### Special Relativity - Velocity

A π0 meson (rest-mass energy 135 MeV) is moving with velocity 0.8$\hat{\mathbf{k}}$ in the laboratory rest frame when it decays into two photons γ1 and γ2. In the π0 rest frame, γ1 is emitted forward and γ2 is emitted backward relative to the direction of flight. The velocity of γ2 in the laboratory rest frame is

A. −1.0$\hat{\mathbf{k}}$
B. −0.2$\hat{\mathbf{k}}$
C. +0.8$\hat{\mathbf{k}}$
D. +1.0$\hat{\mathbf{k}}$
E. +1.8$\hat{\mathbf{k}}$
(GR9277 #37)
Solution:

Photon moves with the speed of light → B, C, E are FALSE.
Since γis emitted backward, the velocity of γ2 is −1.0c

### Special Relativity - Time Dilation

Tau leptons are observed to have an average half-life of Δt1 in the frame S1 in which the leptons are at rest. In an inertial frame S2, which is moving at a speed v12 relative to S1, the leptons are observed to have an average half-life of Δt1. In another inertial reference frame S3, which is moving at a speed v13 relative to S1 and v23  relative to  S2, the leptons have an observed half-life of Δt3. Which of the following is a correct relationship among two of the half lives, Δt1, Δt2, and Δt3.

A. $\Delta&space;t_2=&space;\Delta&space;t_1\sqrt{1-(v_{12})^2/c^2}$

B. $\Delta&space;t_1=&space;\Delta&space;t_3\sqrt{1-(v_{13})^2/c^2}$

C. $\Delta&space;t_2=&space;\Delta&space;t_3\sqrt{1-(v_{23})^2/c^2}$

D. $\Delta&space;t_3=&space;\Delta&space;t_2\sqrt{1-(v_{23})^2/c^2}$

E. $\Delta&space;t_1=&space;\Delta&space;t_2\sqrt{1-(v_{23})^2/c^2}$

(GR9277 #38)
Solution:

Time Dilation:

\begin{aligned}&space;\Delta&space;t'&space;&=\gamma&space;\Delta&space;t_0&space;\\&space;\gamma&space;&=\sqrt{1-\frac{v^2}{c^2}}&space;\end{aligned}

Δt' time in a moving frame
Δttime in a rest frame

In this problem, Δttime in a rest frame, and Δt2, and Δt3 = time in a moving frame → the right answers are:

\begin{aligned}&space;\Delta&space;t_2&space;&=\gamma_{12}&space;\Delta&space;t_1&space;\\&space;\Delta&space;t_3&space;&=\gamma_{13}&space;\Delta&space;t_1&space;\end{aligned}

Answer A → $\Delta&space;t_1&space;=&space;\gamma_{12}\Delta&space;t_2$ → FALSE

Answer B → $\Delta&space;t_3&space;=&space;\gamma_{13}\Delta&space;t_1$ → TRUE

Answer C → $\Delta&space;t_3&space;=&space;\gamma_{23}\Delta&space;t_2$ → FALSE

Answer D → $\Delta&space;t_2&space;=&space;\gamma_{23}\Delta&space;t_3$ → FALSE

Answer E → $\Delta&space;t_2&space;=&space;\gamma_{23}\Delta&space;t_1$ → FALSE

### Electromagnetism - Fourier Series

If n is an integer ranging from 1 to infinity, is an angular frequency, and t is time, then the Fourier series for a square wave, as shown above, is given by which of the following?

A. $V(t)=\frac{4}{\pi}\sum_{1}^{\infty}\frac{1}{n}\sin(n\omega&space;t)$

B. $V(t)=\frac{4}{\pi}\sum_{0}^{\infty}\frac{1}{(2n+1)}\sin(\left(2n+1\right)\omega&space;t)$

C. $V(t)=\frac{4}{\pi}\sum_{1}^{\infty}\frac{1}{n}\cos(n\omega&space;t)$

D. $V(t)=\frac{4}{\pi}\sum_{0}^{\infty}\frac{1}{(2n+1)}\cos(\left(2n+1\right)\omega&space;t)$

E.  $V(t)=-\frac{4}{\pi}+\frac{4}{\pi}\sum_{1}^{\infty}\frac{1}{n^2}\cos(n\omega&space;t)$
(GR9277 #39)
Solution:

The sine function is an odd function.
The cosine function is an even function.
The function in this problem is an odd square wave function.
→ C, D, E are FALSE.

Both A and B are the same, thus the right solutions. However, A is more generalized and B is the simplified answer.

For Answer A, the sine term vanishes when n is even (n = 0, 2, 4, ...) → rewrite n for odd term as 2n +1 with n starts at 0 → the answer can be simplified to B.

