Classical Mechanics - Circular Motion

A car travels with constant speed on a circular road on level ground. In the diagram above, F_air is the force of air resistance on the car. Which of the other force shown best represents the horizontal force of the road on the car's tires?

A. F_A 
B. F_B
C. F_C
D. F_D
E. F_E

(GR9677 #05)

Solution:

F_A = Centripetal force
F_C = Frictional force of the road, equals to force exerted by tires exert in the backward direction so that the car moves in the forward direction (Newton's third law Action-Reaction).

The horizontal force on the car’s tires, F_A + F_C = F_B

Answer: B

Classical Mechanics - Friction Force

A block of mass m sliding down an incline at constant speed is initially at height h above the ground as shown in the figure. The coefficient of kinetic friction between the mass and the incline is µ. If the mass continues to side down the incline at a constant speed, how much energy is dissipated by friction by the time the mass reached the bottom of the incline?

A. mgh/µ
B. mgh
C. µmgh/sin θ 
D. mgh sin θ 
E. 0

(GR9677 #06)

Solution:

mg sin θ − F_r  = ma
At a constant speed, a = 0
F_r  = mg sin θ 

Energy dissipated = Work done by the frictional force
W = F_r ⋅ s

s = length of the inclined surface = h/sin θ

W = mg sin θ (h/sin θ) = mgh

Answer: B

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 - Harmonic Oscillator

A particle of mass m undergoes harmonic oscillation with period T₀. A force f proportional to the speed v of the particle, f = −bv, is introduced. If the particle continues to oscillate, the period with f acting is

A. Larger than T₀
B. Smaller than T₀
C. Independent of b
D. Dependent linearly on b
E. Constantly changing

(GR9677 #08)

Solution:

 f = −bv → minus sign means f is restoring force (damped oscillation).
The oscillation is getting slower (larger period) before it finally comes to stop.

Answer: A

Classical Mechanics - Linear Momentum

A helium atom, mass 4u travels with non relativistic speed v normal to the surface of a certain material, makes an elastic collision with an (essentially free) surface atom, and leaves in the opposite direction with speed 0.6v. The atom on the surface must be an atom of

A. Hydrogen, mass 1u
B. Helium, mass 4u
C. Carbon, mass 12u
D. Oxygen, mass 16u
E. Silicon, mass 28u

(GR9677 #20)

Solution:

m_a = 4u
v_a  = v
v_b  = 0
v_a' = − 0.6v

Conservation of momentum of the system:

m_av_a + m_bv_b = m_av_a' + m_bv_b' 
4uv = 4u(− 0.6v) + m_bv_b' 
4uv = − 2.4uv  + m_bv_b' 
m_bv_b' = 6.4uv (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
4uv² = 4u(− 0.6v)² + m_bv_b'²
4uv² = 4u(0.36v²) + m_bv_b'²
m_bv_b'² =  4uv²  − 1.44uv²
m_bv_b'² =  2.56uv² (Eq.2)

(Eq.1) → (Eq.2)
6.4 (v_b') = 2.56v
v_b' =  (2.56/6.4)v = 0.4v  (Eq.3)

(Eq.3) → (Eq.1)
m_b = 6.4uv / 0.4v = 16u

Answer: D

Classical Mechanics - Moment Inertia

The period of physical pendulum is 2π√(I/mgd), where I is the moment of inertia about the pivot point and d is the distance from the pivot to the center of mass. A circular hoop hangs from a nail on a barn wall. The mass of the hoop is 3 kilograms and its radius is 20 centimeters. If it is displaced slightly by a passing breeze, what is the period of the resulting oscillations?

A. 0.63 s
B. 1.0 s
C. 1.3 s
D. 1.8 s
E. 2.1 s

(GR9677 #21)

Solution:

T = 2π√(I/mgd)
m = 3 kg
r = 20 cm = 0.2 m
In this case d = r

To find total I:
Parallel axis theorem: I = ml² + I_CM
I_CM  = I_loop = mr² 

In this case l = r
→ I = mr² + mr² = 2mr²

T = 2π√(2mr² / mgr) 
= 2π√(2r/g) 
= 2π√[2(0.2)/(10)] 
= 2π(0.2)
= 1.25 s

Answer: C

Classical Mechanics - Curved Trajectory

The curvature of Mars is such that its surface drops a vertical distance of 2.0 meters for every 3600 meters tangent to the surface. In addition, the gravitational acceleration near its surface is 0.4 times that near the surface of Earth. What is the speed a golf ball would need to orbit Mars near the surface, ignoring the effects of air resistance?

A. 0.9 km/s
B. 1.8 km/s
C. 3.6 km/s
D. 4.5 km/s
E. 5.4 km/s

(GR9677 #22)

Solution:

y = ½ g_marst²

v_x = x/t 

Given:
y = 2
g_mars = 0.4 g 
x = 3600

2 = ½(0.4)(10)(t)²
t = 1 s
v_x = 3600/1 = 3600 m/s = 3.6 km/s

Answer: C

Classical Mechanics - Conservative Force

Suppose that the gravitational force law between two massive objects were

F₁₂ = r̂₁₂ Gm₁m₂/r₁₂^(2+ɛ) 

where ɛ is a small positive number. Which of the following statements would be FALSE?

