Physics Problems & Solutions

Electromagnetism - Capacitor

2010-04-15T07:55:00.000-07:00

Two capacitors of capacitance 1.0 microfarad and 2.0 microfarad are each charged by being connected across a 5.0-volt battery. They are disconnected from the battery and then connected to each other with resistive wires so that plates of opposite charge are connected together. What will be the magnitude of the final voltage across the 2.0-microfarad capacitor?
A. 0V
B. 0.6V
C. 1.7V
D. 3.3V
E. 5.0V

(GR9677 #62)

Solution:
Fig 1: Initial configuration.

The charges stored at each capacitor when they are connected across 5.0-V battery:

Fig 2: At the instant after reconnection (the plates of opposite charge are connected together)

Fig 3: In equilibrium

A short time later after reconnection, the net charge is conserved but redistributed between the two capacitors until the potential across them reaches equilibrium (the same).

Thus, the net charges on the upper and lower plates are:

And,

To find ,take the upper plate:

Answer: C

Electromagnetism - Electric Potential

2010-04-14T16:51:00.000-07:00

A cube has a constant electric potential $V$ on its surface. If there are no charges inside the cube, the potential at the center of the cube is
A. zero
B. $V/8$
C. $V/6$
D. $V/2$
E. $V$

(GR8677 #52)

Solution:
Gauss’ Law: $\oint \bar{E} \cdot \hat{n} dA=\frac{q_{\textup{enc}}}{\varepsilon_0}$
With $q_{\textup{enc}}=0 \rightarrow E=0$

The potential is related to the electric field by $E=\nabla\varphi$

Since $E=0 \rightarrow \varphi$ must be constant, which is given in the problem by $V$

Since the potential function has to remain continuous its value everywhere, inside or at the surface the cube $\rightarrow \varphi$ at the center of the cube is also $V$

Answer: E

Electromagnetism - Gauss' Law

2010-04-14T15:54:00.000-07:00

A sphere of radius $R$ carries charge density proportional to the square of the distance from the center: $\rho=Ar^2$, where $A$ is a positive constant. At a distance of $R/2$ from the center, the magnitude of the electric field is:
A. $A/4\pi\varepsilon_0$
B. $AR^3/40\varepsilon_0$
C. $AR^3/24\varepsilon_0$
D. $AR^3/5\varepsilon_0$
E. $AR^3/3\varepsilon_0$

(GR9677 #61)

Solution:

Gauss’ Law: $\oint \bar{E} \cdot \hat{n} dA=\frac{q_{\textup{enc}}}{\varepsilon_0}$
$\rightarrow E\cdot 4\pi \bigg(\frac{R}{2}\bigg)^2=\frac{q_{\textup{enc}}}{\varepsilon_0}$
$\rightarrow E=\frac{q_{\textup{enc}}}{\pi R^2 \varepsilon_0}$
The net charge within the Gussian surface with $\rho$ as a function of $r$ (non-uniformly charged sphere):
$dq_{\textup{enc}}=\rho dV=Ar^2 d\bigg(\frac{4}{3} \pi r^3 \bigg)$
$dq_{\textup{enc}}=Ar^2 \frac{4}{3} \pi 3r^2 dr=A4 \pi r^4 dr$
$q_{\textup{enc}}=\int_0^{R/2} A4\pi r^4 dr=A4 \pi \frac{1}{5} \bigg(\frac{R}{2}\bigg)^5=\frac{A\pi R^5}{40}$
$\rightarrow E=\frac{1}{\pi R^2 \varepsilon_0}\cdot \frac{A \pi R^5}{40}=\frac{AR^3}{40\varepsilon_0}$
Answer: B

Electromagnetism - Gauss' Law

2010-04-13T18:15:00.000-07:00

If an electric field is given in a certain region by $E_x=0,E_y=0,E_z=kz$, where $k$ is a nonzero constant, which of the following is true?
A. There is a time-varying magnetic field.
B. There is charge density in the region.
C. The electric field cannot be constant in time.
D. The electric field is impossible under any circumstances.
E. None of the above.

(GR9277 #64)

Solution:
Gauss’ Law: $\nabla \cdot \textbf{E}= \frac{\rho}{\varepsilon_0}$
For $E_z=kz$,
$\nabla \cdot \textbf{E}= \frac{\partial}{\partial x}E_x+\frac{\partial}{\partial y} E_y+\frac{\partial}{\partial z} E_z=0+0+\frac{\partial}{\partial z} kz=k$
Since $\nabla \cdot \textbf{E}\not=0 \rightarrow$ there is charge density in the region.

Answer: B

Electromagnetism - Gauss' Law

2010-04-13T15:18:00.000-07:00

An isolated sphere of radius $R$ contains a uniform volume distribution of positive charge. Which of the curves on the graph above correctly illustrated the dependence of the magnitude of the electric field of the sphere as a function of the distance $r$ from its center?
A. A
B. B
C. C
D. D
E. E

(GR8677 #10)

Solution:
Gauss’ Law: $\oint \bar{E} \cdot \hat{n} dA=\frac{q_{\textup{enc}}}{\varepsilon_0}$
$\rightarrow E(4\pi r^2)=\frac{\rho V}{\varepsilon_0}=\frac{\rho}{\varepsilon_0}\frac{4}{3}\pi r^3$
$\rightarrow E=\frac{\rho r}{3\varepsilon_0}\propto kr$
Answer: C

Electromagnetism - Gauss' Law

2010-04-13T14:51:00.000-07:00

An infinite, uniformly charged sheet with surface charged density $\sigma$ cuts through a spherical Gaussian surface of radius $R$ at a distance $x$ from its center, as shown in figure. The electric flux $\Phi$ through the Gaussian surface is
A. $\frac{\pi R^2 \sigma}{\varepsilon_0}$
B. $\frac{2 \pi R^2 \sigma}{\varepsilon_0}$
C. $\frac{\pi (R-x)^2 \sigma}{\varepsilon_0}$
D. $\frac{\pi (R^2-x^2) \sigma}{\varepsilon_0}$
E. $\frac{2 \pi (R^2-x^2) \sigma}{\varepsilon_0}$

(GR0177 #60)

Solution:

The gaussian surface marks a circle around the plane with radius $r^2=R^2-x^2$

The net charge within the Gussian surface: $q_{\textup{enc}}=\sigma A=\sigma \pi r^2=\sigma \pi (R^2-x^2)$

Electric Flux:
$\Phi=\frac{q_{\textup{enc}}}{\varepsilon_0}=\frac{\sigma \pi(R^2-x^2)}{\varepsilon_0}$

Answer: D

Electromagnetism - Gauss' Law

2010-04-13T14:17:00.001-07:00

Which of the following electric fields could exist in a finite region of space that contains no charges? (in these expressions, $A$ is a constant, and $\textbf{i}$, $\textbf{j}$, and $\textbf{k}$ are unit vectors pointing in the $x$, $y$, and $z$ directions, respectively.)
A. $A(2xy\textbf{i}-xz\textbf{k})$
B. $A(-xy\textbf{j}+xz\textbf{k})$
C. $A(xz\textbf{i}+xz\textbf{j})$
D. $Axyz(\textbf{i}+\textbf{j})$
E. $Axyz\textbf{i}$

(GR8677 #80)

Solution:
With no charges $\rightarrow$ Gauss’ Law: $\nabla \cdot \textbf{E}=0$
$\rightarrow \frac{\partial}{\partial x} E_x + \frac{\partial}{\partial x} E_y+\frac{\partial}{\partial z} E_z=0$

Option A:
$\nabla \cdot \textbf{E}= \frac{\partial}{\partial x} 2xy-\frac{\partial}{\partial z}xz=2y-x\not=0$

Option B:
$\nabla \cdot \textbf{E}=-\frac{\partial}{\partial y}xy+\frac{\partial}{\partial z} xz=-x+x=0$

Answer: B

Electromagnetism - Capacitor

2010-04-13T13:38:00.000-07:00

A 3-microfarad capacitor is connected in series with 6-microfarad capacitor. When a 300-volt potential difference is applied across this combination, the total energy stored in the two capacitor is
A. 0.09 J
B. 0.18 J
C. 0.27 J
D. 0.41 J
E. 0.81 J

(GR0177 #10)

Solution:
$C$ in Series :
$\frac{1}{C_{net}}=\frac{1}{C_1}+\frac{1}{C_2}=\bigg(\frac{1}{3}+\frac{1}{6}\bigg)\frac{1}{10^{-6}}=\frac{1}{2\cdot 10^{-6}}$
$\rightarrow C_{net} = 2\cdot 10^{-6}$ F

$U=\frac{1}{2}QV$
$U=\frac{1}{2}CV^2=\frac{1}{2}(2\cdot 10^{-6})300^2=0.09$ J

Answer: A

Classical Mechanics - Rotational Motion

2010-03-16T17:00:00.000-07:00

A satellite of mass m orbits a planet of mass $M$ in a circular orbit of radius $R$. The time required for one revolution is
A. independent of $M$
B. proportional to $\sqrt{m}$
C. linear in $R$
D. proportional to $R^{3/2}$
E. proportional to $R^2$

(GR0177 #3)

Solution:
$F_c=F_G$
$\frac{mv^2}{R}=\frac{GmM}{R^2}$
$v^2=\frac{GM}{R}$
$\bigg(\frac{2\pi R}{T}\bigg)^2=\frac{GM}{R}$
$\rightarrow T=\sqrt{\frac{4\pi^2 R^3}{GM}}=\frac{2\pi R^{3/2}}{\sqrt{GM}}$
Answer: D

Classical Mechanics - Rotational Motion

2010-03-16T16:51:00.000-07:00

The coefficient of static friction between a small coin and the surface of a turntable is 0.30. The turntable rotates at 33.3 revolutions per minute. What is the maximum distance from the center of the turntable at which the coin will not slide?
A. 0.024 m
B. 0.048 m
C. 0.121 m
D. 0.242 m
E. 0.484 m

(GR0177 #2)

Solution:
The coin will not slide if $F_c=F_f$
With
$F_c=\frac{mv^2}{r}=\frac{m}{r} \bigg(\frac{2\pi r}{T}\bigg)^2=\frac{4\pi^2 mr}{T^2}$
$F_f=\mu_s F_N=\mu_s mg$
$\rightarrow \frac{4\pi^2 mr}{T^2} =\mu_s mg$
Therefore, the maximum distance from the equilibrium:
$r=\frac{u_s gT^2}{4\pi^2}=0.24$ m
Answer: D

Classical Mechanics - Rotational Motion

2010-03-16T16:43:00.000-07:00

A particle constrained to move in a circle with 10-meter radius. At one instant, the particle’s speed is 10 meters per second and is increasing at a rate of 10 meters per second squared. The angle between the particle’s velocity and acceleration vector is
A. $0^o$
B. $30^o$
C. $45^o$
D. $60^o$
E. $90^o$

(GR0177 #23)

Solution:

The speed is increasing $\rightarrow \vec{a}_{tan}//\vec{v}$
$a_c=\frac{v^2}{r}=\frac{10^2}{10}=10$ m/s²
$\tan \theta= \frac{a_{tan}}{a_c} =\frac{10}{10}=1$
$\rightarrow \theta=45^o$
Answer: C

Classical Mechanics - Fluid Dynamics

2010-03-15T11:04:00.000-07:00

A stream of water of density $\rho$, 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. $\rho v^2 A$
B. $\frac{\rho vA}{2}$
C. $\rho ghA$
D. $\frac{v^2 A}{\rho}$
E. $\frac{v^2 A}{2\rho}$

(GR0177 #57)

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

Classical Mechanics - Fluid Dynamics

2010-03-15T11:01:00.000-07:00

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)

The system is floating: $B=W$

The magnitude of the buoyant force, $B$ is equal to the magnitude of the weight of the displaced air. The magnitude of the weight of the system, $W$ is equal to the magnitude of the weight of Helium and the object:
$\rho_a Vg=\rho_h Vg+m_o g$

$\rightarrow (\rho_a-\rho_h )V=m_o$
$\rightarrow V=\frac{m_o}{\rho_a-\rho_h}=\frac{300}{1.29-0.18}=270$ m³

Answer: D

Classical Mechanics - Hamiltonian

2010-03-14T19:06:00.000-07:00

Two small equal masses $m$ are connected by an ideal massless spring that has equilibrium lengths $l_0$ and force constant $k$. The system is free to move without friction in the plane of the page. If $p_1$ and $p_2$ represent the magnitude of the momenta of the two masses, a Hamiltonian for this system is
A. $\frac{1}{2}\bigg[\frac{p_1^2}{m}+\frac{p_2^2}{m}-2k(l-l_0)\bigg]$
B. $\frac{1}{2}\bigg[\frac{p_1^2}{m}+\frac{p_2^2}{m}-2k(l-l_0)^2\bigg]$
C. $\frac{1}{2}\bigg[\frac{p_1^2}{m}+\frac{p_2^2}{m}-k(l-l_0)\bigg]$
D. $\frac{1}{2}\bigg[\frac{p_1^2}{m}+\frac{p_2^2}{m}-k(l-l_0)^2\bigg]$
E. $\frac{1}{2}\bigg[\frac{p_1^2}{m}+\frac{p_2^2}{m}+k(l-l_0)^2\bigg]$

(GR0177 #92)

