angular momentum nħ, where n is an integer
2. Electrons are associated with waves of wavelength λ = h/p, where p is momentum
3. Light is emitted only when electrons jump between orbits
4. Light is absorbed in quanta of energy E = hv
5. Light behaves like a wave

(GR8677 #32)

Solution:

According to the classical Maxwell wave theory of light, the average energy carried by an emitted electron should increase with the intensity of the incident light.

However, in photoelectric case, the energies of the emitted electrons are independent of the intensity of the incident radiation.

Einstein resolved this paradox by proposed that the incident light consisted of individual quanta, called photons, that interacted with the electrons in the metal like discrete particles, rather than as continuous waves.

Answer: D

Special Relativity - Relativistic Momentum

If the total energy of a particle of mass m is equal to twice its rest energy, then the magnitude of the particle’s relativistic momentum is

A. mc/2
B. mc/√2
C. mc
D. √3mc
E. 2mc

(GR0177 #32)

Solution:

Rest Energy: E₀ = m₀c²

Relativistic Energy:  E² = p²c² + m₀²c⁴

Given: m₀ = m and E = 2E₀

p²c² + m²c⁴ = 4E₀² = 4m²c⁴
p² = 3m²c²
p = √3mc

Answer: D