Classical Mechanics - Linear Momentum

As shown in the picture, a ball of mass m suspended on the end of a wire, is released from height h and collides elastically, when it is at its lowest point, with a block of mass 2m at rest on a frictionless surface. After the collision, the ball rises to a final height equal to

A. 1/9 h
B. 1/8 h
C. 1/3 h
D. 1/2 h
E. 2/3 h

(GR9677 #07)

Solution:

Conservation of momentum of the system:

m_av_a + m_bv_b = m_av_a' + m_bv_b' 

Given:
m_a = m
m_b = 2m
v_b = 0

mv_a + 0 = mv_a' + 2mv_b'  
v_a  = v_a' + 2v_b'    (Eq.1)

Conservation of kinetic energy of the system:

½ m_av_a² + ½ m_bv_b² = ½ m_av_a'² + ½ m_bv_b'²
mv_a² + 0 = mv_a'² + 2mv_b'²
v_a² = v_a'² + 2v_b'²  (Eq.2)

(Eq.1) → (Eq.2)
(v_a' + 2v_b')² = v_a'² + 2v_b'²
v_a'² + 4v_a'v_b' + 4v_b'² = v_a'² + 2v_b'²
4v_a'v_b'  = 2v_b'² − 4v_b'²
2v_a' = − v_b'  
v_b'  = −2v_a'  (Eq.3)

(Eq.3) →  (Eq.1)
v_a  = v_a' + 2v_b' 
v_a  = v_a' + 2(−2v_a')
v_a  = − 3v_a'  (Eq. 4)

For the pendulum, conservation of energy:

at the moment of the collision
U = T → m_agh = ½ m_av_a² → v_a = (2gh)^½

after the collision
U'  = T' → m_agh' = ½ m_av_a'² → v_a' = (2gh')^½

Thus, (Eq. 4):
v_a  = − 3v_a'  
(2gh)^½  = − 3(2gh')^½ 
[(2gh)^½]² = [− 3(2gh')^½]²

h  = 9h'
h' = ¹⁄₉h 

Answer: A

Classical Mechanics - Pendulum

A cylindrical tube of mass M can slide on a horizontal wire. Two identical pendulum, each of mass m and length l, hang from the ends of the tube, as shown above. For small oscillations of the pendulums in the plane of the paper, the eigenfrequencies of the normal modes of oscillation of the system are 0,,   and …

A.

B.

C.

D.

E.

(GR9277 #07)

Solution:

2 modes in which the system can oscillate:

1. The 2 pendulum oscillate out of phase so there is torsional effect on the tube M → there is M in the equation.

2. In phase → no torsional effect on M → No M in the equation.

Answer: A

Classical Mechanics - Static Equilibrium

A 2-kilogram box hangs by a massless rope from a ceiling. A force slowly pulls the box horizontally to the side until the horizontal force is 10 Newtons. The box is then in equilibrium as shown above. The angle that the rope makes with the vertical is closest to

A. arctan 0.5
B. arcsin 0.5
C. arctan 2.0
D. arcsin 2.0
E. 45^o

(GR8677 #07)

Solution:

F_x = F − T sin θ = 0
T sin θ = F 

F_y = T cos θ − mg = 0
T cos θ = mg

T sin θ / T cos θ = tan θ = F/mg =  10/(2 × 10) = 0.5
θ = arctan 0.5

Answer: A

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