Solution:
The Hamiltonian $H=T+U$
with
$T_i=\frac{p_i^2}{2m}$
and,
$U=\frac{1}{2} k(\Delta l)^2=\frac{1}{2} k(l-l_0)^2$
$H=\frac{p_1^2}{2m}+\frac{p_2^2}{2m}+\frac{1}{2}k(l-l_0)^2$
$H=\frac{1}{2} \bigg[\frac{p_1^2}{m}+\frac{p_2^2}{m}+k(l-l_0)^2\bigg]$

Answer: E

Classical Mechanics - Lagrangian

2010-03-14T09:08:00.000-07:00

The Lagrangian for a mechanical system is $L=a\dot q^2+bq^4$ where $q$ is a generalized coordinate and $a$ and $b$ are constants. The equation of motion for this system is
A. $\dot q=\sqrt{\frac{b}{a}}q^2$
B. $\dot q=\frac{2b}{a} q^3$
C. $\ddot q=\frac{-2b}{a} q^3$
D. $\ddot q=+\frac{2b}{a} q^3$
E. $\ddot q=\frac{b}{a} q^3$

(GR0177 #74)

Solution:
The Lagrangian equation of motion for the generalized coordinate $q$:
$\frac{\partial L}{\partial q}=\frac{\partial}{\partial t}\frac{\partial L}{\partial \dot q}$
$\frac{\partial L}{\partial q}=\frac{\partial}{\partial q} (a\dot q^2+bq^4 )=0+\frac{\partial}{\partial q} bq^4=4bq^3$
$\frac{\partial L}{\partial \dot q}=\frac{\partial}{\partial \dot q} (a\dot q^2+bq^4 )=\frac{\partial}{\partial \dot q} a\dot q^2+0=2a\dot q$
$\frac{\partial}{\partial t}\frac{\partial L}{\partial \dot q}=\frac{\partial}{\partial t} 2a\dot q=2a \ddot q$

$\frac{\partial L}{\partial q}=\frac{\partial}{\partial t}\frac{\partial L}{\partial \dot q} \rightarrow 4bq^3=2a\ddot q \rightarrow \ddot q=\frac{2b}{a} q^3$

Answer: D

Classical Mechanics - Hamiltonian

2010-03-14T08:40:00.000-07:00

(Question GR8677 #34-36) The potential energy of a body constrained to move on a straight line is $kx^4$ where $k$ is a constant. The position of the body is $x$, its speed $v$, its linear momentum $p$, and its mass $m$.

The body moves from $x_1$ at time $t_1$ to $x_2$ at time $t_2$. Which of the following quantities is an extremum for the $x-t$ curve corresponding to this motion, if end points are fixed?
A. $\int_{t_1}^{t_2} \bigg(\frac{1}{2} mv^2-kx^4\bigg) dt$
B. $\int_{t_1}^{t_2} \bigg(\frac{1}{2} mv^2\bigg) dt$
C. $\int_{t_1}^{t_2} (mxv) dt$
D. $\int_{x_1}^{x_2} \bigg(\frac{1}{2} mv^2+kx^4\bigg) dx$
E. $\int_{x_1}^{x_2} (mv) dx$

(GR8677 #36)

Solution:
Lagrangian $L=T-U$
Hamiltonian’s principle:
$H=\int_{t_1}^{t_2} L dt$
$U=kx^4$
$T=\frac{1}{2} mv^2$
$\rightarrow H=\int_{t_1}^{t_2}\bigg(\frac{1}{2} mv^2-kx^4\bigg)dt$

Answer: A

Classical Mechanics - Hamiltonian

2010-03-14T08:28:00.000-07:00

(Question GR8677 #34-36) The potential energy of a body constrained to move on a straight line is $kx^4$ where $k$ is a constant. The position of the body is $x$, its speed $v$, its linear momentum $p$, and its mass $m$.

The Hamiltonian function for this system is
A. $\frac{p^2}{2m}+kx^4$
B. $\frac{p^2}{2m}-kx^4$
C. $kx^4$
D. $\frac{1}{2}mv^2-kx^4$
E. $\frac{1}{2}mv^2$

(GR8677 #35)

Solution:
Hamiltonian: $H=T+U$
$U=kx^4$
$T=\frac{1}{2}mv^2=\frac{p^2}{2m}$
$\rightarrow H=\frac{p^2}{2m}+kx^4$

Answer: A

Classical Mechanics - Newton's Law

2010-03-13T17:51:00.000-08:00

For the system consisting of the two blocks shown in the figure, the minimum horizontal force $F$ is applied so that block B does not fall under the influence of gravity. The masses of A and B are 16.0 kilograms and 4.00 kilograms, respectively. The horizontal surface is frictionless and the coefficient of friction between the two blocks is 0.50. The magnitude of $F$ is most nearly
A. 50 N
B. 100 N
C. 200 N
D. 400 N
E. 1,600 N

(GR0177 #73)

Solution:

$F=(m_A+m_B)a \rightarrow a=\frac{F}{m_A+m_B}$

If B does not fall under the influence of gravity:
$\mu N=m_B g \rightarrow N=\frac{m_B g}{\mu}$
And,
$F=F_A+N \rightarrow N=F-F_A$
$N=F-m_A a$
$N =F-m_A \bigg(\frac{F}{m_A+m_B}\bigg)$
$N=F\bigg(1-\frac{m_A}{m_A+m_B} \bigg)=\frac{Fm_B}{m_A+m_B}$

Therefore,
$\frac{m_B g}{\mu}=\frac{Fm_B}{m_A+m_B}$
$F=\frac{(m_A+m_B)g}{\mu}=\frac{(20)(10)}{0.5}=400$ N

Answer: D

Classical Mechanics - Conservation of Momentum

2010-03-13T17:46:00.000-08:00

A uniform stick of length $L$ and mass $M$ lies on a frictionless horizontal surface. A point particle of mass $m$ approaches the stick with speed $v$ on a straight line perpendicular to the stick that intersects the stick at one end, as shown above. After the collision, which is elastic, the particle is at rest. The speed $V$ of the center of mass of the stick after the collision is
A. $\frac{m}{M}v$
B. $\frac{m}{M+m}v$
C. $\sqrt{\frac{m}{M}}v$
D. $\sqrt{\frac{m}{M+m}}v$
E. $\frac{3m}{M}v$

(GR8677 #44)

Solution:
Conservation of momentum (elastic collision):
$m_a v_a+m_b v_b=m_a v^,_a+m_b v^,_b$
$mv+0=0+MV$
$V=\frac{m}{M}v$

Answer: A

Classical Mechanics - Impuls

2010-03-13T17:33:00.000-08:00

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:
$I=\Delta p=F \Delta t=\frac{1}{2}(2)(2)=2$ kgm/s

Answer: C

Classical Mechanics - Conservation of Energy and Momentum

2010-03-13T17:27:00.000-08:00

A particle of mass $m$ is moving along the x-axis with speed $v$ when it collides with a particle of mass $2m$ initially at rest. After the collision, the first particle has come to rest, and the second particle has split into two equal-mass pieces that move at equal angles $\theta>0$ with the x-axis, as shown in the figure. Which of the following statements correctly describes the speeds of the two pieces?
A. Each piece moves with speed $v$
B. One of the pieces moves with speed $v$, the other moves with speed less than $v$.
C. Each piece moves with speed $v/2$
D. One of the pieces moves with speed $v/2$, the other moves with speed greater than $v/2$.
E. Each piece moves with speed greater than $v/2$

(GR0177 #55)

Solution:
Conservation of momentum:
$m_a v_a+m_b v_b=m_a v^,_a+2m_b^, v_b^,$
$mv+0=0+2\bigg(\frac{1}{2}\cdot 2m\bigg) v^,\cos\theta$
$\rightarrow v^,=\frac{1}{2\cos\theta}$

Answer: E

Classical Mechanics - Conservation of Energy and Momentum

2010-03-13T17:17:00.000-08:00

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 #4)

Solution:
Conservation of Momentum (inelastic collision): $m_a v_a+m_b v_b=(m_a+m_b )v^,$
$m_a v_a+0=(m_a+m_b) v^,$
$\rightarrow v^,=\frac{m_a v_a}{m_a+m_b}=\frac{2}{3} v_a$

$KE_i-KE_f$
$=\bigg(\frac{1}{2} m_a v_a^2+\frac{1}{2} m_b v_b^2\bigg)-\frac{1}{2}(m_a+m_b) (v^,)^2$
$=\frac{1}{2} (2m) v_a^2+0-\frac{1}{2} (3m) \bigg(\frac{2}{3} v_a \bigg)^2$
$=mv_a^2-\frac{2}{3} mv_a^2=\frac{1}{3} mv_a^2$

Answer: C

Classical Mechanics - Conservation of Energy and Momentum

2010-03-13T12:26:00.000-08:00

A 10 g bullet is fired into a 2 kg ballistic pendulum as shown in the figure. The bullet remains in the block after the collision and the system rises to a maximum height of 20 cm. Find the initial speed of the bullet.
A. 28.0 m/s
B. 23.8 m/s
C. 3.98 m/s
D. 719 m/s
E. 398 m/s

Solution:
Given:
$m=0.01$ kg
$M=2$ kg
$v_p=0$
$y=0.2$ m

Find: $v_b$

Calculation:
Conservation of Energy: $KE=PE$
($KE$ immediately after the collision changed entirely to $PE$)
$\rightarrow \frac{1}{2} (m+M) (v^,)^2=(m+M)gy$
$\rightarrow v^,=\sqrt{2gy}$

Conservation of Linear Momentum:
$p_i=p_f$
$\rightarrow mv_b+Mv_p=(m+M)v^,$
$v_p=0 \rightarrow mv_b=(m+M)v^,$
$\rightarrow v_b=\bigg(\frac{m+M}{m}\bigg) v^,$
$\rightarrow v_b=\bigg(\frac{m+M}{m}\bigg) \sqrt{2gy}$
$\rightarrow v_b=\bigg( \frac{2.01}{0.01} \bigg) \sqrt{2(9.8)(0.2)}=398$ m/s

Note: some energy has been lost during the collision and converted to heat.

Answer: E

Classical Mechanics - Rotational Motion

2010-03-11T09:46:00.000-08:00

Seven pennies are arranged in a hexagonal, planar pattern so as to touch each neighbor, as shown in the figure. Each penny is a uniform disk of mass $m$ and radius $r$. What is the moment of inertia of the system of seven pennies about an axis that passes through the center of the central penny and is normal to the plane of the pennies?
(A) $\frac{7}{2} mr^2$
(B) $\frac{13}{2} mr^2$
(C) $\frac{29}{2} mr^2$
(D) $\frac{49}{2} mr^2$
(E) $\frac{55}{2} mr^2$

(GR0177 #25)

Solution:
Parallel axis theorem: $I=ml^2+I_{CM}$
Moment inertia of each penny (a uniform disk): $I=\frac{1}{2}mr^2$
With $l=2r$,
$I=m(2r)^2+\frac{1}{2} mr^2=\frac{9}{2} mr^2$
$I_{total}=I_{6 outer pennies}+I_{1 central penny}$
$I_{total}=6\bigg(\frac{9}{2} mr^2\bigg)+\frac{1}{2}mr^2$
$I_{total}=\bigg(\frac{54}{2}+\frac{1}{2}\bigg)mr^2=\frac{55}{2}mr^2$

Answer: E

Classical Mechanics - Rotational Motion

2010-03-11T09:27:00.000-08:00

Uniform disk of radius 0.8 m and mass 3 kg is freely suspended from a horizontal pivot located a radial distance 0.25 m from its center. Find the angular frequency of small amplitude oscillations of the disk.

Solution:
Given:
$r=0.8$ m
$m=3$ kg
$l=0.25$ m

Find: $\omega$

Calculation:
Parallel axis theorem: $I=ml^2+I_{CM}$
Moment inertia for uniform disk: $I_{CM}=\frac{1}{2} mr^2$
$\rightarrow I=ml^2+\frac{1}{2} mr^2=m(l^2+\frac{1}{2}r^2)$

Remember the period of a compound/physical pendulum:
$T=2\pi\sqrt{\frac{I}{mgl}}$ and $\omega =\frac{2\pi}{T}$
$\rightarrow \omega = \sqrt{\frac{mgl}{I}}$
$\rightarrow \omega = \sqrt{\frac{gl}{l^2+\frac{1}{2} r^2}}$
$\rightarrow \omega =\sqrt{\frac{(9.8)(0.25)}{(0.25)^2+\frac{1}{2}(0.8)^2}}=2.53$ rad/s

Classical Mechanics - Rotational Motion

2010-03-11T09:24:00.000-08:00

A skater spins at an initial angular velocity of $\omega_1=11$ rad/s with her arms outstretched. The skater then lowers her arms, thereby decreasing her moment of inertia by a factor 8. What is the skater's final angular velocity? Assume that any friction between the skater's skates and the ice is negligible.

Solution:
Given:
$\omega_1=11$ rad/s
$\frac{I_1}{I_2}=8$

Find: $\omega_2$

Calculation:
Angular momentum: $L=I\omega$
Conservation of angular momentum: $L_1=L_2$
$I_1 \omega_1 =I_2 \omega_2$
$\omega_2=\frac{I_1}{I_2}\omega_1=88$ rad/s

Classical Mechanics - Rotational Motion

2010-03-11T09:20:00.000-08:00

A missile of mass $m=2.3\times 10^4$ kg flies level to the ground at an altitude of $r=10.000$ m with constant speed $v=210$ m/s. What is the magnitude of the missile's angular momentum relative to a point on the ground directly below its flight path?

