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 - Torque

A solid cone hangs from a frictionless pivot at the origin O as shown above. If î, ĵ and k̂ are unit vectors, and a, b, and c are positive constants, which of the following forces F applied to the rim of the cone at point P results in a torque τ on the cone with a negative component τ_z?

A. F = ak̂, P is (0, b, −c)
B. F = −ak̂, P is (0, −b, −c)
C. F = aĵ, P is (–b, 0, −c)
D. F = aĵ, P is (b, 0, −c)
E. F = −ak̂, P is (−b, 0, −c)

(GR9277 #08)

Solution:

Torque:

Result desired: negative component of τ_z so we are looking for minus k̂ component.

For k̂ component: r_xF_y − r_yF_x

Thus, F_x and F_y cannot be zero

(A) FALSE
 F = ak̂ → F_x = 0 and F_y = 0

(B) and (E) FALSE
F = −ak̂ → F_x = 0 and F_y = 0

(C) TRUE
F = aĵ → F_x = 0, F_y = a,  r_x = −b, r_y = 0,
P is (–b, 0, −c) → r_x = −b, r_y = 0
r_xF_y − r_yF_x = −ab − 0 =  −abk̂ (negative k̂ component)

(D) FALSE
F = aĵ → F_x = 0, F_y = a,
P is (b, 0, −c) → r_x = b, r_y = 0,
→ r_xF_y − r_yF_x = ab − 0 =  abk̂ (positive k̂ component)

Answer: C

Lab Methods - Standard Deviation

The magnitude of the force F on an object can be determined by measuring both the mass m of an object and the magnitude of its acceleration a, where F = ma. Assume that these measurements are uncorrelated and normally distributed. if the standard deviations of the measurements of the mass and acceleration are σ_m and σ_a respectively, then σ_F/F is

A.

B.

C.

D.

E.

(GR9277 #48)

Solution:

Multiplying two quantities with uncertainty:

u = x ⋅ y



For F = ma:



Answer: C

Classical Mechanics - 1D Vertical Motion

A rock is thrown vertically upward with initial speed v₀. Assume a friction force proportional to –v, where v is the velocity of the rock, and neglect the buoyant force exerted by air. Which of the following is correct?

1. The acceleration of the rock is always equal to g.
2. The acceleration of the rock is equal to g only at the top of the flight.
3. The acceleration of the rock is always less than g.
4. The speed of the rock upon return to its starting point is v₀.
5. The rock can attain a terminal speed greater than v₀ before it returns to its starting point.

(GR8677 #01)

Solution:

There is a friction force:

• acceleration is not constant → (A) and (C) FALSE
• energy is not conserved so it's initial and final speed is not the same → (D) FALSE
• frictional force slows down the object, so its speed at time t has to be less than its initial speed  → (E) FALSE

Answer: B


Math analysis:

Friction force: F_f  = − kv
The equation of motion with friction force: ma = − mg − kv.

a = − g − (kv/m)
(A) FALSE

At the top of the flight, v = 0
a = − g − (k∙0/m) = − g
(B) TRUE

Moving up: v positive
a = − g − (kv/m) = − g − c
a − g
Moving Down: v negative
a = − g − [k(−v)/m] = − g + c
a − g
(C) FALSE

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

Classical Mechanics - Kinematics

A 5 kilogram stone is dropped on a nail and drives the nail 0.025 meter into a piece of wood. If stone is moving at 10 meters per second when it hits the nail, the average force exerted on the nail by the stone while the nail is going into the wood is most nearly 

A. 10 N
B. 100 N
C. 1000 N
D. 10,000 N
E. 100,000 N

(GR8677 #08)

Solution:

Force: F = ma

The acceleration when the stone hits the nail: 2ay = v² + v₀²

v₀ = 0 → 2ay = v²

a = v²/2y = 10²/2(0.025) = 2 × 10³ m/s²

F = 5 × 2 × 10³ = 10,000 N

Answer: D

Electromagnetism - Coulomb's Law

The exponent in Coulomb’s inverse square law has been found to differ from two by less than one part in a billion by measuring which of the following?

A. The charge on an oil drop in the Millikan experiment
B. The deflection of an electron beam in an electric field
C. The neutrality of charge of an atom
D. The electric force between two charged objects
E. The electric field inside a charged conducting shell

(GR8677 #86)

Solution:

The Coulomb force can be expressed by .

The smallest value for n = (2.7 ± 3.1) × 10⁻¹⁶ has been obtained by Williams, Faller, and Hill experiment (Phys. Rev. Lett. 26, 721, 1971).

The experiment was done by measuring the electric field inside a charged conducting shell.

See: Improved result for the accuracy of Coulombs law: A Review of the Williams, Faller, and Hill experiment.

Answer: E

Classical Mechanics - Impulse

The figure shows a plot of the time dependent force F_x(t) acting on a particle in motion along the x-axis. What is the total impulse delivered to the particle?

