2-2
(a)
Scalar equations can be considered in this case because relativistic and classical
velocities are in the same direction.
p = " mv = 1.90mv =
mv
$1 # ( v c )2 '
&%
)(
12
*
1
12
$1 # (v c) 2 '
&%
)(
$ + 1 .2'
= 1.90 * v = &1 # 0 )
&% , 1.90 / )(
12
c
= 0.85c
(b)
No change, because the masses cancel each other
2-3
As F is parallel to v, scalar equations are used. Relativistic momentum is given by
mv
p = " mv =
1 2 , and relativistic force is given by
$1 # (v c)2 '
%&
()
)
dp d ++
mv
F=
= *
dt dt + #
2 &1 2
1
"
v
c
(
)
+, $%
'(
dp
m
=
dt 1 " v 2 c 2
F=
2-7
(
2
E = " mc2 , p = " mu ; E = " mc
2 2
)
32
[ (
)]
2
2
; p = (" mu) ;
(
E 2 " p2 c2 = # mc 2
2
) " (# mu )2 c 2 = # 2 $%&(mc 2)
2*
u 2 -*
u2 = mc 2 ,1 " 2 /, 1 " 2 /
+
c .+
c .
2 2
( 2)
E = p c + mc
2-8
2-9
(
2
" (mc) u 2 '(
)
2
"1
( )
2
++
.
+
+/
0 dv 3
2 5
1 dt 4
( )
= mc2
2
Q.E.D.
2
2
)(
)
(a)
ER = mc2 = 1.67 " 10 #27 kg 3 " 10 8 m s = 1.503 " 10 #10 J = 939.4 MeV
(Numerical round off gives a slightly larger value for the proton mass)
(b)
E = " mc =
(c)
K = E " mc2 = 4.813 # 10 "10 J " 1.503 # 10 "10 J = 3.31 # 10 "10 J = 2.07 # 103 MeV
(a)
When K = (" # 1)mc = 5mc , " = 6 and E = " mc = 6(0.511 0 MeV) = 3.07 MeV .
(b)
2
1 *, $ v ' -/
= 1 #& )
" ,+ % c ( /.
2
1.503 # 10 $10 J
%1 $ (0.95c c )2 '
&
(
2
12
12
= 4.813 # 10
3
J ) 3.01 # 10 MeV
2
2
* $ 1'2 and v = c,1 " & ) /
,+ % # ( /.
$10
12
1 2
* $1 '2= c,1 " & ) /
,+ % 6 ( /.
= 0.986c
2-12
(a)
2
2
2 12
[
mp c2 = 938 MeV , " e = 1 # ( 0.75)
first equation, we find " p = 1 +
But " p =
(b)
(
)
2
When K e = K p , me c (" e #1) = mp c " p # 1 . In this case me c = 0.511 0 MeV and
1
12
2'
$
&%1 # u p c )(
(
)
]
= 1.511 9 . Substituting these values into the
(0.511 0)(1.511 9 # 1)
me c (" # 1)
=1 +
= 1.000 279 .
2
939
m pc
2
(
"2 1 2
; therefore u p = c 1 " # p
)
= 0.023 6c .
# " e &# me &
When pe = pp , " p mp u p = " emeu e or u p = % (%
( ue ,
$ " p '$ m p '
" 1.511 9 % ( 0.511 0 c 2 +
-(0.75c ) = 6.17 . 10 /4 c .
up = $
'*
# 1.000 279 & *) 939 c2 -,
2-13
(a)
E = 400mc2 = " mc2
$
v2 '
" = &1 # 2 )
%
c (
#1 2
= 400
2
$
v2 ' $ 1 '
1
#
=
)
&
) &
%
c 2 ( % 400 (
12
v *
1 = ,1 #
c + 4002 /.
v = 0.999 997 c
!
2-14
(b)
K = E " mc2 = (400 "1)mc2 = 399mc2 = (399) (938.3 MeV ) = 3.744 # 10 5 MeV
(a)
E = mc2
E
4 " 10 26 J
m= 2 =
c
3.0 " 108 m s
!
(
(b)
!
2-15
(a)
)
2
= 4.4 " 10 9 kg
2.0 " 10 30 ) kg
(
t=
= 4.5 " 10 20 s = 1.4 " 1013 years
4.4 " 109 kg s
2
"2
#
Vq &
v )+ #
Vq & -+
and
K = " mc # mc = Vq and so, " = %1 +
=
1
"
1
+
(
* %
( .
$
c ,+ $
mc2 '
mc2 ' /+
2
2
2
$
v &
1
= %1 "
c & 1 + 5.0 # 10 4 eV 0.511 MeV
'
(
or v = 0.413c .