Solution:
Given:
$m=2.3\times 10^4$ kg
$v=210$ m/s
$r=10.000$ m

Find: Angular momentum $L$

Calculation:
$L=I\omega$
with moment inertia: $I=mr^2$
and angular speed: $v=r\omega \rightarrow \omega= \frac{v}{r}$
$\rightarrow L=mr^2 \frac{v}{r}=mrv$
$\rightarrow L=(2.3 \times 10^4)(10^4)(210)=4.83\times 10^{10}$ kgm²/s

Classical Mechanics - Rotational Motion

2010-03-11T09:18:00.000-08:00

A car engine develops a torque of 500 Nm and rotates at 3000 rev/min. What horsepower does the engine generate? (1 hp = 746 W).

Solution:
Given:
$\tau= 500$ Nm
$\omega=3000$ rev/min $=3000\times \frac{2\pi}{60}=314$ rad/s

Find: $P$ (in hp)

Calculation:
$P=\tau \omega =(500)(314)=15.7 \times 10^4$ W
$P=\frac{15.7\times 10^4}{746}=210.45$ hp

Classical Mechanics - Rotational Motion

2010-03-11T09:13:00.000-08:00

A uniform rod of mass $m = 5.3$ kg and $l = 1.3$ m rotates about a fixed frictionless pivot located at one of its ends. The rod is released from rest at an angle $\theta = 35^o$ beneath the horizontal. What is the angular acceleration of the rod immediately after it is released?

Solution:
Given:
$m=5.3$ kg
$l=1.3$ m
$\theta=35^o$
$r=\frac{1}{2}l=0.65$ m

Find: $\alpha$

Calculation:

Torque: $\tau=I\alpha=rF \rightarrow \alpha=\frac{rF}{I}$
Moment inertia of the uniform rod: $I = \frac{1}{3} ml^2$
with $F=mg\cos\theta$

$\rightarrow \alpha=\frac{rmg\cos\theta}{(1/3)ml^2}=\frac{3rg\cos\theta}{l^2}$
$\rightarrow \alpha=\frac{3(0.65)(9.8)\cos(35^o)}{(1.3)^2}=9.26$ rad/s²

Classical Mechanics - Rotational Motion

2010-03-11T09:09:00.000-08:00

The net work done in accelerating a wheel from rest to an angular speed of 30 rev/min is 5500 J. What is the moment of inertia of the wheel?

Solution:
Given:
$\omega_0 = 0$
$\omega = 30$ rev/min $= 30\times \frac{2\pi}{60} = 3.14$ rad/s
$W=5500$ J

Find: $I$

Calculation:
Assuming that all of the work performed on the wheel goes to increase its rotational kinetic energy, we have:
$W=KE=\frac{1}{2}I\omega^2$
$\rightarrow I=\frac{2W}{\omega^2}=\frac{2(5500)}{(3.14)^2}=1114.6$ kgm²

Classical Mechanics - Rotational Motion

2010-03-11T09:05:00.000-08:00

A tire placed on a balancing machine in a service station starts from rest and turns through 5.3 revolutions in 2.3 s before reaching its final angular speed. Calculate:
(a) the angular acceleration of the tire (assuming that this quantity remains constant)
(b) the final angular speed of the tire

Solution:

Given:
$\theta = 5.3 \times 2\pi$ rad $= 33.3$ rad
$t = 2.3$s
$\omega_0 = 0$
$\alpha$ constant

Find: $\alpha$ and $\omega$

Calculation:
(a) For $\alpha$ constant:
$\rightarrow \theta=\omega_0t+\frac{1}{2}\alphat^2$
$\rightarrow \theta=0+\frac{1}{2}\alpha t^2$
$\rightarrow \alpha=\frac{2\theta}{t^2}=\frac{(2)(33.3)}{(2.3)^2}=12.58$ rad/s²
(b) $\omega=\alpha t = (12.58)(2.3) = 28.9$ rad/s

Classical Mechanics - Rotational Motion

2010-03-11T08:52:00.000-08:00

A horizontal beam of length 10 m and weight 200 N is attached to a wall as shown. The far end is supported by a cable which makes an angle of 60^o with respect to the beam. A 500 N person stands 2 m from the wall. Determine the tension in the cable.
A. 0 N
B. 700 N
C. 500 N
D. 231 N
E. 808 N

Solution:

The system is in equilibrium $\rightarrow \Sigma \tau=0$ (the sum of the torques with respect to any point must be zero).
$\tau=rF \sin \alpha$

Choose the point of the beam to the wall as the reference:
$(500)(2)(\sin 90)+(200)(5)(\sin 90)-(10)T(\sin 120)=0$
$\rightarrow T=231$ N

Answer: D

Classical Mechanics - Rotational Motion

2010-03-11T08:40:00.000-08:00

A wheel 4 m in diameter rotates on a fixed frictionless horizontal axis, about which its moment of inertia is $10$ kgm². A constant tension of $40$ N is maintained on a rope wrapped around the rim of the wheel. If the wheel starts from rest at $t=0$s, find the length of rope unwound at $t=3$ s.
A. 36 m
B. 72 m
C. 18 m
D. 720 m
E. 180 m

Solution:
Given:
$d=4$ m $\rightarrow r=2$ m
$I=10$ kgm²
$T=40$ N
$\Delta t=3$s

Find: the length of rope unwound, $l$

Calculation:
Torque: $\tau=rF=I\alpha \rightarrow rT=I\alpha$
$\rightarrow \alpha=\frac{rT}{I}=\frac{2(40)}{10}=8$ rad/s²

By definition: $\Delta \theta= \frac{l}{r}$
For $T$ constant $\rightarrow \alpha$ constant:
$\Delta \theta=\omega_0 t+\frac{1}{2} \alpha t^2$
$\rightarrow \omega_0 t+\frac{1}{2} \alpha t^2=l/r$
The wheel starts from rest $\rightarrow \omega_0=0$
$\rightarrow \frac{1}{2} \alpha t^2=\frac{l}{r}$
$\rightarrow l=\frac{1}{2} r \alpha t^2=\frac{1}{2} (2)(8)(3)^2=72$ m

Answer: D

Classical Mechanics - Rotational Motion

2010-03-11T08:09:00.000-08:00

A uniform cylinder of radius $r=0.25$ m is given an angular speed of $\omega_0=35$ rad/s about an axis, parallel to its length, which passes through its center. The cylinder is gently lowered onto a horizontal frictional surface, and released. The coefficient of friction of the surface is $\mu=0.15$. How long does it take before the cylinder starts to roll without slipping? What distance does the cylinder travel between its release point and the point at which it commences to roll without slipping?

Solution:
Given:
$r=0.25$ m
$\omega_0=35$ rad/s
$\mu=0.15$
Moment inertia of uniform cylinder: $I=\frac{1}{2} mr^2$

Find: $t$ and $s$

Calculation:

Let $f$ be the frictional force.

When the cylinder is slipping:
$F=f$
$\rightarrow ma=\mu mg$
$\rightarrow a=\mu g$
$\rightarrow \frac{v-v_0}{t}=\mu g$

Cylinder initially at rest: $v_0=0$
$\rightarrow v=\mu gt$ … (Eq.1)

$f$ decelerates cylinder’s rotational motion,
Torque: $\tau=I \alpha =-rf$
$\rightarrow \frac{1}{2}mr^2 \alpha=-r \mu mg$
$\rightarrow r\alpha=-2 \mu g$
$\rightarrow r\bigg(\frac{\omega-\omega_0}{t}\bigg)=-2 \mu g$
$\rightarrow r\omega=r\omega_0-2 \mu gt$ … (Eq.2)

(Eq.1)-(Eq.2): $v-r\omega=-r\omega_0+3 \mu gt$

At no slip condition, linear speed: $v=r\omega$
$\rightarrow r\omega_0=3 \mu gt$
$\rightarrow t=\frac{r\omega_0}{3 \mu g}=\frac{(0.25)(35)}{3(0.15)(9.8)}=1.98$ s

$\rightarrow s=v_0 t+\frac{1}{2}at^2=0+\frac{1}{2}\mu gt^2$
$s=\frac{1}{2}(0.15)(9.8)(1.98)^2=2.82$ m

Classical Mechanics - Rotational Motion

2010-03-11T08:04:00.000-08:00

A weight of mass 2.6 kg is suspended via a light inextensible cable which is wound around a pulley of mass 6.4 kg and radius 0.4 m. Treating the pulley as a uniform disk, find the downward acceleration of the weight and the tension in the cable. Assume that the cable does not slip with respect to the pulley.

Solution:
Given:
$m_{object}= m = 2.6$ kg
$m_{pulley}= M = 6.4$ kg
$r=0.4$ m

Find: $a$ and $T$

Calculation:

Torque: $\tau=I\alpha=rT \rightarrow T=\frac{I\alpha}{r}$

Linear acceleration: $a=r\alpha \rightarrow \alpha=\frac{a}{r}$

Moment inertia of pulley as uniform disk: $I=\frac{1}{2}Mr^2$

$\rightarrow T=\frac{(1/2)Mr^2(a/r)}{r}=\frac{1}{2}Ma$

$\sum F=mg-T \rightarrow ma=mg-\frac{1}{2}Ma$
$\rightarrow a=\frac{mg}{m+0.5M}=\frac{(2.6)(9.8)}{(2.6)+(0.5)(6.4)}=4.39$ m/s²

$T=\frac{1}{2}Ma=0.5(6.4)(4.39)=14.04$ N

Classical Mechanics - Rotational Motion

2010-03-10T15:53:00.000-08:00

A rod of mass $m=3$kg and length $l=1.2$m pivots about an axis, perpendicular to its length, which passes through one of its ends. What is the moment of inertia of the rod? Given that the rod's instantaneous angular velocity is $60$ deg/s, what is its rotational kinetic energy?

Solution:
Given:
$m=3$ kg
$l=1.2$ m
$\omega=60$ deg/s $=60\times \frac{\pi}{180}$ rad/s $=1.047$rad/s

Find: $I$ and $KE$

Calculation:
Moment of inertia of the rod about a perpendicular axis passing through its centre is: $I_{CM}=\frac{1}{12} ml^2$

Using the parallel axis theorem, the moment of inertia about a parallel axis passing through one of the ends of the rod is:
$I=mr^2+I_{CM}=m\bigg(\frac{l}{2}\bigg)^2+\frac{1}{12} ml^2$
$I=ml^2 \bigg(\frac{1}{4}+\frac{1}{12}\bigg)=\frac{1}{3} ml^2$
$I=\frac{1}{3}(3)(1.2)^2=1.44$ kgm²
$KE=\frac{1}{2} I\omega^2=\frac{1}{2} (1.44)(1.047)^2=0.78$ J

Classical Mechanics - Rotational Motion

2010-03-10T15:40:00.000-08:00

A particle is moving along a circle of radius $r$. The linear and angular velocities at an instant during the motion are $v$ and $\omega$ respectively. Then, the product $v\omega$ represents:
A. centripetal acceleration
B. tangential acceleration
C. angular acceleration divided by radius
D. none of the above

Solution:

Linear velocity:
$v=\frac{\Delta l}{\Delta t}=r\frac{\Delta \theta}{\Delta t}$
$\rightarrow v=r\omega$
$\rightarrow \omega=\frac{v}{r}$
$\rightarrow v\omega=\frac{v^2}{r}=a \rightarrow$ centripetal acceleration

Answer: A

Classical Mechanics - Rotational Motion

2010-03-10T14:35:00.000-08:00

A thin uniform rod of mass $M$ and length $L$ is positioned vertically above an anchored frictionless pivot point, as shown in the figure, and then allowed to fall to the ground. With what speed does the free end of the rod strike the ground?
A. $\sqrt{\frac{1}{3}gL}$
B. $\sqrt{gL}$
C. $\sqrt{3gL}$
D. $\sqrt{12gL}$
E. $12\sqrt{gL}$

(GR0177 #26)

Solution:
Conservation of energy:
The initial total energy (before the rod begins to fall) = the final total energy (equals its kinetic energy, when the rod falls onto the ground):
$Mgy=\frac{1}{2} I\omega^2$
with $y=\frac{L}{2}$ and moment inertia of uniform rod: $I=\frac{1}{3}ML^2$
$\rightarrow Mg \frac{L}{2} = \frac{1}{2} \bigg(\frac{1}{3} ML^2 \bigg) \omega^2=\frac{1}{6} ML^2 \omega^2$
$\rightarrow \omega=\sqrt{\frac{3g}{L}}$
$\rightarrow v=\omega L=\sqrt{\frac{3g}{L}}L=\sqrt{3gL}$

Answer: C

Classical Mechanics - Rotational Motion

2010-03-10T14:18:00.000-08:00

A wheel $4$ m in diameter rotates with a constant angular acceleration $a= 4$ rad/s². The wheel starts from rest at $t=0$ s where the radius vector to point $P$ on the rim makes an angle of $45^o$ with the $x$-axis. Find the angular position of point $P$ at arbitrary time $t$.
A. $45^o$
B. $45+2t^2$ degrees
C. $45+114.6 t^2$ degrees
D. $4 t^2$ degrees
E. $229.2 t^2$ degrees

