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Instructor: Dr. Alexey Belyanin http://faculty.physics.tamu.edu/belyanin/ Office: MIST 426 Office Phone: (979) 845-7785 Email: [email protected] Office Hours: MWF 1:40-2:40 pm or by appointment

Instructor: Dr. Alexey Belyanin Office: MIST 426 Office Phone: (979) 845-7785 Email: [email protected] Office

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Instructor: Dr. Alexey Belyanin http://faculty.physics.tamu.edu/belyanin/

Office: MIST 426 Office Phone: (979) 845-7785 Email: [email protected]

Office Hours: MWF 1:40-2:40 pm or by appointment

Momentum vmp

N

iitotal pP

1

For N particles:

Why bother introducing one more definition?

Fdt

pd

Most general form of the Newton’s law:• Valid when mass is changing• Valid in relativistic physics

N

iexti

total Fdt

Pd

1,

Total momentum is conserved if the sum of external forces equals to zero!

Nuclear reactions• Fission: decay of heavy nuclei into lighter fragments

•Fusion: synthesis of light nuclei into a heavier nucleus

Energy released per proton is ~10-20 MeV!!

Energy is released in fission reaction if the mass of an initial nucleus is larger that the sum of masses of all final fragments

MU > MRb + MCs + 3 mn

Rubidium and Cesium are more tightly bound, or have larger binding energy than Uranium.

It is energetically favorable for Uranium to split.

When is the energy released in fission reactions?

M = MU – (MRb + MCs + 3 mn)

Energy released E = M c2

Proton-proton cycle: four hydrogen nuclei fuse to form one helium nucleus

Hydrogen Fusion

Einstein’s relation: E = mc2

!04 mmm Hep

J103.4MeV8.26 122 cmE

Energy released in one reaction:

(Binding energy)

kg10048.0 27m

Hans Bethe 1939

0.007, or 0.7% of the rest energy of protons (4mpc2) is released

This is 107 times more efficient than chemical reactions!

There is more than enough nuclear fuel for 1010 years!

years10310104 10

5612

Lt

Does nuclear fusion provide enough energy to power the Sun?

Assume 1056 protons in the core:

Center of Mass

i

N

iicm rm

Mr

1

1

N

iimM

1

i

N

iicm xm

Mx

1

1i

N

iicm ym

My

1

1

Motion of the Center of Mass

N

iiicm vm

Mv

1

1

N

iiixcm x

vmM

v1

1

N

iiiycm y

vmM

v1

1

i

N

iicm am

Ma

1

1

N

iicm FaM

1

N

iicm FaM

1

The center of mass of a system moves as if all of the mass of the system were concentrated at that point and as if all of the forces were acting at that point

For internal forces

external

N

iexternalicm FFaM

1

jiij FF

Only external forces affect the motion of the center of mass

Momentum is a vector!

vmp

cmi

N

ii

N

iitotal vMvmpP

11

N

iiicm vm

Mv

1

1

cmtotal vMP

Vector equation!

dt

Pd

dt

vdMFaM totalcm

externalcm

externaltotal F

dt

Pd

If ,0externalF

0dt

Pd total

ConstPtotal

Conservation of Momentum

If there is no net external force acting on a system, then the total momentum of the system is a constant

ConstPtotal

r P (before)

r P (after)

Problem Solving

For Conservation of Momentum problems:

1. BEFORE and AFTER

2. Do X and Y Separately

True in X and Y directions separately!

externalx

totalx Fdt

dP external

y

totaly Fdt

dP

,0externalxFIf only then consttotal

xP