1 A proton moving at speed
v
= 0
.
999
c
will appear like a moving disk.
(a)
Compute the thickness of the disk assuming that at rest the proton’s diameter
is 10

15
m. (b) Compute the proton’s kinetic energy in MeV/c, assuming its
mass is 938 MeV/c
2
. Suppose the proton collides and fuses with a stationary
proton to create a new particle. (c) What is the mass of the new particle?
(a) Let
d
be the diameter of the proton at rest in some frame of reference.
When the proton moves relative to this frame, its diameter will be Lorentz
contracted in the direction of motion. Therefore, the proton’s thickness will
be
t
=
d/γ
, where
γ
= 1
/
1

β
2
and
β
=
v/c
.
Setting
β
= 0
.
999 and
d
= 10

15
m gives
t
= 4
.
5
×
10

17
m = 4
.
5
×
10

2
fm.
(b) The total energy of a particle is given by
E
=
γmc
2
, therefore, its kinetic
energy is
T
=
E

mc
2
, that is,
T
= (
γ

1)
mc
2
. Therefore, since
γ
= 22
.
4,
T
= (22
.
4

1)
×
(938MeV/c
2
)
×
c
2
= 20
,
041 MeV, that is, 20.04 GeV.
(c) In collisions, in a given frame of reference, energy and momentum are
conserved, or, equivalently, 4momentum
P
= (
E, p
) is conserved. That is,
E
1
+
E
2
=
E
F
,
and
p
1
+
p
2
=
p
F
.
Since
E
1
=
γmc
2
and
E
2
=
mc
2
, the total initial energy is
γmc
2
+
mc
2
=
21
,
917 MeV, which must equal the total final energy.
Likewise, the total
initial momentum is
p
1
+ 0 =
γmv
+ 0 =
γmc
2
β/c,
= 20
,
958 MeV/c
,
which must be equal to the total final momentum.
Now that we have the
energy and momentum of the final state, we can compute its mass using the
general relation
E
2
= (
pc
)
2
+ (
mc
2
)
2
, that is, using
m
=
E
2

(
pc
)
2
/c
2
=
21
,
917
2

20
,
958
2
/c
2
= 6
,
400 MeV/c
2
.