!
!
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
2-9
(b)
No change, because the masses cancel each other.
(a)
When K = (" # 1)mc = 5mc , " = 6 and E = " mc = 6(0.511 0 MeV) = 3.07 MeV .
(b)
2
1 *, $ v ' -/
= 1 #& )
" ,+ % c ( /.
(a)
2
When K e = K p , me c (" e #1) = mp c " p # 1 . In this case me c = 0.511 0 MeV and
2
12
2-12
2
2
* $ 1'2 and v = c,1 " & ) /
,+ % # ( /.
2
2
[
mp c2 = 938 MeV , " e = 1 # ( 0.75)
first equation, we find " p = 1 +
But " p =
(b)
1
12
2'
$
&%1 # u p c )(
(
)
(
2 12
]
12
1 2
* $1 '2= c,1 " & ) /
,+ % 6 ( /.
= 0.986c
)
= 1.511 9 . Substituting these values into the
(0.511 0)(1.511 9 # 1)
me c2 (" # 1)
=1 +
= 1.000 279 .
2
939
m pc
(
"2 1 2
; therefore u p = c 1 " # p
)
= 0.023 6c .
# " &# m &
e
e
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
#
=
&
)
&
2)
%
c ( % 400 (
12
1 v *
= ,1 #
c + 4002 /.
v = 0.999 997 c
!
2-15
(b)
2
2
2
5
K = E " mc = (400 "1)mc = 399mc = (399) (938.3 MeV ) = 3.744 # 10 MeV
(a)
2
"2
#
Vq &
Vq & -+
v )+ #
and
K = " mc # mc = Vq and so, " = %1 +
=
1
"
1
+
(
* %
( .
2
$
c ,+ $
mc '
mc2 ' /+
2
2
2
12
$
v &
1
= %1 "
c & 1 + 5.0 # 10 4 eV 0.511 MeV
'
(
)
12
(
&
2)
&
*
= 0.412 7
or v = 0.413c .
(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%
[
2-19
3
" 100%
40
or about
]
"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
Conservation of mass-energy requires K + 2mc2 = 2E where K is the electron’s
kinetic energy, m is the electron’s mass, and E is the gamma ray’s energy.
!
E=
K
2
+ mc = ( 0.500 + 0.511) MeV = 1.011 MeV .
2
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)
Ee " = pe " c + mc
E
= 1.011 MeV c .
c
where Ee " is the electron’s total energy, Ee " = K + mc2 , yields
Using
pe " =
Finally, cos " =
(1.00) 2 + 2 (1.00) (0.511) MeV
1
2
2
K + 2 Kmc =
c
pe #
2p
c
= 1.422 MeV c .
= 0.703 ; " = 45.3° .
2-23 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
2
1 " v2l c2
+
mh c
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 =
(0.987 )(6.22 )
= 0.284m h
Substituting in this value and numerical quantities in the mass-energy
conservation equation, one obtains 3.34 " 10 #27 kg = 6.22m l + 2.01mh which in turn gives
3.34 " 10
#27
#27
kg
3.34 " 10
3.78
kg = (6.22 )( 0.284 )ml + 2.01mh or mh =
= 8.84 " 10 #28 kg and
ml = (0.284 ) mh = 2.51 " 10 #28 kg .
2-26
3 108 kg 1 "
!
!
1 452 kg 1 "
2-31
1
Energy conservation:
v2
2
c
v2
c
2
= M.
=
Mv
c
2
1 " 02
1 400 kgc +
900 kgc
2
1 " 0.852
Momentum conservation: 0 +
Mc
=
2
1 " v2 c2
900 kg (0.85c)
1 " 0.85
2
=
;
Mv
1 " v2 c2
.
v 1 452
=
= 0.467
c 3 108
v = 0.467 c .
(a)
Dividing gives
(b)
Now by substitution 3 108 kg 1 " 0.467 2 = M = 2.75 # 10 3 kg .
Conservation of momentum " mu :
mu
1 "u
2
2
c
+
m( "u)
3 1 "u
Conservation of energy " mc2 :
2
2
c
=
Mv f
1 " vf2
c
2
=
2mu
3 1 " u 2 c2
.
;
mc
2
1 " u 2 c2
+
mc
2
3 1 " u 2 c2
To start solving we can divide: vf =
M
2
1 " u 4c
2
=
M=
3 1 "u
=
4mc
2
3 1 " u 2 c2
.
Then
4m
2
2
1 " vf2 c 2
2u u
= .
4
2
2
c
=
M
4 " u 2 c2
(1 2 )
2m 4 " u 2 c2
3 1 " u2 c 2
Note that when v << c , this reduces to M =
result.
2-33
Mc
=
4m
3
, in agreement with the classical
The energy that arrives in one year is
(
)(
)
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 .