### Classical Mechanics - Circular Motion

A rigid cylinder rolls at constant speed without slipping on top of a horizontal plane surface. The acceleration of a point on the circumference of the cylinder at the moment when the point touches the plane is

A. directed forward
B. directed backward
C. directed up
D. directed down
E. zero
(GR9277 #40)
Solution:

Constant speed → no tangential acceleration → Uniform Circular Motion

$\vec{a}=\vec{a}_\text{centripetal}$

When the point touches the plane,

$\vec{a}$ is directed up

### Classical Mechanics - Rotational Motion

Questions 41-42

A cylinder with moment inertia 4 kgm² about a fixed axis initially rotates at 80 radians per second about this axis. A constant torque is applied to slow it down to 40 radians per second. The kinetic energy lost by the cylinder is

A. 80 J
B. 800 J
C. 4000 J
D. 9600 J
E. 19,200 J
(GR9277 #41)

Solution:

Rotational Kinetic Energy:  $T_\text{rot}&space;=&space;\frac{1}{2}I\omega^2$

$\Delta&space;T_\text{rot}&space;=&space;\frac{1}{2}I\left&space;(\omega_f^2-\omega_i^2&space;\right&space;)&space;=&space;\frac{1}{2}\cdot&space;4\left&space;(40^2-80^2&space;\right&space;)&space;=&space;-9600&space;\text{&space;J}$

### Classical Mechanics - Rotational Motion

Questions 41-42

If the cylinder takes 10 seconds to reach 40 radian per second, the magnitude of the applied torque is

A. 80 Nm
B. 40 Nm
C. 32 Nm
D. 16 Nm
E. 8 Nm
(GR9277 #42)
Solution:

$\tau&space;=&space;I\alpha&space;=&space;I&space;\frac{\left&space;(&space;\omega_f&space;-&space;\omega_i&space;\right&space;)}{t}=4\frac{\left&space;(&space;40&space;-&space;80&space;\right&space;)}{10}=-16&space;\text{&space;Nm}$

### Classical Mechanics - Lagrangian

If  ∂L/∂qn = 0, where L is the Lagrangian for a conservative system without constraints and qn is a generalized coordinate, then the generalized momentum pn  is

A. An ignorable coordinate
B. Constant
C. Undefined
D. Equal to  $\frac{d}{dt}&space;\left&space;(\frac{\partial&space;L}{\partial&space;q_n}&space;\right&space;)$
E. Equal to the Hamiltonian for the system
(GR9277 #43)
Solution:

Lagrangian Equation of Motion for the generalized coordinate q:

$\frac{\partial&space;L}{\partial&space;q}=\frac{\partial}{\partial&space;t}\frac{\partial&space;L}{\partial&space;\dot&space;q}$

Momentum,

\begin{aligned}&space;p&=mv=m\dot{x}&space;\\L&=T-U=\frac{1}{2}m\dot{x}^2-mgy\\&\rightarrow&space;\frac{\partial&space;L}{\partial&space;\dot{x}}=m\dot{x}=p&space;\end{aligned}

For the generalized coordinate q:

If  $\frac{\partial&space;L}{\partial&space;q_n}=0&space;\rightarrow&space;\frac{\partial&space;L}{\partial&space;\dot{q_n}}=p_n=&space;\text{constant}$

### Classical Mechanics - Energy

A ball is dropped from a height h. As it bounces off the floor, its speed is 80% of what it was just before it hit the floor. The ball will then rise to a height of most nearly

A. 0.94 h
B. 0.80 h
C. 0.75 h
D. 0.64 h
E. 0.50 h
(GR9277 #45)
Solution:

Conservation of energy before and when the ball hits the ground:

\begin{aligned}&space;\frac{1}{2}m&space;\cdot&space;0+mgh&=\frac{1}{2}mv^2+mg\cdot&space;0&space;\\mh&=\frac{1}{2}v^2&space;\\v^2&space;&=&space;2gh&space;\end{aligned}

Conservation of energy when and after the ball hits the ground

\begin{aligned}&space;\frac{1}{2}mv'^2+mg\cdot&space;0&=\frac{1}{2}m\cdot&space;0+mgh'&space;\\\frac{1}{2}mv'^2&=mgh'&space;\\v'^2&space;&=&space;2gh'&space;\\(0.8v)^2&=0.64v^2&space;=&space;0.64\times&space;2gh&space;=&space;2gh'&space;\\h'&=0.64h&space;\end{aligned}

### Thermal Physic - Critical Isotherm

Questions 46-47.

Isotherms and coexistence curves are shown in the PV diagram above for a liquid-gas system. The dashed lines are the boundaries of the labeled regions. Which numbered curve is the critical isotherm?

A. 1
B. 2
C. 3
D. 4
E. 5
(GR9277 #46)
Solution:

Critical isotherm (critical temperature) → dP/dV = 0
• the derivative of the curve is zero
• the tangent to the curve results in a horizontal line
• the point where the vertical and horizontal dashed lines cross