Solution:
Given:
$\theta_0=45^o$
$t_0=0$
$\alpha = 4 \times \frac{180^o}{\pi rad}=229.2$ deg/s ²
Constant angular acceleration $\rightarrow \Delta \theta=\omega_0 t+\frac{1}{2} \alpha t^2$

Find: $\theta$

Calculation:
$\Delta \theta=\theta-\theta_0=\omega_0 t+\frac{1}{2} \alpha t^2$
$\theta=\theta_0+\omega_0 t+\frac{1}{2} \alpha t^2$
The wheel starts from rest $\omega_0=0$
$\theta=\theta_0+\frac{1}{2} \alpha t^2= 45^o+\frac{1}{2}229.2 t^2=45^o+114.6^o t^2$

Answer: C

Classical Mechanics - Rotational Motion

2010-03-10T14:11:00.000-08:00

A cylinder with a moment of inertia $I_0$ rotates with angular velocity $\omega_0$. A second cylinder with moment of inertia $I_1$ initially not rotating drops onto the first cylinder and the two reach the same final angular velocity $\omega_f$. Find $\omega_f$.
A. $\omega_f=\omega_0$
B. $\omega_f=\omega_0 \frac{I_0}{I_1}$
C. $\omega_f=\frac{I_0\omega_0}{I_0+I_1}$
D. $\omega_f=\frac{\omega_0I_1}{I_0}$
E. $\omega_f=\omega_0\frac{I_0+I_1}{I_0}$

Solution:
Conservation of angular momentum:
$\Sigma L_0=\Sigma L_1$
$I_0 \omega_0=(I_0+I_1 )\omega_f \rightarrow \omega_f=\frac{I_0 \omega_0}{I_0+I_1}$

Answer: C

Classical Mechanics - Rotational Motion

2010-03-10T13:55:00.000-08:00

A child is standing on the edge of a merry-go-around that has the shape of a solid disk, as shown in the figure. The mass of the child is 40 kilograms. The merry-go-around has a mass 200 kilograms and a radius of 2.5 meters and it is rotating with an angular velocity of 2.0 radians per second. The child then walk slowly toward the center of the merry-go-around. What will the final angular velocity of the merry-go-around when the child reaches the center? (The size of the child can be neglected).

A. 2.0 rad/s
B. 2.2 rad/s
C. 2.4 rad/s
D. 2.6 rad/s
E. 2.8 rad/s

(GR0177 #89)

Solution:
Given:
$m_c=40$ kg
$m_m=200$ kg
$\omega_1=2$ rad/s
$r=2.5$ m

Find: $\omega_2$

Calculation:
Conservation of momentum: $\Sigma L_1 = \Sigma L_2$
$(I_m+I_c)\omega_1=I_m \omega_2$
$\omega_2=\frac{(I_m+I_c)}{I_m}\omega_1$

with moment inertia for uniform disk:
$I_m=\frac{1}{2} m_m r^2=\frac{1}{2} (200)(2.5)^2=625$ kgm²
and
$I_c=m_c r^2=(40)(2.5)^2=250$ kgm²

$\omega_2=\bigg(\frac{625+250}{625}\bigg)2=2.8$ rad/sec

Answer: E

Classical Mechanics - Rotational Motion

2010-03-10T13:45:00.000-08:00

30 kg child stands on the edge of a stationary merry-go-round of mass 100 kg and radius 2.0 m. The rotational inertia of the merry-go-round about its axis of rotation is 150 kgm². The child catches a ball of mass 1.0 kg thrown by a friend. Just before the ball is caught, it has a horizontal velocity of 12 m/s that makes an angle of 37° with a line tangent to the outer edge of the merry-go-round, as shown in the overhead view of the Fig. What is the angular speed of the merry-go-round just after the ball is caught?

Solution:
Given:
$m_c=30$ kg
$m_m=100$ kg
$r=2$ m
$I_m=150$ kgm²
$m_b=1$ kg
$v_b=12$ m/s
$\theta=37^o$

Calculation:
Conservation of momentum: $\Sigma L_1 =\Sigma L_2$

After the boy on the merry-go-around catches the ball, that momentum is transferred to the moments of inertia of the merry-go-around, plus the boy and the ball:
$m_b v_b=I_{total} \omega_2$

$\omega_2=\frac{m_b v_b}{I_{total}} =\frac{m_b v_b}{I_m+I_b+I_c}$

with
$I_b=m_b r_b^2=(1)(2 \sin 37)^2=1.45$ kgm²
$I_c=m_c r^2=(30)(2)^2=120$ kgm²

$\omega_2=\frac{(1)(12)}{(150+1.45 +120)}=\frac{12}{271.45}=0.0442$ rad/sec

or 0.42 revolutions per minute

Classical Mechanics - Rotational Motion

2010-03-10T12:53:00.000-08:00

The matrix shown above transforms the components of a vector in one coordinate frame $S$ to the components of the same vector in a second coordinates frame $S$'. This matrix represents a rotation of the reference frame $S$ by
A. $30^o$ clockwise about the $x$-axis
B. $30^o$ counterclockwise about the $z$-axis
C. $45^o$ clockwise about the $z$-axis
D. $60^o$ clockwise about the $y$-axis
E. $60^o$ counterclockwise about the $z$-axis

(GR0177 #75)

Solution:
Rotation about $z$-axis has the form:

Since $\sin 60=\frac{1}{2}\sqrt3$ and $\cos 60=\frac{1}{2} \rightarrow$ E is the right answer.

Answer: E

Classical Mechanics - Conservation of Energy and Momentum

2010-02-27T11:00:00.000-08:00

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 $h_0$ 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. $\frac{1}{16} h_0$
B. $\frac{1}{8} h_0$
C. $\frac{1}{4} h_0$
D. $\frac{1}{3} h_0$
E. $\frac{1}{2} h_0$

(GR8677 #5)

Solution:

Conservation of energy of Putty A at $h_0$ and $h=0$ (when A hits B):
$Mgh_0=\frac{1}{2} Mv_A^2$
$v_A=\sqrt{2gh_0}$

Conservation of Momentum of Putty A and B when and after A hits B (inelastic collision):
$m_A v_A+m_B v_B=(m_A+m_B)v_{AB}$
$M\sqrt{2gh_0}=4Mv_{AB}$
$v_{AB}=\frac{\sqrt{2gh_0}}{4}$

Conservation of energy of Putty A and B at $h=0$ and $h_{max}$:
$\frac{1}{2} m_{AB}v_{AB}^2=m_{AB}gh_{max}$
$\frac{1}{2} 4M \bigg(\frac{\sqrt{2gh_0}}{4}\bigg)^2=4Mgh_{max}$
$h_{max}=\frac{1}{16} h_0$

Answer: A

Classical Mechanics - Oscillatory Motion

2010-02-27T10:46:00.001-08:00

A particle of mass $m$ that moves along the $x$-axis has potential energy $V(x)=a+bx^2$, where $a$ and $b$ are positive constants. Its initial velocity is $v_0$ at $x=0$. It will execute simple harmonic motion with a frequency determined by the value of
A. $b$ alone
B. $b$ and $a$ alone
C. $b$ and $m$ alone
D. $b$,$a$ and $m$ alone
E. $b$,$a$,$m$ and $v_0$

(GR8677 #60)

Solution:
$F=-\frac{\partial V}{\partial x}=-\frac{\partial}{\partial x}(a+bx^2 )=-2bx$
$F=ma=-m\omega^2 x=-m(2\pi f)^2 x=-2bx$
We can see that $f$ depends on $b$ and $m$ alone

Answer: C

Classical Mechanics - Oscillatory Motion

2010-02-27T10:34:00.000-08:00

Which of the following best illustrates the acceleration of a pendulum bob at points a through e?

(GR0177 #1)

Solution:
The acceleration of pendulum is the sum of two components: $a=a_{cent}+a_{tan}$

Centripetal (radial) acceleration, to keep the particle moving on circular arc:
$a_{cent}=\frac{v^2}{r}=\omega^2 r$
with angular speed $\omega=\frac{d\theta}{dt}$

Tangential (linear) acceleration, to return the particle to the equilibrium position:
$a_{tan}=\alpha r$
with angular acceleration $\alpha=\frac{d\omega}{dt}$

At equilibrium (position c):
$\omega=$ constant $\rightarrow \alpha=0 \rightarrow a=a_{cent}$
The string provides the acceleration equal the magnitude but opposite direction to acceleration due to gravity

At maximum amplitude (position e and a):
$v=0 \rightarrow a=a_{tan}$

At other positions (d and b):
$a=a_{cent}+a_{tan}$

$a_{tan}$ has component $g \sin \theta$
$a_{cent}$ has component $g \cos \theta$

Answer: C

Classical Mechanics - Oscillatory Motion

2010-02-11T15:55:00.000-08:00

Two identical springs with spring constant $k$ are connected to identical masses of mass $M$, as shown in the figures. The ratio of the period for the springs connected in parallel (Fig. 1) to the period for the springs connected in series (Fig.2) is
A. $1/2$
B. $1/\sqrt2$
C. $1$
D. $\sqrt2$
E. $2$

(GR0177 #90)

Solution:
Parallel: $k_{eff}=k_1+k_2$
Series: $\frac{1}{k_{eff}}=\frac{1}{k_1}+\frac{1}{k_2}$
$T=2\pi \sqrt{\frac{m}{k}}$

Since $k_1=k_2=k$
Parallel: $k_p=2k$
Series: $\frac{1}{k_s}=\frac{2}{k}\rightarrow k_2=\frac{k}{2}$

The ratio of the period
$\frac{T_p}{T_s}=\frac{2\pi \sqrt{m/k_p}}{2\pi \sqrt{m/k_s}}=\sqrt{\frac{k_s}{k_p}}=\sqrt{\frac{k}{2\cdot2k}}=\frac{1}{2}$

Answer: A

Classical Mechanics - Oscillatory Motion

2010-02-10T09:39:00.000-08:00

Having landed on a newly discovered planet, an astronaut sets up a simple pendulum of length 0.6 m, and finds that it makes 51 complete oscillations in 1 minute. The amplitude of the oscillations is small compared to the length of the pendulum. What is the surface gravitational acceleration on the planet?

Solution:

Given:
$l = 0.6$ m
$f = 51/60 = 0.85$ Hz

Find: $a$

Formula:
$\omega = 2\pi f$
$a = A\omega^2 = l\omega^2$

Calculation:
$\omega = 2(3.14)(0.85) = 5.338$ rad/s
$a = (0.6)(5.338)^2 = 17.09$ m/s²

Classical Mechanics - Oscillatory Motion

2010-02-10T09:31:00.000-08:00

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) $\sqrt2 g$
(E) $2g$

(GR0177 #72)

Solution:
In equilibrium (before the string is cut):
Block 1:
$\sum F_1 = 0 \rightarrow T-m_1g-kx=0$ …(Eq.1)

Block 2:
$\sum F_2 = 0 \rightarrow kx-m_2g=0 \rightarrow kx=m_2g$ … (Eq.2)

After the string is cut:
$T=0, \sum F_1 = -m_1a_1$ and $a_2=0$
(minus sign $-m_1a$ is because the direction is in downward or parallel with $m_1g$)

(Eq.1) becomes:
$-m_1a_1 = -m_1g-kx$
$m_1a_1 = m_1g+kx$ ...(Eq.3)

Submit (Eq.2) to (Eq.3): $m_1a_1 = m_1g+m_2g$
Since $m_1= m_2 = m \rightarrow a_1 = 2g$

Answer: E

Classical Mechanics - Oscillatory Motion

2010-02-10T09:20:00.001-08:00

Classical Mechanics - Oscillatory Motion

2010-02-10T09:15:00.001-08:00

A block of mass $m = 3$kg is attached to two springs and slides over a horizontal frictionless surface. Given that the force constants of the two springs are $k_1 = 1200$ N/m and $k_2 = 400$ N/m, find the period of oscillation of the system.

Solution:

Given:
$m = 3$kg
$k_1 = 1200$ N/m
$k_2 = 400$ N/m

Find: $T$

Formula:
$k_{eff}=\frac{k_1 k_2}{k_2+k_1}$
$T=2\pi\sqrt{\frac{m}{k}}$

Calculation:
$k_{eff}=\frac{k_1 k_2}{k_2+k_1}=\frac{1200\cdot 400}{1200+400}=300$ N/m
$T=2\pi\sqrt{\frac{m}{k}}=6.28\sqrt{\frac{3}{300}}=0.628$ s

Classical Mechanics - Oscillatory Motion

2010-02-09T17:21:00.000-08:00

For an inductor and capacitor connected in series, the equation describing the motion of charge is
$L\frac{d^2Q}{dt^2}+\frac{1}{C}Q=0$
where $L$ is the inductance, $C$ is capacitance, and $Q$ is the charge. An analogous equation can be written for a simple harmonic oscillator with position $x$, mass $m$, and spring constant $k$. Which of the following correctly lists the mechanical analogs of $L$, $C$, and $Q$?

