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Relativistic Physics: Calculating Momentum and Energy of Moving Particles, Exercises of Physics

Solutions to various problems in relativistic physics, focusing on the calculation of momentum and energy of moving particles. It includes problems involving electrons and protons, and demonstrates the use of relativistic equations to find their velocities and total energies.

Typology: Exercises

2021/2022

Uploaded on 09/27/2022

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Recitation 8
Chapter 9
Problem 9.23. Calculate the momentum of an electron moving with a speed of (a) 0.0100c,(b) 0.500c,(c) 0.900c.
The momentum (using relativistic mass) is given by
p=mv =γm0v(1)
(a)
γ=1.0000500 (2)
p=5.11 keV/c (3)
(b)
γ=1.155 (4)
p=295 keV/c (5)
(c)
γ=2.294 (6)
p=1.06 MeV/c (7)
Problem 9.30. Show that, for any object moving at less than one-tenth the speed of light, the relativistic kinetic energy
agrees with the result of the classical equation K=1
2mv2to within less than 1%. Therefore, for most purposes the classical
equation is good enough to describe these objects, whose motion we call nonrelativistic.
The kinetic energy is the energy that is due to the objects motion. In other words, the increase in the total energy over the
rest mass (in the absence of potential energies etc.). In math
K=Em0c2=mc2m0c2= (γ1)m0c2=
1
q1v2
c2
1
m0c2="1v2
c20.5
1#m0c2(8)
Using the Taylor series expansion around x= 0,
(1 + x)b1 + bx +O(x2) (9)
we see that for small x=v2/c2, the kinetic energy looks like
K=1+0.5v2
c2+O(v4/c4)1m0c2= 0.5m0v2+O(m0v4/c2)1
2m0v2(10)
with the relative error on the order of v2/c2<0.01 = 1% for v < 0.1c, which is what we set out to show.
Alternatively, we can compare the exact kinetic energy with the nonrelativistic form for v= 0.1c. From the above analysis,
we see that the relative error will be less for slower v.
γ=1
10.12= 1.00503782 (11)
Krel = (γ1)m0c2= 0.00503782γm0c2(12)
Knon =1
2m0v2= 0.5·m0·0.01c2= 0.005m0c2(13)
Knon
Krel
=0.005
0.00503782 = 0.992494 (14)
for a 7.51% relative underestimate.
Problem 9.31. An electron has a kinetic energy five times greater than its rest energy. Find (a) its total energy and (b) its
speed.
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Recitation 8

Chapter 9

Problem 9.23. Calculate the momentum of an electron moving with a speed of (a) 0. 0100 c, (b) 0. 500 c, (c) 0. 900 c.

The momentum (using relativistic mass) is given by

p = mv = γm 0 v (1)

(a)

γ = 1. 0000500 (2) p = 5.11 keV/c (3)

(b)

γ = 1. 155 (4) p = 295 keV/c (5)

(c)

γ = 2. 294 (6) p = 1.06 MeV/c (7)

Problem 9.30. Show that, for any object moving at less than one-tenth the speed of light, the relativistic kinetic energy agrees with the result of the classical equation K = 12 mv^2 to within less than 1%. Therefore, for most purposes the classical equation is good enough to describe these objects, whose motion we call nonrelativistic.

The kinetic energy is the energy that is due to the objects motion. In other words, the increase in the total energy over the rest mass (in the absence of potential energies etc.). In math

K = E − m 0 c^2 = mc^2 − m 0 c^2 = (γ − 1)m 0 c^2 =

 √^1

1 − v 2 c^2

 (^) m 0 c^2 =

[(

v^2 c^2

]

m 0 c^2 (8)

Using the Taylor series expansion around x = 0,

(1 + x)b^ ≈ 1 + bx + O(x^2 ) (9)

we see that for small x = −v^2 /c^2 , the kinetic energy looks like

K =

[(

v^2 c^2

  • O(v^4 /c^4 )

]

m 0 c^2 = 0. 5 m 0 v^2 + O(m 0 v^4 /c^2 ) ≈

m 0 v^2 (10)

with the relative error on the order of v^2 /c^2 < 0 .01 = 1% for v < 0. 1 c, which is what we set out to show. Alternatively, we can compare the exact kinetic energy with the nonrelativistic form for v = 0. 1 c. From the above analysis, we see that the relative error will be less for slower v.

γ =

Krel = (γ − 1)m 0 c^2 = 0. 00503782 γm 0 c^2 (12)

Knon =

m 0 v^2 = 0. 5 · m 0 · 0. 01 c^2 = 0. 005 m 0 c^2 (13) Knon Krel

for a 7.51% relative underestimate.

Problem 9.31. An electron has a kinetic energy five times greater than its rest energy. Find (a) its total energy and (b) its speed.

The total energy is (a) E = K + m 0 c^2 = (5 + 1)m 0 c^2 = 6 · 511 keV = 3.07 MeV (15)

(b)

E = mc^2 = γm 0 c^2 = 6m 0 c^2 (16)

γ =

1 − v 2 c^2

v^2 c^2

v c

v = 0. 986 c (20)

Problem 9.35. The rest energy of an electron is 0. 511 MeV. The rest energy of a proton is 938 MeV. Assume that both particles have kinetic energies of 2. 00 MeV. Find the speed of (a) the electron and (b) the proton. (c) By how much does the speed of the electron exceed that of the proton? (d) Repeat the calculations assuming that both particles have kinetic energies of 2 , 000 MeV.

First we’ll work out the solution symbolically, since we’ll need it twice. The total energy yields γ, which in turn yields v.

E = K + m 0 c^2 = γm 0 c^2 (21)

γ =

K

m 0 c^2

γ =

1 − v 2 c^2

γ^2

v^2 c^2

v c

γ^2

v =

γ^2

· c (26)

Applying these formula to our various situations, we get (a)

γa =

2 .00 MeV 0 .511 MeV

va = 0. 979 c (28)

(b)

γb =

2 .00 MeV 938 MeV

vb = 0. 0652 c (30)

(c) va − vb = 0. 914 c (31)

(d)

γe =

2 .00 GeV 0 .511 MeV

  • 1 = 3. 91 e 3 (32)

ve = (1 − 3. 26 · 10 −^8 )c (33)

γp =

2 .00 GeV 938 MeV

vp = 0. 948 c (35) ve − vp = 0. 0523 c (36)