1. The total mechanical energy of the planet-Sun system would be conserved.
2. The angular momentum of a single planet moving about the Sun would be conserved.
3. The periods of planets in circular orbits would be proportional to the (3+ɛ)/2 power of their respective orbital radii.
4. A single planet could move in a stationary non circular elliptical orbit about the Sun.
5. A single planet could move in a stationary circular orbit about the Sun.

(GR9677 #23)

Solution:

(A) TRUE.
Gravitational force is a conservative force.
In conservative field, the total mechanical energy is conserved.

(B) TRUE
In conservative field, angular momentum, L is conserved.

(C) TRUE
F_g = F_c
GMm/r^(2+ɛ) = mrω²
GMm/r^(2+ɛ) = mr(2π/T)²
GM/r^(3+ɛ) = 4π²/T²
T² = 4π²r^(3+ɛ)/GM
T ∝ r^(3+ɛ)/2 

(D) FALSE
Central force = centripetal force (F_g = F_c) produces circular orbit.
Non central forces do not produce circular orbit.

(E) TRUE
See (D)

Answer: D

Notes:

Central force:

1. It is a force whose magnitude depends only on the distance between the object and the origin.
2. It is a conservative field, can be expressed as F = − ∇V (the negative gradient of a potential energy).
3. Gravitational force, Coulomb force, and Elastic Force (Harmonic Oscillator) are examples of central (conservative) forces.
4. In conservative field, the net work done by the force is zero, W = ∮_c F ∙ dr = 0 → the total mechanical energy is conserved.
5. Conservative force is irrotional (torque = 0), since curl ∇V or ∇ × ∇V = 0.
6. Torque, τ = dL/dT = 0 → angular momentum, L is conserved
7. Central force = centripetal force (F_g = F_c) produces circular orbit.

Classical Mechanics - Fluid

An open-ended U-tube of uniform cross-sectional area contains water (density 1.0 gram/cm³) standing initially 20 cm from the bottom in each arm. An immiscible liquid of density 4.0 gr/cm³ is added to one arm until a layer 5 cm high forms. What is the ratio h₂/h₁ of the heights of the liquid in the two arms?

A. 3/1
B. 5/2
C. 2/1
D. 3/2
E. 1/1

(GR9677 #30)

Solution:

Given:
ρ_water = 1 gr/cm³
ρ_liquid = 4 gr/cm³
h_a = 5 cm

P₁ = P₂
ρ_lgh_a = ρ_wgh_b 
h_b = ρ_lh_a / ρ_w = (4)(5) / (1) = 20

h₁ = h + h_a  
h₂ = h + h_b 
-------------- −
h₁ − h₂ = h_a − h_b  = 5 − 20 = − 15 (Eq.1)

Initially, h₁ + h₂ = 20 + 20 = 40
Finally, h₁ + h₂ = 40 + 5 = 45 (Eq.2)

(Eq.1) → h₁ − h₂ = − 15
(Eq.2) → h₁ + h₂ = 45
------------------------------ +
2h₁ = − 15 + 45 = 30
h₁ = 15

h₂ = 45 − 15 = 30

h₂/h₁ = 30/15 = 2

Answer: C

Classical Mechanics - Fluid

A sphere of mass m is released from rest in a stationary viscous medium. In addition to the gravitational force of magnitude mg, the sphere experiences a retarding force of magnitude bv, where v is the speed of the sphere and b is a constant. Assume that the buoyant force is negligible. Which of the following statements about the sphere is correct?

1. Its kinetic energy decreases due to the retarding force.
2. Its kinetic energy increases to a maximum, then decreases to zero due to the retarding force.
3. Its speed increases to a maximum, then decreases back to a final terminal speed.
4. Its speed increases monotonically, approaching a terminal speed that depends on b but not on m.
5. Its speed increases monotonically, approaching a terminal speed that depends on b and m.

(GR9677 #31)

Solution:

ma = mg − bv

Terminal speed, a = 0

mg = bv
v = mg/b → (E) TRUE

(A) FALSE.
m is released from rest v_i = 0 → KE_i = 0

(B) FALSE.
Terminal speed released from rest v_f  = mg/b ≠ 0 → KE_f  ≠ 0

(C) FALSE.
v_i = 0 to v_f  = mg/b, no v_max

(D) FALSE.
 v_f  = mg/b, depend on both m and b

Answer: E

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 

Classical Mechanics - Acceleration

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x) = βx⁻ⁿ where β and n are constants and x is the position of the particle. What is the acceleration of the particle as a function of x?

