Nuclear & Particle Physics - Standard Model

According to the Standard Model of elementary particles, which of the following is NOT a composite object?

A. Muon
B. Pi-meson
C. Neutron
D. Deuteron
E. Alpha particle
(GR9677 #63)
Solution:

A.TRUE
Muon is a lepton. Leptons, along with quarks, are considered the fundamental particles.

B. FALSE
Pi-Meson consists of a quark and its antiparticle. (Contribution to this part of the solution is due to user danty.) Moreover, a pi-meson is a hadron. Hadrons interact with the strong-force, and all of them are composed of combinations of quarks. (The fundamental particles are classified as quarks and leptons.)

C. FALSE
A neutron is made up of 3 quarks.

D. FALSE
A deuteron consists of a proton and a neutron. (tritium is two neutrons and a proton, while regular Hydrogen is just an electron and proton).

E. FALSE
An alpha particle consists of electrons and protons and neutrons.

Nuclear & Particle Physics - Binding Energy

The binding energy of a heavy nucleus is about 7 million electron volts per nucleon, whereas the binding energy of a medium-weight nucleus is about 8 million electron volts per nucleon. Therefore, the total kinetic energy liberated when a heavy nucleus undergoes symmetric fission is most nearly

A. 1876 MeV
B. 938 MeV
C. 200 MeV
D. 8 MeV
E. 7 MeV
(GR9677 #64)
Solution:

KEEf   Ei

Symmetric fission: the splitting of the nucleus into two fragments of approximately equal mass.

AX  →  A1Y  + A2Z
with A1 =  A= A/2 and Y = Z  (for symmetric fission)

A Ei  → A1 Ef 1 + AEf = 2(A/2) Ef  = A Ef

KE = A Ef  − A E = A (8  − 7) MeV/nucleon = A MeV/nucleon

For heavy nucleus, A ≈ 200 nucleons. Example: 238U → A ≈ 238 nucleons

→ KE ≈ 200  MeV

Classical Mechanics - Conservation of Momentum

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

A. ½ Mv²
B. ½ mv²
C. ½ (m)v²
D. ½ (M²/m)v²
E. ½ [Mm/(m)]v²
(GR9677 #65)
Solution:

Conservation of momentum:

$mv_0=Mv&space;\rightarrow&space;v_0=\frac{M}{m}v$

The man does work on both himself and the boat:

\begin{aligned}&space;\\W&=KE_{\text{total}}&space;\\&=\frac{1}{2}mv_0^2+\frac{1}{2}Mv^2&space;\\&=\frac{1}{2}m\left&space;(&space;\frac{M}{m}v&space;\right&space;)^2+\frac{1}{2}Mv^2&space;\\&=\frac{1}{2}\left&space;(&space;\frac{M^2}{m}+M&space;\right&space;)v^2&space;\end{aligned}

Classical Mechanics - Orbital Path

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

A. Spiral
B. Circle
C. Ellipse
D. Parabola
E. Hyperbola
(GR9677 #66)
Solution:

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

If vescape = vcircular → Parabola
If ve $>$ √2 v→ Hyperbola
Since ve = 1.5 $>$ √2 → Hyperbola

Note:
To find escape velocity:
Fc = F
→ mv2r = GMm r
→ vcircular (GM r)

KE = PEgravity
→ ½ mvGMm r
→ vescape (2GM r) = (GM r)
vescape = vcircular

Special Relativity - Black Hole

A black hole is an object whose gravitational field is so strong that even light cannot escape. To what approximate radius would Earth (mass = 5.98 × 1024 kilograms) have to be compressed in order to become a black hole?