	$L$	$C$	$Q$
A.	$m$	$k$	$x$
B.	$m$	$1/k$	$x$
C.	$k$	$x$	$m$
D.	$1/k$	$1/m$	$x$
E.	$x$	$1/k$	$1/m$

(GR0177 #59)

Solution:

Hooke's Law:
$F=-kx \Rightarrow m \ddot{x}+kx=0$

LC circuit equation:
$L\ddot{Q}+Q/C=0$

Compare both equations:
$L = m$, $C = 1/k$, $Q=x$

Answer: B

Classical Mechanics - Oscillatory Motion

2010-02-09T17:14:00.000-08:00

A spring stretches 0.150 m when a 0.3 kg mass hung from it. The spring is then stretched an additional 0.1 m from this equilibrium point, and released. Determine
(A) the spring constant
(B) the amplitude of the oscillation
(C) the maximum velocity
(D) the magnitude of velocity when the mass is 0.05 m from equilibrium
(E) the magnitude of the maximum acceleration of the mass

Solution:

Given:
$x_1 = 0.15$ m
$m = 0.3$ kg
$x_2 = 0.1$ m

Find:
(A) $k$
(b) $A$
(c) $v_0$
(d) $v$ when $x = 0.05$ m
(e) $a_{max}$

Formula:
$F = -kx$
$v_0=A\sqrt{\frac{k}{m}}$
$v=v_0\sqrt{1-\frac{x^2}{A^2}}$
$a_{max} = a_0 = kA/m$

Calculation:
(A) $mg = kx_1 \Rightarrow k = \frac{mg}{x_1} = \frac{(0.3)(9.8)}{0.15} = 19.6$ N/m
(B) Since the spring stretched 0.1 m from equilibrium and is given no initial speed, $A = 0.1$ m
(C) $v_0=A\sqrt{\frac{k}{m}}=0.1\sqrt{\frac{19.6}{0.3}}=0.808$m/s
(D) when $x = 0.05$ m $\Rightarrow v=v_0\sqrt{1-\frac{x^2}{A^2}}=0.808\sqrt{1-\frac{0.05^2}{0.1^2}}=0.7$m/s
(E) $a_0 = kA/m = (19.6)(0.1)/(0.3) = 6.53$ m/s²

Classical Mechanics - Oscillatory Motion

2010-02-09T17:01:00.000-08:00

When a 4.0 kg mass is hung vertically on a light spring that obeys Hooke's law, the spring stretches 2.0 cm. How much work must an external agent do to stretch the spring 4.0 cm from its equilibrium position?
A. 1.57 J
B. 0.39 J
C. 0.20 J
D. 0.78 J
E. 3.14 J

Solution:

1. In equilibrium: $x=2$ cm $=2\cdot 10^{-2}$ m
$m=4$ kg
$F=kx=mg\Rightarrow k=\frac{mg}{x}=\frac{4\cdot10}{2\cdot10^{-2}}=2000$ N/m

2. When there's an external agent: $x=4$cm$=4\cdot10^{-2}$m
$W=\int^x_0{F ds}=\int^x_0{kx dx}$
$\Rightarrow = \frac{1}{2}kx^2\bigg|_0^{x_0}=\frac{1}{2}kx^2=\frac{1}{2}\cdot 2000 \cdot (4\cdot10^{-2})^2=1.6$ J

Classical Mechanics - Oscillatory Motion

2010-02-09T16:55:00.000-08:00

A $100$g mass attached to a spring moves on a horizontal frictionless table in simple harmonic motion with amplitude $16$cm and period $2$s. Assuming that the mass is released from rest at $t=0$s and $x=-16$ cm, find the displacement as a function of time.

A. $x=16\cos(\pi t)$
B. $x=-16\cos(\pi t+\pi)$
C. $x=16\cos(\pi t+\pi)$
D. $x=-16\cos(2\pi t+\pi)$
E. $x=-16\cos\bigg(\frac{\pi t}{2}\bigg)$

Solution:

Given:
$m=100$g
$A=16$cm
$T=2$s
at $t=0$s, $x=-16$cm

Find: $x$ as a function of $t$

Formula:
$x=A\cos(\omega t+\phi)$
$f=1/T$
$\omega=2\pi f$

Calculation:
$f=1/T=1/2$ Hz
$\omega=2\pi f=\pi$ rad/s

$x=A\cos(\omega t+\phi)$

At $t=0$s $x=-16$cm
$\Rightarrow -16=16\cos\phi \Rightarrow \cos\phi=-1$
$\Rightarrow \phi=180^o = \pi$ rad
$\Rightarrow x=16 \cos (\pi t+\pi)$

Answer: C

Classical Mechanics - Oscillatory Motion

2010-02-09T16:26:00.000-08:00

A block of mass $m = 4$ kg is attached to a spring, and undergoes simple harmonic motion with a period of $T = 0.35$s. The total energy of the system is $E = 2.5$ J. What is the force constant of the spring? What is the amplitude of the motion?

Given:
$m=4$ kg
$T=0.35$ s
$E=2.5$ J

Find:
$k$ and $x$

Formula:
$\omega=\frac{2\pi}{T}=\sqrt\frac{k}{m}$
$E=\frac{1}{2}kx^2$

Calculation:
$\frac{2\pi}{T}=\sqrt\frac{k}{m}\Rightarrow k=m\bigg(\frac{2\pi}{T}\bigg)^2$
$k=4\bigg(\frac{2\cdot3.14}{0.35}\bigg)^2=1288$ N/m
$E=\frac{1}{2}kx^2\Rightarrow x=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2\cdot 2.5}{1288}}=0.062$ m

Classical Mechanics - Oscillatory Motion

2010-02-09T16:01:00.001-08:00

A block attached to a spring executes simple harmonic motion in a horizontal plane with amplitude of 0.25 m. At a point 0.15 m away from the equilibrium position, the velocity of the block is 0.75 m/s. What is the period of oscillation of the block?

Solution:

Given:
$A=0.25$ m
at $t=0$ (say) $\Rightarrow x=0.15$m
and $v=0.75$ m/s

Find: $T$

Formula:
$v=v_0\sqrt{1-\frac{x^2}{A^2}}$
$v_0=\frac{2\pi A}{T}$

Calculation:
$v=\frac{2\pi A}{T}\sqrt{1-\frac{x^2}{A^2}}$
$\Rightarrow T=\frac{2\pi A}{v}\sqrt{1-\frac{x^2}{A^2}}$
$T=\frac{(6.28)(0.25)}{0.75}\sqrt{1-\frac{0.15^2}{0.25^2}}=1.67$ s

Classical Mechanics - Oscillatory Motion

2010-02-09T15:55:00.000-08:00

A piston in a stream engine executes simple harmonic motion. Given that the maximum displacement of the piston from its center-line is $\pm7$ cm, and that the mass of the piston is $4$ kg, find the maximum velocity of the piston when the steam engine is running at 4000 rev./min. What is the maximum acceleration?

Solution:

Given:
$A=7$ cm $=0.07$ m
$m=4$ kg
$f=4000$ rev./min $=4000/60$ rev/s $=66.6$ Hz

Find:
$v_{max}$ and $a_{max}$

Formula:
$\omega=2\pi f$
$v_{max}=v_0=A\omega$
$a_{max}=a_0=A\omega^2$

Calculation:
$\omega=2\pi f=(6.28)(66.6)=418.66$ rad/s
$v_0=A\omega=(0.07)(418.25)=29.3$ m/s
$a_0=A\omega^2=(0.07)(418.25)^2=1.224\times10^4$ m/s$^2$

Classical Mechanics - Work and Energy

2010-01-26T08:57:00.000-08:00

(GR8677 #34)

Solution:
$\vec{F}=-\nabla U \Rightarrow F_x=-\frac{\partial U}{\partial x}$
$U=kx^4$
$F_x=-\frac{\partial kx^4}{\partial x}=-4kx^3$

Answer: B

Classical Mechanics - Newton's Law

2010-01-26T08:50:00.000-08:00

A rock is thrown vertically upward with initial speed $v_0$. 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?
A. The acceleration of the rock is always equal to $g$.
B. The acceleration of the rock is equal to $g$ only at the top of the flight.
C. The acceleration of the rock is always less than $g$.
D. The speed of the rock upon return to its starting point is $v_0$.
E. The rock can attain a terminal speed greater than $v_0$ before it returns to its starting point.

(GR8677 #1)

Solution:
The equation of motion with no friction force:
$ma=-mg$
With friction force, $F_f \propto -v$:
$ma=-mg-kv$

(A) FALSE
$ma=-mg-kv \Rightarrow a=-g-\frac{kv}{m}$

(B) TRUE
At the top of the flight, $v=0$, therefore
$a=-g-\frac{k \cdot 0}{m}=-g$

(C) FALSE
$v>0$ (moving up) $\Rightarrow a=-g-\frac{kv}{m}<-g$
$v<0$ (moving down)$ \Rightarrow a=-g-\frac{k(-v)}{m}>-g$

(D) FALSE
Since there is a friction force, energy is not conserved, so it’s initial and final speed is not the same.

(E) FALSE
The frictional force slows down the object, so its speed at time $t$ has to be less than its initial speed, $v_0$.

Answer: B

Classical Mechanics - Newton's Law

2010-01-26T04:14:00.000-08:00

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 #8)

Solution:
$2as=v^2_t- v^2_0 $
$v_0=0 \Rightarrow a=\frac{v^2_t}{2s}$
$F=ma=m\frac{v^2_t}{2s}=5\frac{10^2}{2 \cdot 0.025}=10,000$ N

Answer: D

Classical Mechanics - Kinematics

2010-01-26T04:04:00.000-08:00

A stone is thrown at an angle $45^o$ above the horizontal x-axis in the +x-direction. If air resistance is ignored, which of the velocity versus time graphs shown above best represent $v_x$ versus $t$ and $v_y$ versus $t$, respectively?

     $v_x$ versus $t$       $v_y$ versus $t$
A.          I                                  IV
B.          II                                   I
C.          II                                  III
D.          II                                  V
E.         IV                                  V

(GR0177 #24)

Solution:
An object such as a cannonball is fired at an angle to the ground:
- The shape of the path or its trajectory is parabolic
- In the x-direction, the object travels at constant speed
- In y-direction, we treat the object as if it were tossed directly upward (positive velocity) and then returned to the ground (negative velocity $\rightarrow $ the object reverses the direction of its motion).

Answer: C

Classical Mechanics - Kinematics

2010-01-26T03:43:00.000-08:00

A ball is thrown horizontally from the top of a tower 40 m high. The ball strikes the ground at a point 80 m from the bottom of the tower. Find the angle that the velocity vector makes with the horizontal just before the ball hits the ground.
A. $315^o$
B. $41^o$
C. $0^o$
D. $90^o$
E. $82^o$

(GRE REA #23)

Solution:

Given:
$y=40$m
$x=80$m

Find: $\alpha$

Formula:
$2as=v^2_t-v^2_0$
$s=v_0t+\frac{1}{2}at^2$

Calculation:
Vertical components (y-axis):
$s=y, a=g, v_{0y}=0, v_t=v_y$
$2gy=v^2_y \Rightarrow v_y=\sqrt{2\cdot10\cdot40}=10\sqrt8$
$y=\frac{1}{2}gt^2 \Rightarrow t=\sqrt{2\cdot40/10}=\sqrt8$

Horizontal components (x-axis):
$v_x=x/t \Rightarrow v_x=80/\sqrt8$
$\tan\alpha=\frac{v_y}{v_x}=10\sqrt8\frac{80}{\sqrt8}=1 \Rightarrow \alpha=315$

Answer: A

Classical Mechanics - Kinematics

2010-01-26T03:36:00.000-08:00

Suppose that a man jumps off a building 202 m high onto cushions having a total thickness of 2 m. If the cushions are crushed to a thickness of 0.5 m, what is the man's average acceleration as he slows down?
A. $g$
B. $133 g$
C. $5 g$
D. $2 g$
E. $266 g$

(GRE REA #22)

Solution:
Given:
$h_B=202$m
$h_C=2$ m
$h_T=0.5$ m

Find: $a$

Formula: $2ah=v^2_t-v^2_0$

Calculation:
$2g(h_B-h_C)=v^2_t-v^2_0$
$2g(200)=v^2_t-0 \Rightarrow v^2_t=400g=v^2_0$ when he reached the cushion
$2a(h_T-h_C)=-v^2_0$
$2a(0.5-2)=-400g \Rightarrow =133g$

Answer: B

Classical Mechanics - Kinematics

2010-01-26T03:31:00.000-08:00

Consider that a coin is dropped into a wishing well. You want to determine the depth of the well from the time $T$ between releasing the coin and hearing it hit the bottom. If the speed of sound is $v_s = 330$ m/s, and $T = 2.059$ s, what is the depth $h$ of the well?
A. 20.77 m
B. 19.60 m
C. 23,564 m
D. 18.43 m
E. 39.20 m

(GRE REA #5)

Solution:
$T=t_1+t_2$
$t_1=$ the time for the coin to reach the bottom
$t_2=$ the time for the sound to travel
$h=\frac{1}{2}gt^2_1$
$v_s=\frac{h}{t_2}\Rightarrow t_2=\frac{h}{v_s}$
$h=\frac{1}{2}g(T-t_2)^2=\frac{1}{2}g\bigg(T-\frac{h}{v_s}\bigg)^2$
$h=\frac{1}{2}gT^2+\frac{1}{2}g\bigg(\frac{h}{v_s}\bigg)^2-gT\frac{h}{v_s}$
$h^2-\bigg(2v_sT+\frac{2v^2_s}{g}\bigg)h+v^2_sT^2=0$