A. 0
B. 1 kg m/s
C. 2 kg m/s
D. 3 kg m/s
E. 4 kg m/s

(GR0177 #54)

Solution:

Impulse: I = ∆p = F ∆t
Area =  ½(2)(2) = 2 kg m/s

Answer: C

Fluid - Buoyancy

A balloon is to be filled with helium and used to suspend a mass of 300 kilograms in air. If the mass of the balloon is neglected, which of the following gives the approximate volume of helium required? (The density of air is 1.29 kilograms per cubic meter and the density of helium is 0.18 kilogram per cubic meter.)

A. 50 m³
B. 95 m³
C. 135 m³
D. 270 m³
E. 540 m³

(GR0177 #56)

Solution:

ΣF = 0
ρ_agV =  ρ_hgV + m₀g

(ρ_a -  ρ_h)V = m₀
V = m₀ / (ρ_a - ρ_h) = 300 / (1.29 - 0.18) = 270 m³

Answer: D

Fluid - Force

A stream of water of density ρ, cross-sectional area A and speed v strikes a wall that is perpendicular to the direction of the stream, as shown in the figure. The water then flows sideways across the wall. The force exerted by the stream on the wall is

A. ρv²A
B. ρvA/2
C. ρghA
D. v²A/ρ
E. v²A/2ρ

(GR0177 #57)

Solution:

Dimensional analysis:
Force has unit Newton = kg m/s²
The only choice has the right unit of force is choice A = (kg/m³)(m/s)²(m²) = kg m/s²

Answer: A

Electromagnetism - Lorentz Force

A proton moves in the +z direction after being accelerated from rest through a potential difference V. The proton then passes through a region with a uniform electric field E in the +x direction and a uniform magnetic field B in the +y direction, but the proton’s trajectory is not affected. If the experiment were repeated using a potential difference of 2V, the proton would then be

A. Deflected in the +x-direction
B. Deflected in the -x-direction
C. Deflected in the +y-direction
D. Deflected in the -y-direction
E. Undeflected

(GR0177 #58)

Solution:

Lorentz Force: 

Since it is not deflected, 

Potential energy transforms into kinetic energy, qV = ½ mv²

Thus, velocity is  

For 2V, velocity is  

  

Therefore,

Particle is deflected in −x direction

Answer: B

Electromagnetism - Cyclotron

A nonrelativistic particle with a charge twice that of an electron moves through a uniform magnetic field. The field has a strength of π/4 tesla and is perpendicular to the velocity of the particle. What is the particle’s mass if it has a cyclotron frequency of 1,600 hertz?

A. 2.5 × 10^-23 kg
B. 1.2 × 10^-22 kg
C. 3.3 × 10^-22 kg
D. 5.0 × 10^-21 kg
E. 7.5 × 10^-21 kg

(GR0177 #62)

Solution:





Angular velocity: 





Answer: A

Classical Mechanics - Harmonic Oscillator

Two identical blocks are connected by a spring. The combination is suspended, at rest, from a string attached to the ceiling, as shown in the figure. The string breaks suddenly. Immediately after the string breaks, what is the downward acceleration of the upper block?

A. 0
B. g/2
C. g
D. √2 g
E. 2g

(GR0177 #72)

Solution:

In equilibrium (before the string is cut):
Block 1:

                                    …(Eq.1)

Block 2:
                  … (Eq.2)

After the string is cut:
, and
(minus sign    is because the direction is in downward or parallel with  )

(Eq.1) becomes:

                                                                   ...(Eq.3)

Submit (Eq.2) to (Eq.3):   

Since  

Answer: E

Classical Mechanics - Potential Energy

A Particle of mass m moves in a one-dimensional potential V(x) = −ax² + bx⁴, where a and b are positive constants. The angular frequency of small oscillations about the minima of the potential is equal to

A. π(a/2b)^1/2
B. π(a/m)^1/2
C. (a/mb)^1/2
D. 2(a/m)^1/2
E. (a/2m)^1/2

(GR9677 #92)

Solution:

The minima of the potential (most probable value x or the equilibrium position of the mass):





Angular Frequency:

Conservative force:




Thus, the angular frequency about the minima of the potential:



Answer: D

Condensed Matter - Effective mass

Lattice forces affect the motion of electrons in a metallic crystal, so that the relationship between the energy E and wave number k is not the classical equation E = ħ²k²/2m, where m is the electron mass. Instead, it is possible to use an effective mass m* given by which of the following?

A.

B.

C.

D.

E.

(GR9277 #97) 

Solution: 

F = m*a

F = dp/dt = ħ dk/dt     (momentum, p = ħk)

a = dv_g/dt

Group velocity, v_g = dω/dk

Energy of electron, E = hf = ħω 
ω = E/ħ





F = m*a