Electromagnetism - Oscillation

Question 3-4: refer to a thin, nonconducting ring of radius R, as shown below, which has a charge Q uniformly spread out on it.

A small particle of mass m and charge –q is placed at point P and released. If R ≫ x, the particle will undergo oscillations along the axis of symmetry with an angular frequency that is equal to:

(GR9677 #04)

Solution:

F_electric = kqQ/r²
F_centripetal = mv²/r = mω²r

F_e = F_c
kqQ/r² = mω²r
ω² = kqQ/mr³

with
k = 1/4πɛ₀
r²  = R² + x²
R ≫ x
r² ∼ R²  → r³ ∼ R³

ω = √(^qQ/_4πɛ0mR³)

Answer: A

Notes: see problem GR9277 #65

Classical Mechanics - Earth’s gravitational force

Questions 4-5

The magnitude of the Earth’s gravitational force on a point mass is F(r), where r is the distance from the Earth’s center to the point mass. Assume the Earth is a homogeneous sphere of radius R.

What is ? 

A. 32
B. 8
C. 4
D. 2
E. 1

(GR9277 #04)

Solution:

Gravity force: F = GMm/R² → F ∝ 1/R²

F(R) ∝ 1/R²
F(2R) ∝ 1/4R²

F(R) / F(2R) = 4R²/ R² = 4

Answer: C

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 - 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