A. −nβ²x⁻²ⁿ⁻¹
B. −nβ²x⁻ⁿ⁻¹
C. −nβ²x⁻ⁿ
D. −β²x⁻ⁿ⁺¹
E. −β²x⁻²ⁿ⁺¹

(GR9677 #44)

Solution:

a = dv/dt = (dv/dx) (dx/dt) =  (dv/dx) v

Given: v(x) = βx⁻ⁿ
dv/dx =  −nβx⁻ⁿ⁻¹

Thus, a = (−nβx⁻ⁿ⁻¹) βx⁻ⁿ = −nβ²x⁻²ⁿ⁻¹

Answer: A

Classical Mechanics - Conservation of Momentum

A man of mass m on an initially stationary boat gets off the boat by leaping to the left in an exactly horizontal direction. Immediately after the leap, the boat of mass M is observed to be moving to the right at speed v. How much work did the man do during the leap (both on his own body and on the boat)?

A. ½ Mv²
B. ½ mv²
C. ½ (M + m)v²
D. ½ (M + M²/m)v²
E. ½ [Mm/(M + m)]v²

(GR9677 #65)

Solution:

Conservation of momentum:



The man does work on both himself and the boat:



Answer: D

Classical Mechanics - Orbital Path

When it is about the same distance from the Sun as is Jupiter, a spacecraft on a mission to the outer planets has a speed that is 1.5 times the speed of Jupiter in its orbit. Which of the following describes the orbit of the spacecraft about the Sun?

A. Spiral
B. Circle
C. Ellipse
D. Parabola
E. Hyperbola

(GR9677 #66)

Solution:

The mission is to outer planets, so the path should not be bounded (no longer a close path): A, B, C are FALSE.

If v_escape = √2 v_circular → Parabola
If v_e  √2 v_c → Hyperbola
Since v_e = 1.5  √2 → Hyperbola

Answer: E

Notes:

To find escape velocity:
F_c = F_g 
mv²/ r = GMm / r² 
v_circular = √(GM / r)

KE = PE_gravity 
½ mv² = GMm / r  
v_escape = √(2GM / r) = √2 √(GM / r)
v_escape = √2 v_circular

Classical Mechanics - Lagrangian

A bead is constrained to slide on a frictionless rod that is fixed at an angle θ with a vertical axis and is rotating with angular frequency ω about the axis, as shown above. Taking the distance s along the rod as the variable, the Lagrangian for the bead is equal to

A. ½ mṡ ² − mgs cos θ 
B. ½ mṡ ² + ½ m(ωs)² − mgs 
C. ½ mṡ ² + ½ m(ωs cos θ)² + mgs cos θ
D. ½ m(ṡ sin θ)² − mgs cos θ 

E. ½ mṡ ² + ½ m(ωs sin θ)² − mgs cos θ

(GR9677 #68)

Solution:

Lagrangian: L = T − U

Potential energy: U = mgh = mgs cos θ 
Kinetic Energy:  T_kin  =  ½ mṡ ²
Rotational kinetic energy:  T_rot =  ½ Iω²
with moment inertia: I = mr²  = m(s sin θ)²
→ T_rot = ½ m(ωs sin θ)²

L = T_kin + T_rot − U =  ½ mṡ ² + ½ m(ωs sin θ)² − mgs cos θ

Answer: E

Classical Mechanics - Pendulum

Two pendulums are attached to a massless spring, as shown above. The arms of the pendulums are of identical lengths l, but the pendulum balls have unequal masses m₁ and m₂. The initial distance between the masses is the equilibrium length of the spring, which has spring contant K. What is the highest normal mode frequency of this system?

A.

B.

C.

D.

E.

(GR9677 #84)

Solution:

A. FALSE
ω = √(g/l) is angular frequency for single pendulum

B. and C are FALSE
Both answers do not depend on g/l

D. TRUE
If there is no dependence on K → ω = √(g/l)
And if m₂ → ∞,  m₁ will still oscillate with spring constant K but has no dependence on m_2, as if it were connected to a stationary object.

E. FALSE
If there is no dependence on K → ω = √(2g/l), not angular frequency for single pendulum
And if m₂ → ∞,  ω = √(2g/l) has no dependence on K and m₁

Answer: D 

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 - Gravitation

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. 

Suppose there is a very small shaft in the Earth such that the point mass can be placed at a radius of R/2. What is ?

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

(GR9277 #05)

Solution:

Newton's law of gravity becomes a linear law inside a body → F ∝ r
F(R) ∝ R
F(R/2) ∝ R/2



Answer: C

Classical Mechanics - Friction Force

Two wedges, each of mass m, are placed next to each other on a flat floor. A cube of mass M is balanced on the wedges as shown above. Assume no friction between the cube and the wedges, but a coefficient of static friction between the wedges and the floor. What is the largest M that can be balanced as shown without motion of the wedges?

A.

B.

C.

D.

E. All M will balance

(GR9277 #06)

Solution:

Answer: D

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