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Ponderable: it's Rocket Science | Introduction to Physics Science | PHYS 0475, Exams of Physics

University of Pittsburgh (Pitt) - Medical Center-Health System Physics

Material Type: Exam; Class: INTRO PHYS SCIENCE & ENGRG 1 / RECITATION; Subject: Physics; University: University of Pittsburgh; Term: Fall 2007;

Typology: Exams

Pre 2010

Uploaded on 09/17/2009

koofers-user-6d9 🇺🇸

10 documents

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1

Tuesday 10/16/07

Ponderable

Neutron stable inside nucleus. But by itself, it will decay in about 15 minutes. Get proton +

electron + antineutrino. Energy principle still applies.

Initially Ei = mnc2

Finally Ef = mpc2 + Kp + mec2 + Ke + Kantineutrino (mantineutrino very small but not quite zero)

Since the K’s aren’t zero, that says mn > mp + me so less mass than when you started.

mpc2 + Kp + mec2 + Ke + Kantineutrino = mnc2 + W

(938.3 MeV) + Kp +(0.511 MeV) + Ke + Kantineutrino = (939.6 MeV) + 0

(938.8 MeV) + Kp + Ke + Kantineutrino = (939.6 MeV) + 0

Kp + Ke + Kantineutrino = 0.8 MeV = 1.28e-13 J

That’s what happens in fission and fusion reactions. Reaction products less massive than

starting particles. Lots of energy output.

At first scientists didn’t see antineutrino. They thought energy principle didn’t work. Pauli

predicted, but not seen for 30 years.

Ponderable: It’s rocket science!

Consider a rocket blasting off from a planet. (Assume the launch happens very quickly and

then the rocket coasts away from the planet.) If we know the launch speed, what is the rocket’s

final speed for some distance r from the planet’s center? We could use the momentum

principle, but the force varies—we’d have to integrate or do it on a computer. Since distance is

involved, we should more likely use the energy principle. We will pick the system to be both

the planet and rocket so no external work is done (and thus we don’t have to worry about

integrating the force.)

m

M

R

Partial preview of the text

Download Ponderable: it's Rocket Science | Introduction to Physics Science | PHYS 0475 and more Exams Physics in PDF only on Docsity!

Tuesday 10/16/

Ponderable Neutron stable inside nucleus. But by itself, it will decay in about 15 minutes. Get proton +

electron + antineutrino. Energy principle still applies.

Initially Finally EEi = m n c^2

Since the fK^ = m ’s aren’t zero, that saysp c^2 + K p^ + m e c^2 + K m ne > + K m p antineutrino+ m e so less mass than when you started.^ ( m antineutrino^ very small but not quite zero)

m (938.3 MeV)p c^2 + K p + m + K e c^2 p + K + (0.511 MeV)e + K antineutrino + K = m e n + Kc^2 + (^) antineutrino W = (939.6 MeV) + 0 (93 K p + K 8.8 MeV)e + K antineutrino + K p + K = 0.8 MeV = 1.28ee + K antineutrino = (939.6 MeV) + 0-13 J

That’s what happens in fission and fusion reactions. Reaction products less massive than starting particles. Lots of energy output.

At first scientists d predicted, but not seen for 30 years.idn’t see antineutrino. They thought energy principle didn’t work. Pauli

Ponderable: It’s rocket science!

Consider a rocket blasting off from a planet. then the rocket coasts away from the planet.) If we know the launch speed, what is the rocket’s (Assume the launch happens very quickly and

final speed for some distance principle, but the force varies— we’d have to integrate or do it on a computer. Since distanc r from the planet’s center? We could use the momentume is

involved, we should more likely use the energy principle. We will pick the system to be both the planet and rocket so no external work is done (and thus we don’t have to worry about

integrating the force.)

M m R

( ) ( )

= + $^ $ %^ &^ '!^! "^

f i f i

f f i i

m Mm f m Mm i

M m Mm f M m Mm i

f i i

v v GM r r

mv G^ Mm r mv G^ Mm r v c

K U K U

Mc K mc K U Mc K mc K U

E E W E

21 21 since

2

(^22)

2 2 2 2

We are assuming th calculation above was straightforward, so let’s also consider how things look graphically.e planet’s kinetic energy KM doesn’t change much. The mathematical

Discussion

Plot U vs. r^!

Plot K vs. r^!

K rocket never actually stops moving. As the gravitational force gets very small, the rate of is decreasing due to the planet pulling on the rocket. The way we’ve drawn this graph, the

change of the rocket’s momen and so does K. tum becomes negligible—the speed becomes nearly constant

Because there are no external forces acting on the system, the total energy That tells us the shape of the K curve is just like that of U , except “flipped over” so K + U is constant. that the

two curves added up will produce a horizontal line corresponding to the constant energy of the system.

As! r! ", U = # GM r! m , which is negative, gets closer to zero

r!