$v_s=330$ m/s
$T=2.059$ s

$h^2-23,583.4h+461,679.5=0$
$h=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=19.6$ m

Answer: B

Classical Mechanics - Newton's Law

2010-01-26T03:25:00.000-08:00

The figure above represents a point mass $m$ attached to the ceiling by a cord of fixed length $l$. If the point mass moves in a horizontal circle of radius $r$ with uniform angular velocity $\omega$, the tension in the cord is
A. $mgr/l$
B. $mg\cos(\theta/2)$
C. $\frac{m\omega r}{\sin(\theta/2)}$
D. $m(\omega^2r^2+g^2)^{1/2}$
E. $m(\omega^4r^2+g^2)^{1/2}$

(GR8677 #37)

Solution:

Horizontal components:
$T\sin(\theta/2)=F_a=mv^2/r=m\omega^2r$

Vertical components:
$T\cos(\theta/2)=mg$

The magnitude of $T$:
$T=\sqrt{T^2\cos^2(\theta/2)+T^2\sin^2(\theta/2)}$
$=m(\omega^4r^2+g^2)^{1/2}$

Answer: E

Classical Mechanics - Newton's Law

2010-01-25T19:13:00.000-08:00

Find the tension $T_2$ in cord 2 for the system drawn below. The system is in equilibrium.
A. 19.6 N
B. 39.2 N
C. 0 N
D. 17.0 N
E. 33.9 N

Solution:
$T_1=mg=2\cdot9.8=19.6 N$
$T_1-T_3\cos60=0 \Rightarrow T_3=T_1/\cos60=39.2 N$
$T_2-T_3\sin60=0 \Rightarrow T_2=T_3\sin60=33.9 N$

Answer: E

Classical Mechanics - Newton's Law

2010-01-25T18:57:00.000-08:00

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 #7)

Solution:
$F_x=F-T\sin\theta=0\Rightarrow T\sin\theta =F$
$F_y=T\cos\theta-mg=0\Rightarrow T\cos\theta =mg$
$\frac{T\sin\theta}{T\cos\theta}=\tan\theta=\frac{F}{mg}=\frac{10 N}{1\cdot 10 N}=0.5$
$\theta=\arctan 0.5$

Answer: A

Quantum Mechanics - Bohr Theory

2010-01-22T20:49:00.000-08:00

The energy required to remove both electrons from the helium atom in its ground state is 79.0 eV. How much energy is required to ionize helium (i.e. to remove one electron)?
A. 24.6 eV
B. 39.5 eV
C. 51.8 eV
D. 54.4 eV
E. 65.4 eV
GR0177 #18

(GR0177 #18)

Solution:

Helium has 2 protons, so its atomic number $\Rightarrow Z=2$
In ground state $\Rightarrow n=1$

The ground state energy of the helium atom (i.e. the minimum energy required to remove both electrons leaving behind He++ ion) is $E=79$ eV

Since He+ is indeed a one-electron atom, just like Hydrogen, the ionization potential is given by
$E=Z^2E_1=2^2(13.6)$ eV $=54.4$ eV
where $E_1$ is the energy of the ground state of Hydrogen

Thus, the ionization potential of neutral He (i.e. the energy required to remove a single electron from a helium atom in the ground state) is $79-54.4 = 24.6$ eV, and this is confirmed experimentally.

Answer: A

Quantum Mechanics - Free Particle

2010-01-20T19:22:00.000-08:00

A free particle with initial kinetic energy $E$ and de Broglie wavelength $\lambda$ enters a region in which it has potential energy $V$. What is the particle’s new de Broglie wavelength?
A. $\lambda(1+E/V)$
B. $\lambda(1-V/E)$
C. $\lambda(1-E/V)^{-1}$
D. $\lambda(1+V/E)^{1/2}$
E. $\lambda(1-V/E)^{-1/2}$

(GR0177 #46)

Solution:

The initial kinetic energy of free particle:
$E=\frac{p^2}{2m}\Rightarrow p=\sqrt{2mE}$

and, de Broglie wavelength:
$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2mE}}$

Energy of the particle when it enters the region:
$E_f=E-V$

So, its wavelength becomes:
$\lambda_f=\frac{h}{\sqrt{2mE_f}}=\frac{h}{\sqrt{2m(E-V)}}$

$\lambda_f^2=\frac{h^2}{2mE}\bigg(\frac{E}{E-V}\bigg)=\frac{h^2}{2mE}\bigg(\frac{1}{1-V/E}\bigg)$

$\lambda_f=\frac{h^2}{\sqrt{2mE}}\bigg(\frac{1}{1-V/E}\bigg)^{1/2}=\lambda(1-V/E)^{-1/2}$

Answer: E

Atomic Physics - Atomic Spectra

2010-01-18T11:52:00.001-08:00

In the hydrogen spectrum, the ratio of the wavelength for Lyman-radiation ($n = 2$ to $n = 1$) to Balmer-radiation ($n = 3$ to $n = 2$) is
A. 5/48
B. 5/27
C. 1/3
D. 3
E. 27/5

(GR0177 #21)

Solution:
Rydberg formula for Hydrogen
$\frac{1}{\lambda_{vac}}=R_H\bigg(\frac{1}{n^2_1}-\frac{1}{n^2_2}\bigg)$

where
$\lambda_{vac}=$ the wavelength of the light emitted in vacuum
$R_H=$ Rydberg constant for Hydrogen
$n_1$ and $n_2$ are integers such that $n_1$<$n_2$

For Lyman-radiation $(n = 2 \rightarrow n = 1)$:
$\frac{1}{\lambda_L}=R_H\bigg(1-\frac{1}{4}\bigg)=\frac{3}{4}R_H$

For Balmer-radiation $(n = 3 \rightarrow n = 2)$:
$\frac{1}{\lambda_B}=R_H\bigg(\frac{1}{4}-\frac{1}{9}\bigg)=\frac{5}{36}R_H$

THe Ratio
$\frac{\lambda_L}{\lambda_B}=\frac{4}{3}\cdot\frac{5}{36}=\frac{5}{27}$

Answer: B

Quantum Mechanics - Bohr Theory

2010-01-18T10:58:00.000-08:00

The energy levels of the hydrogen atom are given in terms of the principal quantum number $n$ and a positive constant $A$ by the expression
A. $A\bigg(n+\frac{1}{2}\bigg)$
B. $A(1-n^2)$
C. $A\bigg(-\frac{1}{4}+\frac{1}{n^2}\bigg)$
D. $An^2$
E. $-\frac{A}{n^2}$

(GR8677 #19)

Solution:

Energy level of hydrogen atom is $E_n=-\frac{Z^2}{n^2}13.6$ eV

Answer: E

Atomic Physics - Quantum Number

2010-01-18T10:17:00.000-08:00

The ground state electron configuration for Phosphorus, which has 15 electrons, is
A. $1s^2 2s^2 2p^6 3s^1 3p^4$
B. $1s^2 2s^2 2p^6 3s^2 3p^3$
C. $1s^2 2s^2 2p^6 3s^2 3d^3$
D. $1s^2 2s^2 2p^6 3s^1 3d^4$
E. $1s^2 2s^2 2p^6 3p^2 3d^3$

(GR0177 #17)

Solution:

For phosphorus, which has 15 electrons, the right configuration $1s^2 2s^2 2p^6 3s^2 3p^3$

Answer: B

Atomic Physics - Quantum Number

2010-01-18T10:01:00.000-08:00

The configuration of the Potassium atom in its ground state is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1$. Which of the following statement about Potassium is true:
A. Its $n=3$ shell is completely filled
B. Its $4s$ subshell is completely filled
C. Its least tightly bound electron has $l=4$
D. Its atomic number is 17
E. Its electron charge distribution is spherically symmetrical

(GR8677 #30)

Solution:

A. FALSE.
$n=3 \Rightarrow l=0,1,2 \Rightarrow s,p,d$-subshells, but the configuration does not have $d$ subshell.

B. FALSE.
$4s$ subshell $\Rightarrow$ for $s$-subshell total electron should be 2 not 1.

C. FALSE
$l=4$ corresponds to $g$-subshell, but the configuration does not have $g$-subshell.

D. FALSE
The number of electrons in the configuration is 2+2+6+2+6+1=19 not 17.

E. TRUE
The single $s$-orbital $(l=0)$ are shaped like spheres. For $n=1$ the sphere is "solid" (it is most dense at the center and fades exponentially outwardly), but for $n=2$ or more, each single $s$-orbital is composed of spherically symmetric surfaces which are nested shells (i.e., the "wave-structure" is radial, following a sinusoidal radial component as well). Since the outermost electron shell (valence shell) of the Potassium atom is $4s^1 \Rightarrow n>1$ and $l=0 \Rightarrow$ spherically symmetric.

Also, note that Potassium has one outer electron, like Hydrogen. So, it will also have a spherically symmetrical charge distribution

Answer: E

Atomic Physics - Quantum Number

2010-01-18T09:42:00.000-08:00

In a $^3S$ state of the helium atom, the possible values of the total electronic angular momentum quantum number are
A. 0 only
B. 1 only
C. 0 and 1 only
D. 0, 1/2, and 1
E. 0, 1, and 2

(GR9277 #31)

Solution:
Spectroscopic notation: $N^{2s+1}L_j$
$N$ = principal quantum number and will often be omitted
$s$ = total spin quantum number $\Rightarrow 2s+1$ = the number of spin states
$L$ = orbital angular momentum quantum number, $l$ but is written as $S$, $P$, $D$, $F$, … for $l = 0,1,2,3, …$
$j$ = total angular momentum quantum number, $j = l + s$

For $^3S$:
$S \Rightarrow l = 0$
$3 \Rightarrow 3 = 2s+1 \Rightarrow s = 1$
$j = l + s = 0 + 1 = 1$

Answer: B

Quantum Mechanics - Bohr Theory

2009-10-19T14:38:00.001-07:00

Given that the binding energy of the hydrogen atom ground state is $E_0=13.6$eV, the binding energy of the $n=2$ state of positronium (positron-electron system) is
A. $8E_0$
B. $4E_0$
C. $E_0$
D. $E_0/4$
E. $E_0/8$

(GR9277 #30)

Solution:

Positronium: an exotic atom consisting of an electron (e-) and its anti-particle of equal mass, a positron (e+). Theoretical physicists like Positronium as a tool to model atomic interactions and test quantum theories.

Hydrogen (H) vs Positronium (Ps)

Positron Research Group St. Olaf College, Northfield, MN

The orbit of the Ps atom (e- and e+) and the set of energy levels is similar to that of the H atom. However, Ps is less massive than H.

H consists of electron travels around a much heavier proton and its center of mass, as in our standard atom, is somewhere close to the center of mass of the proton. Ps' center of mass is somewhere in between e- and e+.

To Calculate Ps effective/reduced mass:

where
= reduced mass
and = masses of the positon and electron respectively.

Since

Therefore, the frequencies associated with the spectral lines are less than half of those of the corresponding H lines.

Bohr's Theory of H atom:
eV

For H atom at ground state, eV

For Positronium at $n=2$ energy level

Notes:

- $Z$ = the number of positive charges in the nucleus. In typical cases, this represents the number of protons. In Ps, it is a positron that contributes the charge ($Z=1$).

- For any value of the radius, the electron and the positron are each moving at half the speed around their common center of mass, and each has only one fourth the kinetic energy.

- The most exciting feature of Ps is the fact that it can spontaneously explode or self-annihilate, converting the mass of the e+ and e- into pure energy in the form of gamma rays. If we had one milligram of Ps, the energy released from its annihilation would be equal to 25 kilotons of dynamite. This self-annihilation feature of Ps means it is extremely short-lived, with an average lifetime on the order of nanoseconds (billionths of a second).

Answer: E

Quantum Mechanics - Finite Potential Well

2009-10-05T16:01:00.000-07:00

An attractive, one-dimensional square well has depth as shown above. Which of the following best shows a possible wave function for a bound state?

(GR9277 #29)

Solution:
1D square well has depth finite potential well.

Wave function of a particle in finite potential well should have these following properties:
as or get further into regions where the classical particle cannot penetrate at all due to its inadequate energy
is always a decaying exponential function
must be continuous and differentiable

Therefore:

(A) FALSE
does not go to zero as

(C) FALSE
is not continuous

(D) and (E) FALSE
are not decaying exponential functions

Answer: B

Quantum Mechanics - Probability

2009-10-05T14:59:00.000-07:00

A system is known to be in the normalized state described by the wave function
$\psi(\theta,\phi)=\frac{1}{\sqrt{30}}(5Y^3_4+Y^3_6-2Y^0_6)$, where the $Y^m_l(\theta,\phi)$ are the spherical harmonics. The probability of finding the system in a state with azimuthal orbital quantum number $m=3$ is
A. 0
B. 1/15
C. 1/6
D. 1/3
E. 13/15

(GR9277 #28)

Solution:
For the state $|\alpha\rangle = \sum_i{c_i}|a^{(i)}\rangle$
Probability is simply $P= \sum_i|{c_i}|^2$

States with $m=3 \Rightarrow Y^3_l \Rightarrow Y^3_4$ and $Y^3_6$
with $c_1=\frac{5}{\sqrt{30}}$ and $c_2=\frac{1}{\sqrt{30}}$
Therefore,
$P=\bigg(\frac{5}{\sqrt{30}}\bigg)^2+\bigg(\frac{1}{\sqrt{30}}\bigg)^2=\frac{26}{30}=\frac{13}{15}$
Answer: E

Optic - Telescope

2009-09-11T05:26:00.000-07:00

Two stars are separated by an angle of $3\times{10^{-5}}$ radian. What is the diameter of the smallest telescope that can resolve the two stars using visibe light ($\lambda\cong{600}$ nm)?
A. 1 mm
B. 2.5 cm
C. 10 cm
D. 2.5 m
E. 10 m

(GR0177 #13)

Solution:

Rayleigh telescope resolution limit:
$\sin{\theta} = 1.22\frac{\lambda}{D}$
where
$\theta =$ the angular resolution
$\lambda =$ the wavelength of light
$D =$ the diameter of the lens' aperture.

