Physics Problems & Solutions: #05

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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 - 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 $\frac{F(R)}{F\left (\frac{R}{2} \right )}$ ?

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

$\frac{F(R)}{F\left (\frac{R}{2} \right )} = \frac{R}{R/2} =2$

Answer: C

Classical Mechanics - Pendulum

Two small spheres of putty, A and B of mass M and 3M, respectively, hang from the ceiling on strings of equal length l. Sphere A is drawn aside so that it is raised to a height h₀ as shown above and then released. Sphere A collides with sphere B; they stick together and swing to a maximum height h equal to

A. (1/16) h₀
B. (1/8) h₀
C. (1/4) h₀
D. (1/3) h₀
E. (1/2) h₀

(GR8677 #5)

Solution:

Conservation of energy of A before and when it hits B:
$\begin{aligned} Mgh_0&=\frac{1}{2} Mv_A^2\\v_A&=\sqrt{2gh_0}\end{aligned}$

Conservation of momentum when and after collision:
$\begin{aligned} m_A v_A+m_B v_B&=(m_A+m_B)v_{AB}\\M\sqrt{2gh_0}+0&=4Mv_{AB}\\v_{AB}&=\frac{\sqrt{2gh_0}}{4}\end{aligned}$

Conservation of energy of A and B at h = 0 and h_max:
$\begin{aligned} \frac{1}{2} m_{AB} v_{AB}^2&=m_{AB} gh_{\max}\\\frac{1}{2} 4M\left (\frac{\sqrt{2gh_0}}{4} \right )^2&=4Mgh_{\max}\\h_{\max}&=\frac{1}{16} h_0\end{aligned}$

Answer: A

Thermal Physics - Equipartition Law

A three-dimensional harmonic oscillator is in thermal equilibrium with a temperature reservoir at temperature T. The average total energy of oscillator is

A. ½kT
B. kT
C. ³⁄₂kT
D. 3kT
E. 6kT

(GR0177 #05)

Solution:

Equipartition of Energy: E = ½ fkT

f = Degree of Freedom (DoF)

3D harmonic oscillator has 6 DoF = 3 components of momentum (kinetic energy) and 3 components of position (potential energy)
→ E = ⁶⁄₂ kT = 3kT

Answer: D