U

U =! GM r! m

R

R is planet’s radius

Fr =! dU dr = G M r 2 m , which is negative of the slope

What hap distance from the planet? Since we knowpens if the launch speed is “just right” so that U goes to zero when we are infinitely far from the K approaches 0 at a very, very large

planet, so must K + U. Since this is a constant value, the total energy is zero for any

distance r^! from the planet.

v^ GMR

mv G^ Mm R

K U

escape

i

i i 2

2

In this borderline case, where the initial speed is this is not a bound state) the initial speed is called the “es just big enough to makecape speed.” For Earth, K + U not negative (so v

km/s. escape^ = 11

R^ K^ +^ U^ = 0^ r! U =! GM r! m

K

Consider two protons that are initially a distance r^! apart.

The electrical force is very similar in form to the gravitational force:

F! elec = 1 4 !" 0^ Q^! r q^2 =^9 #^109 Nm

2 $ %& C 2 ' ()

(^ 1.6 #^10! *^19 C)^2

r^2

This gives us the electrical potential energy as:

U = + 4 !"^10^ Qqr!

Since there are again no external forces, the total system energy is constant. Start by graphing the U function, which is positive this time. It still approaches zero as the separation increases.

When the separation is very small, the potential energy is very large. As the protons move away from each other, their kinetic energies rise, and their interaction energy (electric potential

energy) falls. Algebraically E

mf^ =^ Ei^ +^ W K^1 c^2 +^ K^1 f^ +^ m^2 c^2 +^ K^2 f^ +^ U^ f^ =^ m^1 c^2 +^ K^1 i^ +^ m^2 c^2 +^ K^2 i^ +^ Ui^ +^0 K^1 f^ +^ K^2 f^ +^ U^ f^ =^ Ui 1 f +^ K 2 f =^ K^ f =^ Ui

! r

K + U = constant

U = + (^4) !"^10^ Q r! q

K

r^!^ System is both charges

U (^) f! 0

Ponderable: it's Rocket Science | Introduction to Physics Science | PHYS 0475, Exams of Physics

Related documents

Partial preview of the text

Download Ponderable: it's Rocket Science | Introduction to Physics Science | PHYS 0475 and more Exams Physics in PDF only on Docsity!

Tuesday 10/16/

Ponderable Neutron stable inside nucleus. But by itself, it will decay in about 15 minutes. Get proton +

electron + antineutrino. Energy principle still applies.

Initially Finally EEi = m n c^2

Since the fK^ = m ’s aren’t zero, that saysp c^2 + K p^ + m e c^2 + K m ne > + K m p antineutrino+ m e so less mass than when you started.^ ( m antineutrino^ very small but not quite zero)

That’s what happens in fission and fusion reactions. Reaction products less massive than starting particles. Lots of energy output.

At first scientists d predicted, but not seen for 30 years.idn’t see antineutrino. They thought energy principle didn’t work. Pauli

Ponderable: It’s rocket science!

Consider a rocket blasting off from a planet. then the rocket coasts away from the planet.) If we know the launch speed, what is the rocket’s (Assume the launch happens very quickly and

final speed for some distance principle, but the force varies— we’d have to integrate or do it on a computer. Since distanc r from the planet’s center? We could use the momentume is

involved, we should more likely use the energy principle. We will pick the system to be both the planet and rocket so no external work is done (and thus we don’t have to worry about

integrating the force.)

= + $^ $ %^ &^ '!^! "^

K U K U

E E W E

We are assuming th calculation above was straightforward, so let’s also consider how things look graphically.e planet’s kinetic energy KM doesn’t change much. The mathematical

Discussion

Plot U vs. r^!

Plot K vs. r^!

K rocket never actually stops moving. As the gravitational force gets very small, the rate of is decreasing due to the planet pulling on the rocket. The way we’ve drawn this graph, the

change of the rocket’s momen and so does K. tum becomes negligible—the speed becomes nearly constant

Because there are no external forces acting on the system, the total energy That tells us the shape of the K curve is just like that of U , except “flipped over” so K + U is constant. that the

two curves added up will produce a horizontal line corresponding to the constant energy of the system.

U

R

What hap distance from the planet? Since we knowpens if the launch speed is “just right” so that U goes to zero when we are infinitely far from the K approaches 0 at a very, very large

planet, so must K + U. Since this is a constant value, the total energy is zero for any

distance r^! from the planet.

K U

2

In this borderline case, where the initial speed is this is not a bound state) the initial speed is called the “es just big enough to makecape speed.” For Earth, K + U not negative (so v

km/s. escape^ = 11

K

Consider two protons that are initially a distance r^! apart.

The electrical force is very similar in form to the gravitational force:

(^ 1.6 #^10! *^19 C)^2

This gives us the electrical potential energy as:

Since there are again no external forces, the total system energy is constant. Start by graphing the U function, which is positive this time. It still approaches zero as the separation increases.

When the separation is very small, the potential energy is very large. As the protons move away from each other, their kinetic energies rise, and their interaction energy (electric potential

energy) falls. Algebraically E