From the problem:
$\lambda = 600$ nm = $600\times{10^{-9}}$ m
$\theta = 3\times{10^{-5}}$ rad
Since $\theta\ll$ assume $\sin{\theta}\approx\theta$
$\Rightarrow D=1.22\frac{\lambda}{\theta}$
$D=\frac{1.22\times{6}}{3}{10^{-2}}$
$D\approx{2.5}\times10^{-2}}$ m $\approx 2.5$ cm

Answer: B

Optic - Telescope

2009-09-10T19:53:00.000-07:00

A collimated laser beam emerging from a commercial HeNe laser has a diameter of about 1 millimeter. In order to convert this beam into a well-collimated beam of diameter 10 millimeters, two convex lenses are to be used. The first lens is of focal length 1.5 centimeters and is to be mounted at the output of the laser. What is the focal length, $f$, of the second lens and how far from the first lens should it be placed?
A. f = 4.5 cm, Distance = 6.0 cm
B. f = 10 cm, Distance = 10 cm
C. f = 10 cm, Distance = 11.5 cm
D. f = 15 cm, Distance = 15 cm
E. f = 15 cm, Distance = 16.5 cm

(GR9677 #58)

Solution:
Since two convex lenses are involved, we can treat this system as a telescope.

The magnification of the telescope:
$M=\frac{f_o}{f_e}$
The optical path length between objective and eyepiece:
$d=f_o+f_e$
where:
$f_o =$ focal length of the objective
$f_e =$ focal length of the eyepiece

From the problem:
$M=10$ and the first lens to be mounted at the OUTPUT of the laser $\Rightarrow f_e=1.5$ cm (beside, from the equation: $f_o > f_e$. There is no way $f_e > f_o$)

Therefore,
$f_o=M\times{f_e}=10\times{1.5}=15$ cm
$d=f_e+f_o=1.5+15=16.5$ cm

Answer: E

Optic - Interference

2009-09-04T07:51:00.000-07:00

Light from a laser falls on a pair of very narrow slits separated by 0.5 micrometer, and bright fringes separated by 1.0 millimeter are observed on a distant screen. If the frequency of the laser light is doubled, what will be the separation of bright fringes?
A. 0.25 mm
B. 0.5 mm
C. 1.0 mm
D. 2.0 mm
E. 2.5 mm

(GR0177 #70)

Solution:
Given
$d$ = 0.5 micrometer
$x_1$ = 1.0 millimeter
$f_2 = 2f_1$
Find $x_2$

Double-slit interference equation: $d\sin{\theta}=m\lambda$
with $\sin{\theta} = \frac{x}{L}$ and $\lambda=\frac{c}{f}$
$\Rightarrow d\frac{x}{L}=m\frac{c}{f}$
$\Rightarrow x \propto \frac{1}{f}$
$\Rightarrow \frac{x_1}{x_2} = \frac{f_2}{f_1}$
$\Rightarrow \frac{1}{x_2} = \frac{2f_1}{f_1} = 2$
$\Rightarrow x_2 = \frac{1}{2}$
Answer: B

Optic - Telescope

2009-09-03T09:16:00.000-07:00

The angular separation of the two components of a double star is 8 microradians, and the light from the double star has a wavelength of 5500 A. The smallest diameter of a telescope mirror that will resolve the double star is most nearly
A. 1 mm
B. 1 cm
C. 10 cm
D. 1 m
E. 100 m

(GR9277 #68)

Solution:
Rayleigh telescope resolution limit:
$\sin{\theta} = 1.22\frac{\lambda}{D}$
where
$\theta$ = the angular resolution
$\lambda$ = the wavelength of light
$D =$ the diameter of the lens' aperture.

From the problem:
$\lambda = 5500\times{10^{-10}}$ m
$\theta = 8\times{10^{-6}}$ rad
Since $\theta\ll$, assume $\sin{\theta}\approx\theta$
$\Rightarrow D=1.22\frac{\lambda}{\theta}$
$D=\frac{1.22\times{5500}}{8}{10^{-4}}$
$D\approx{8}\times10^{-2}}$ m $\approx 10$ cm

Answer: C

Quantum Mechanics - Infinite Potential Well

2009-09-02T08:31:00.001-07:00

Calculate the probability of finding the ground state particle in a box between $x$ and $x+\frac{1}{3}L$.

Solution:

The wavefunction for 1D infinite potential box is

$\psi_n(x)=\sqrt{\frac{2}{L}}\sin\bigg(\frac{n\pi{x}}{L}\bigg)$

Probability
$P=\int|\psi|^2dx$

$P=\int_{0}^{\frac{L}{3}}\bigg|\sqrt{\frac{2}{L}}\sin\bigg(\frac{n\pi{x}}{L}\bigg)\bigg|^2dx=\frac{2}{L}\int_{0}^{\frac{L}{3}}\sin^2\bigg(\frac{n\pi{x}}{L}\bigg)dx$

Double angle formula:
$\cos2\alpha=1-2\sin^2{\alpha}$

And let $\alpha=\frac{n\pi}{L}, n=1$ (for ground/the lowest energy state)

$P=\frac{2}{L}\int_{0}^{\frac{L}{3}}\frac{1}{2}(1-\cos2\alpha) dx$
$=\frac{x}{L}\bigg|_0^{\frac{L}{3}}-\frac{1}{L}\int_{0}^{\frac{L}{3}}\cos2\alpha{x} dx$
$=\frac{1}{3}-\frac{1}{2\pi}\sin\frac{2\pi}{3}=0.19$

Optics - Thin Film

2009-08-31T05:17:00.001-07:00

Blue light of wavelength 480 nanometers is most strongly reflected off a thin film of oil on a glass slide when viewed near normal incidence. Assuming that the index of refraction of the oil is 1.2 and that of the glass is 1.6, what is the minimum thickness of the oil film (other than zero)?
A. 150 nm
B. 200 nm
C. 300 nm
D. 400 nm
E. 480 nm

(GR0177 #69)

Solution:

Air to Oil, $n_a < n_g \Rightarrow \Delta{a} = \lambda/2$
Oil to Glass, $n_o < n_g \Rightarrow \Delta{b} = 2t+\frac{\lambda}{2}$
$\Delta = \Delta{b}-\Delta{a} = 2t+\frac{\lambda}{2}-\frac{\lambda}{2}=2t$

Blue Light $\Rightarrow$ Constructive interference:
$\Delta = m\lambda$
$\Rightarrow m\lambda = 2t$

with $\lambda_{oil}= \frac{\lambda_{air}}{n_{oil}}=\frac{480}{1.2}=400$ nm
and choose $m=1$
$\Rightarrow t = \frac{\lambda_{oil}}{2}=\frac{400}{2}=200$ nm

Click HERE for more info on Thin Film.

Answer: B

Optics - Thin Film

2009-08-27T20:17:00.001-07:00

Consider two horizontal glass plate with a thin film of air between them. For what values of the thickness of the film of air will the film, as seen by reflected light, appear bright if it is illuminated normally from above by blue light of wavelength 488 nanometers?
A. 0, 122 nm, 244 nm
B. 0, 122 nm, 366 nm
C. 0, 244 nm, 488 nm
D. 122 nm, 244 nm, 366 nm
E. 122 nm, 366 nm, 610 nm

(GR9677 #82)

Solution:
Glass to Air, $n_g > n_a \Rightarrow \Delta{a} = 0$
Air to Glass, $n_a < n_g \Rightarrow \Delta{b} = 2t+\frac{\lambda}{2}$
$\Delta = \Delta{b}-\Delta{a} = 2t+\frac{\lambda}{2}-0$

Blue Light $\Rightarrow$ Constructive interference:
$\Delta = m\lambda$
$\Rightarrow m\lambda = 2t+\frac{\lambda}{2}$
$\Rightarrow 2t = \bigg(m-\frac{1}{2}\bigg)\lambda$
$\Rightarrow t = \frac{1}{2}\bigg(m-\frac{1}{2}\bigg)\lambda$
with $\lambda_{air} = 488$ nm and $m=0,1,2,3...$
$m=1 \Rightarrow t = \frac{1}{4}\lambda = 122$ nm
$m=2 \Rightarrow t = \frac{3}{4}\lambda = 366$ nm
$m=3 \Rightarrow t = \frac{5}{4}\lambda = 610$ nm

Click HERE for more info on Thin Film.

Answer: E

Optic - Thin Film

2009-08-27T18:59:00.001-07:00

Lenses are often coated with thin films of transparent substances like MgF2 ($n=1.38$) in order to reduce the reflection from the glass surface, using interference. How thick a coating is needed to produce a minimum reflection at the center of the visible spectrum ($550$ nm)?

Solution:

Air to Coating Layer, $n_a < n_l \Rightarrow \Delta{a} = \lambda/2$
C.Layer to Glass, $n_l < n_g \Rightarrow \Delta{b} = 2t+\frac{\lambda}{2}$
$\Delta = \Delta{b}-\Delta{a} = 2t+\frac{\lambda}{2}-\frac{\lambda}{2}=2t$

To reduce reflection $\Rightarrow$ Destructive interference:
$\Delta = \bigg(m+\frac{1}{2}\bigg)\lambda$
$\Rightarrow 2t = \bigg(m+\frac{1}{2}\bigg)\lambda$
with $\lambda_{layer} = \frac{\lambda_{air}}{n_{layer}}=\frac{550}{1.38}$ nm
Minimum $m=0$
$\Rightarrow t = \frac{1}{2}\bigg(0+\frac{1}{2}\bigg)\frac{550}{1.38} = 100$ nm

Click HERE for more info on Thin Film.

Optic - Thin Film

2009-08-26T19:08:00.001-07:00

White light in air shines on an oil film that floats on water. When looking straight down at the film, the reflected light is red, with a wavelength of 636 nm. What is the minimum possible thickness of the film?

Solution:

Air to Oil, $n_a < n_o \Rightarrow \Delta{a} = \lambda/2$
Oil to Water, $n_o > n_w \Rightarrow \Delta{b} = 2t$
$\Delta = \Delta{b}-\Delta{a} = 2t-\frac{\lambda}{2}$

Red Light $\Rightarrow$ Constructive interference:
$\Delta = m\lambda$
$\Rightarrow m\lambda = 2t-\frac{\lambda}{2}$
$\Rightarrow 2t = \bigg(m+\frac{1}{2}\bigg)\lambda$
with $\lambda_{oil} = \frac{636}{1.5}=424$ nm and $m=0$ (minimum thickness)
$\Rightarrow t = \frac{1}{2}\bigg(0+\frac{1}{2}\bigg)424 = 106$ nm

Click HERE for more info on Thin Film.

Optic - Thin Film

2009-08-26T08:38:00.001-07:00

It is necessary to coat a glass lens with a non-reflecting layer. If the wavelength of the light in the coating is $\lambda$, the best choice is a layer of material having an index of refraction between those of glass and air and a thickness of
A. $\lambda/4$
B. $\lambda/2$
C. $\lambda/\sqrt{2}$
D. $\lambda$
E. $1.5\lambda$

(GR8677 #73)

Solution:

$n_a < n_l < n_g$

Air to Layer, $n_a < n_l \Rightarrow \Delta{a} = \lambda/2$
Layer to Glass, $n_l < n_g \Rightarrow \Delta{b} = 2t+\lambda/2$
$\Delta = \Delta{b}-\Delta{a} = 2t+\frac{\lambda}{2}-\frac{\lambda}{2}=2t$

Non-reflecting $\Rightarrow$ destructive interference:
$\Delta = \bigg(m+\frac{1}{2}\bigg)\lambda$
with $\lambda_{layer} = \lambda$ and $m=0$ (The thinnest the better)
$\Rightarrow 2t=\bigg(0+\frac{1}{2}\bigg)\lambda$
$\Rightarrow t = \frac{1}{4}\lambda$

Click HERE for more info on Thin Film.

Answer: A

Thermodynamics - Carnot Engine/Cycle

2009-08-26T05:55:00.001-07:00

In the cycle shown above, $KL$ and $NM$ represent isotherms, while $KN$ and $LM$ represent reversible adiabats. A system is carried through the Carnot cycle $KLMN$, taking in heat $Q_2$ from the hot reservoir $T_2$ and releasing heat $Q_1$ to the cold reservoir $T_1$. All of the following statements are true, EXCEPT:
A. $Q_1/T_1=Q_2/T_2$
B. The entropy of the hot reservoir decreases
C. The entropy of the system increases
D. The work $W$ done is equal to the net heat absorbed, $Q_2-Q_1$
E. The efficiency of the cycle is independent of the working substance.