A. 1 nm
B. 1 µm
C. 1 cm
D. 100 m
E. 10 km
(GR9677 #67)
Solution:

Schwarzschild radius: GMm/R = ½ mc2
→ = 2GM c

6.67 × 10−11 m3/(kg·s2)
5.98 × 1024 kg
3 × 10m/s

=  2 (6.67 × 10−11)(5.98 × 1024) / (3 × 108)2
= (2·6.67·5.98 / 9) × 10−11× 1024 × 10−16
≈ (2·7·/ 9) × 10−3 ≈ 1 cm

Classical Mechanics - Lagrangian

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

A. ½ mṡ ² − mgs cos θ
B. ½ mṡ ² + ½ m(ωs − mgs
C. ½ mṡ ² + ½ m(ωcos θ + mgs cos θ
D. ½ m(ṡ sin θ)² − mgs cos θ
E. ½ mṡ ² + ½ m(ωsin θ − mgs cos θ
(GR9677 #68)
Solution:

Lagrangian: L = T U

Potential energy: U = mgh = mgs cos θ
Kinetic Energy:  Tkin  =  ½ mṡ ²
Rotational kinetic energy:  Trot =  ½ ²
with moment inertia: = mr²  = m(sin θ
→ Trot = ½ m(ωsin θ

Tkin Trot − U =  ½ mṡ ² + ½ m(ωsin θ − mgs cos θ

Electromagnetism - Conductor

Two long conductors are arranged as shown above to form overlapping cylinders, each of radius r, whose centers are separated by a distance d. Current of density J flows into the plane of the page along the shaded part of one conductor and equal current flows out of the plane of the page along the shaded portion of the other, as shown. What are the magnitude and direction of the magnetic field at point A?

A. (µ0 ⁄ 2π)πdJ, in the +y-direction
B. (µ0 ⁄ 2π)d2⁄ r, in the +y-direction
C. (µ0 ⁄ 2π)4d2⁄ r, in the −y-direction
D. (µ0 ⁄ 2π)J2⁄ d, in the −y-direction
E. There is no magnetic field at A
(GR9677 #69)
Solution:

Using right-hand rule, the B direction is in the +y direction
→ C, D, E are FALSE

For option (B), if r → 0,  → ∞, cannot be infinite.

Electromagnetism - Power

A charged particle, A, moving at a speed much less than c, decelerates uniformly. A second particle, B, has one-half the mass, twice the charge, three times the velocity, and four times the acceleration of particle A. According to classical electrodynamics, the ratio PB/Pof the powers radiated is

A. 16
B. 32
C. 48
D. 64
E. 72
(GR9677 #70)
Solution:

Larmor formula:
Power radiated by an accelerating non-relativistic point charge particle,
P = q²a² ⁄ 6πɛ0c³

does not depend on m and v.

PB/P=  (2q)² (4a)² q²a²  =  64

Electromagnetism - Particle Trajectory

The figure above shows the trajectory of a particle that is deflected as it moves through the uniform electric field between parallel plates. There is potential difference V and distance d between the plates, and they have length L. The particle (mass m, charge q) has non relativistic speed v before it enters the field, and its direction at this time is perpendicular to the field. For small deflections, which of the following expressions is the best approximation to the deflection angle θ?

A. Arctan ( (L/d)(Vq/mv2) )
B. Arctan ( (L/d)(Vq/mv2))
C. Arctan ( (L/d)2 (Vq/mv2) )
D. Arctan ( (L/d)(2Vq/mv2)½ )
E. Arctan ( (L/d)½(2Vq/mv2) )
(GR9677 #71)
Solution:

tan θ vvat /
L/t → t = L/v
F  =  ma = qE = qV → a = qV md

tan θ at / =  (qV md) (L/v)  / = (qVL /mdv2)
→  θ = arctan (qVL /mdv2) = ( (L/d)(Vq/mv2) )

Lab Methods - Voltage Amplifier

In a voltage amplifier, which of the following is NOT usually a result of introducing negative feedback?

A. Increased amplification
B. Increased bandwidth
C. Increased stability
D. Decreased distortion
E. Decreased voltage gain
(GR9677 #72)
Solution:

Negative feedback → noise cancellation.
Answer B, C, D, and E are benefits of negative feedback.

Positive feedback → increased amplification.