)
12
(
&
2)
&
*
= 0.412 7
12
(b)
K=
1
mv2 = Vq
2
" 2Vq %
v =$
# m '&
(c)
12
(
12
) -+
)+2 5.0 ( 10 4 eV
=*
2
,+ 0.511 MeV c
.
/+
= 0.442 c
The error in using the classical expression is approximately
7.5% in speed.
2-18
(a)
3
" 100% or about
40
The mass difference of the two nuclei is
"m = 54.927 9 u # 54.924 4 u = 0.003 5 u
"E = (931 MeV u) (0.003 5 u) = 3.26 MeV.
(b)
The rest energy for an electron is 0.511 MeV. Therefore,
K = 3.26 MeV " 0.511 MeV = 2.75 MeV .
2-19
[
]
"m = 6mp + 6mn # mC = 6(1.007 276 ) + 6(1.008 665) # 12 u = 0.095 646 u ,
"E = (931.49 MeV u)(0.095 646 u) = 89.09 MeV .
Therefore the energy per nucleon =
2-20
89.09 MeV
= 7.42 MeV .
12
"m = m # mp # me = 1.008 665 u # 1.007 276 u # 0.000 548 5 u = 8.404 $ 10 #4 u
(
) (
)
E = c 2 8.404 " 10 #4 u = 8.404 " 10 #4 u ( 931.5 MeV u ) = 0.783 MeV .
E, p
2-21
e(–)
"
e(+)
!
!
K, p(e–) positron
at rest
!
"
E, p
2
Conservation of mass-energy requires K + 2mc = 2E where K is the electron’s kinetic
energy, m is the electron’s mass, and E is the gamma ray’s energy.
K
+ mc2 = ( 0.500 + 0.511) MeV = 1.011 MeV .
2
E=
Conservation of momentum requires that pe " = 2pcos # where pe " is the initial
momentum of the electron and p is the gamma ray’s momentum,
2
2
2
2
( 2)
Using Ee " = pe " c + mc
E
= 1.011 MeV c .
c
2
where Ee " is the electron’s total energy, Ee " = K + mc , yields
pe " =
Finally, cos " =
2-23
(1.00) 2 + 2 (1.00) (0.511) MeV
1
K 2 + 2 Kmc 2 =
c
pe #
2p
c
= 1.422 MeV c .
= 0.703 ; " = 45.3° .
In this problem, M is the mass of the initial particle, ml is the mass of the lighter
fragment, vl is the speed of the lighter fragment, mh is the mass of the heavier fragment,
and vh is the speed of the heavier fragment. Conservation of mass-energy leads to
ml c 2
Mc 2 =
mh c2
+
1 " v2l c2
1 " v 2h c 2
From the conservation of momentum one obtains
(ml )(0.987 c)(6.22 ) = (mh )(0.868c )(2.01)
(mh )(0.868 c)( 2.01)
ml =
= 0.284m h
(0.987 )(6.22 )
Substituting in this value and numerical quantities in the mass-energy conservation
#27
equation, one obtains 3.34 " 10
kg = 6.22m l + 2.01mh which in turn gives
3.34 " 10 #27 kg
= 8.84 " 10 #28 kg and
3.78
3.34 " 10 #27 kg = (6.22 )( 0.284 )ml + 2.01mh or mh =
ml = (0.284 ) mh = 2.51 " 10 #28 kg .
2-31
Conservation of momentum " mu :
mu
1 "u
2
2
c
+
m( "u)
3 1 "u
2
2
c
=
Mv f
1 " vf2
c
2
=
2mu
3 1 " u 2 c2
.
2
Conservation of energy " mc :
!
mc2
!
1 " u 2 c2
+
mc2
3 1 " u 2 c2
To start solving we can divide: vf =
M
2
1 " u 4c
2
=
1 " vf2 c 2
4m
3 1 "u
=
4mc 2
3 1 " u 2 c2
.
The energy that arrives in one year is
2
2
c
=
M
(1 2 )
4 " u 2 c2
2m 4 " u 2 c2
3 1 " u2 c 2
Note that when v << c , this reduces to M =
2-33
Mc 2
2u u
= . Then
4
2
M=
!
=
4m
, in agreement with the classical result.
3
(
)(
)
E = P "t = 1.79 # 1017 J s 3.16 # 10 7 s = 5.66 # 10 24 J .
Thus, m =
E
5.66 " 10 24 J
=
c2
3.00 " 108 m s
(
)
2
= 6.28 " 10 7 kg