(GR8677 #95)

Solution:
The efficiency of Carnot engine:
$\eta = \frac{W}{Q_2}=1-\frac{T_1}{T_2}$
where
$W$ = work done = $Q_2-Q_1 \Rightarrow$ (D) TRUE
$Q_2 = Q$ Hot/input
$T_2 = T$ Hot/high (in Kelvin)
$T_1 = T$ Cold/low (in Kelvin)
$\Rightarrow$ (E) TRUE

Since $W = Q_2-Q_1$
$\Rightarrow \eta = 1-\frac{Q_1}{Q_2}=1-\frac{T_1}{T_2}$
$\Rightarrow \frac{Q_1}{T_1}=\frac{Q_2}{T_2}
\Rightarrow$ (A) TRUE

The hot reservoir has decreasing entropy because it gets cooler as the cycle proceeds $\Rightarrow$ (B) TRUE

In order to approach the Carnot efficiency, the processes involved in the heat engine cycle must be reversible and involve no change in entropy $\Rightarrow$ (C) FALSE

Answer: C

Statistical Mechanics - Equipartition Law

2009-08-25T10:10:00.000-07:00

A three-dimensional harmonic oscillator is in thermal equilibrium with a temperature reservoir at temperature $T$. The average total energy of oscillator is
A. $\frac{1}{2}kT$
B. $kT$
C. $\frac{3}{2}kT$
D. $3kT$
E. $6kT$

(GR0177 #5)

Solution:
The equipartition theorem: each independent degree of freedom (DoF) associated with a quadratic term in the energy possesses an average energy $\frac{1}{2}kT$ in thermal equilibrium at temperature $T$.

3D harmonic oscillator has 6 DoF consist of 3 components of momentum (kinetic Energy) and 3 components of position (potential energy)
$\Rightarrow E=6\times\frac{1}{2}kT=3kT$

Answer: D

Statistical Mechanics - Equipartition Law

2009-08-25T08:19:00.000-07:00

In a gas of $N$ diatomic molecules, two possible models for a classical description of a diatomic molecule are:

Which of the following statements about this gas is true?
A. Model I has a spesific heat $c_v=\frac{3}{2}Nk$
B. Model II has a smaller spesific heat than Model I
C. Model I is always correct
D. Model II is always correct
E. The choice between Models I and II depends on the temperature

(GR8677 #87)

Solution:
From the Equipartition of Energy Law, molar specific heat at constant temperature: $c_v=\frac{f}{2}R$
where $f$ = Degree of Freedom (DoF)

Model I is diatomic molecules in low temperature $\Rightarrow$ has 5 DoF (3 translational + 2 rotational) $\Rightarrow c_v=\frac{5}{2}R$

Model II is diatomic molecules in high temperature $\Rightarrow$ has 7 DoF (3 translational + 2 rotational + 2 vibrational) $\Rightarrow c_v=\frac{7}{2}R$ (see this problem).

Answer: E

Statistical Mechanics - Equipartition Law

2009-08-25T05:19:00.000-07:00

For an ideal diatomic gas in thermal equilibrium the ratio of the molar heat capacity at constant volume at very high temperatures to that at very low temperatures is equal to
A. 1
B. 5/3
C. 2
D. 7/3
E. 3

(GR9677 #79)

Solution:
From the Equipartition of Energy Law, molar specific heat at constant temperature: $c_v=\frac{f}{2}R$
where $f$ = Degree of Freedom (DoF)

At very low temperature, diatomic molecule only has 3 DoF related to translational motion.

At very high temperature, diatomic molecule has 7 DoF related to translational (3 DoF), rotational (2 DoF) and vibrational (2 DoF) motion (see this problem).

$\Rightarrow \frac{c_{v(T\gg)}}{c_{v(T\ll)}}=\frac{7}{3}$
Answer: D

Statistical Mechanics - Blackbody Radiation

2009-08-24T19:14:00.001-07:00

A black body at temperature $T_1$ radiates energy at a power level of 10 milliwatts (mW). The same blackbody, when at a temperature $2T_1$, radiates energy at a power level of
A. 10 mW
B. 20 mW
C. 40 mW
D. 80 mW
E. 160 mW

(GR8677 #46)

Solution:
Stefan–Boltzmann law (energy radiated by a blackbody): $P\propto{T^4}$
$\Rightarrow \frac{P_1}{P_2}=\frac{T_1^4}{T_2^4}$
$\frac{10}{P_2}=\frac{T_1^4}{(2T_1)^4}$
$\frac{10}{P_2}=\frac{1}{16} \Rightarrow P_2=160$
Answer: E

Electromagnetism - Maxwell's Equations

2009-08-23T11:23:00.000-07:00

Starting with the equation for $\nabla\times\vec{H}$, show that, in the absence of sources, $\vec{H}$ satisfies the wave equation.

Solution:
Maxwell's Equation: $\nabla\times\vec{H} = \vec{J} + \frac{\partial\vec{D}}{\partial{t}}$
$\vec{D}=\epsilon_0\vec{E}$ and in the absence of sources $\vec{J}=0$
$\Rightarrow \nabla\times\vec{H} = \epsilon_0\frac{\partial\vec{E}}{\partial{t}}$
$\nabla\times(\nabla\times\vec{H}) = \epsilon_0\frac{\partial}{\partial{t}}(\nabla\times\vec{E})$
$\nabla(\nabla\cdot\vec{H})-\nabla^2\vec{H}=\epsilon_0\frac{\partial}{\partial{t}}\bigg(-\frac{\partial\vec{B}}{\partial{t}}\bigg)=-\mu\epsilon_0\frac{\partial^2\vec{H}}{\partial{t}^2}$
$\nabla\cdot\vec{H}=0$ and $\mu\epsilon_0=1/c^$
$\Rightarrow \nabla^2\vec{H}=\frac{1}{c^2}\frac{\partial^2\vec{H}}{\partial{t}^2}$

Electromagnetism - Maxwell Equation

2009-08-23T08:24:00.000-07:00

Which one of the following Maxwell equations implies that there are no magnetic monopoles?
A. $\nabla\cdot\vec{E} = \rho/\epsilon_0$
B. $\nabla\cdot\vec{B} = 0$
C. $\nabla\times\vec{E} = -\frac{\partial\vec{B}}{\partial{t}}$
D. $\nabla\times\vec{B} = \mu_0\vec{J}+\mu_0\epsilon_0(\partial\vec{E}/\partial{t})$
E. Magnetic monopoles have recently been found

Solution:
$\nabla\cdot\vec{B} = 0$ implies that there are no magnetic monopoles.
For more info, see problem on Electromagnetism-Maxwell'Equation, Magnetic Monopole
Answer: B

Gravitational and Electric Potential

2009-08-22T21:03:00.000-07:00

Show the identical properties of gravitational and electric potential!

Solution:

Click to enlarge.

Electromagnetic - Electric Field and Potential

2009-08-22T20:31:00.000-07:00

A ring of radius $R$ has a uniform charge density with a total charge $Q$. Calculate the electric field and potential along the axis of the ring at a point $P$, a distance $x$ from the center of the ring, at any point $x\gg{R}$, and at the center of the ring.

Solution:

Electric Field: $d\vec{E}=k\frac{dQ}{r^2}\hat{r}$

A. At point $P$:
From the picture above: $d\vec{E}=k\frac{dQ}{r^2}\cos{\theta}\hat{i}$
The coefficient of $\hat{j}$ which is the $y$-components of the field is zero, since $dE$ and $dE$' in $y$-axis vanish by symmetry.
With $\cos{\theta}=x/d$ and $d=\sqrt{x^2+R^2}$
$\rightarrow dE_x=k\frac{xdQ}{(x^2+R^2)^{3/2}}$
$E_x=k\frac{x}{(x^2+R^2)^{3/2}}\int{dQ}$
$E_x=k\frac{Qx}{(x^2+R^2)^{3/2}}$

B. For $x\gg{R}$:
$R\rightarrow{0}, E_x=k\frac{Q}{x^2}$, acts like a single charge.

C. At the center of the ring:
$E=0$, since $q$ spread uniformly in the ring or as mentioned above, since $dE$ and $dE$' in $y$-axis vanish by symmetry.

Electric Potential: $V=k\frac{Q}{r}$ (a scalar quantity)

A. At point $P$:
$V=k\int\frac{dq}{r}=\frac{k}{\sqrt{x^2+R^2}}\int{dq}=\frac{kQ}{\sqrt{x^2+R^2}}$

B. For $x\gg{R}$:
$R\rightarrow{0}, V=k\frac{Q}{x}$, acts like a single charge.

C. At the center of the ring:
$x=0, \rightarrow V=k\frac{Q}{R}$
Remember that $\vec{E}=-\nabla{V} \rightarrow$ if $E=0$, $V$ should be a constant.

Quantum Mechanics - Uncertainty Principle

2009-08-22T05:48:00.000-07:00

If a freely moving electron is localized in space to within $\Delta x_0$ of $x_0$, its wave function can be described by a wave packet $\psi(x,t)=\int_\infty^{-\infty}e^{i(kx-\omega t)}f(k)dk$, where $f(k)$ is peaked around a central value $k_0$. Which of the following is most nearly the width of the peak in $k$?
A. $\Delta k = 1/x_0$
B. $\Delta k = \frac{1}{\Delta x_0}$
C. $\Delta k = \frac{\Delta x_0}{x_0^2}$
D. $\Delta k = k_0\frac{\Delta x_0}{x_0}$
E. $\Delta k = \sqrt{k_0^2+(1/x_0)^2}$

(GR9277 #27)

Solution:
In quantum mechanics, the momentum $(p=\hbar{k})$ and position $(x)$ wave functions are Fourier transform pairs and the relation between $p$ and $x$ representations forms the Heisenberg uncertainty relation:
$\Delta{x}\Delta{k}\geq1 \Rightarrow \Delta k \geq \frac{1}{\Delta x}$
Answer: B

Specialized Topic - Radioactive Decay

2009-08-22T05:45:00.000-07:00

A radioactive nucleus decays, with the activity shown in the graph above. The half-life of the nucleus is
A. 2 min
B. 7 min
C. 11 min
D. 18 min
E. 23 min

(GR9277 #26)

Solution:
In radioactive decay, the half-life is the length of time after which there is a 50% chance that an atom will have undergone nuclear decay.

From the graph we find that at $t=0 \Rightarrow N_o=6\cdot{10^3}$ counts/min
50% of $N_o \Rightarrow N =3\cdot{10^3}$ counts/min

From the graph, $t$ at $N =3\cdot{10^3}$ counts/min is about in the middle of 5 and 10 minutes.

Answer: B

Quantum Mechanics - Probability

2009-08-21T14:47:00.001-07:00

A particle is in a state given at $t=0$ by:
$\psi = \frac{1}{3}u_0{(x)}+\frac{i\sqrt{2}}{3}u_1{(x)}-\frac{\sqrt{6}}{3}u_2{(x)}$
where $u_0, u_1, u_2$ are simple harmonic oscillator eigenfunctions corresponding to the energies $\frac{1}{2}\hbar\omega, \frac{3}{2}\hbar\omega, \frac{5}{2}\hbar\omega$, respectively.

(a) What is the most likely value of the energy that will be found in a single observation on this system? What is the probability of finding this value?

(b) What is the average of the energy that would be obtained if the experiments in part (a) could be repeated many times? What is the probability of getting this value?

(c) A measurement of the energy yields a value $\frac{3}{2}\hbar\omega$. The measurements is immediately repeated. What is the resultant value of the energy? What is the wavefunction immediately after the second measurement?

(d) What is the wavefunction of the undisturbed system after a time $t$ has elapsed?

Solution

First, check if $\psi$ is normalized:
Since $u_0, u_1, u_2$ are normalized and the moduli of their coefficients add up to unity, the whole wavefunction is normalized.

(a) The most likely value of the energy corresponds to the eigenfunction with the largest coefficient ie. $u_2$. This value is $\frac{5}{2}\hbar\omega$ with the probability $P=\bigg|-\frac{\sqrt{6}}{3}\bigg|^2=\frac{2}{3}$.

(b) The average energy is
$\langle{E}\rangle = \bigg[\frac{1}{9}\times\frac{1}{2}\hbar\omega{+}\frac{2}{9}\times\frac{3}{2}\hbar\omega{+}\frac{6}{9}\times\frac{5}{2}\hbar\omega\bigg]=\frac{37}{18}\hbar\omega$

The probability of observing this value is zero, since it is not one of the eigenvalues.

(c) The result energy value obtained is $\frac{3}{2}\hbar\omega$.
The wavefunction is $u_1{(x)}e^{-i3\omega{t}/2}$

(d) The wavefunction is
$\begin{aligned}
\psi &= \frac{1}{3}u_0{(x)}e^{-i\omega{t}/2}\\
&+\frac{i\sqrt{2}}{3}u_1{(x)}e^{-i3\omega{t}/2}\\
&-\frac{\sqrt{6}}{3}u_2{(x)}e^{-i5\omega{t}/2}$