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41. (a) We use the relativistic relationship between speed and momentum:
p
=
γmv
=
mv
p
1
−
(
v/c
)
2
,
which we solve for the speed
v
:
v
c
=
s
1
−
1
(
pc
mc
2
)
2
+1
.
For an antiproton
mc
2
= 938
.
3MeV and
pc
=1
.
19 GeV = 1190 MeV, so
v
=
c
s
1
−
1
(1190 MeV
/
938
.
3MeV)
2
+1
=0
.
785
c.
For the negative pion
mc
2
= 193
.
6MeV, and
pc
is the same. Therefore,
v
=
c
s
1
−
1
(1190 MeV
/
193
.
6MeV)
2
+1
=0
.
993
c.
(b) See part (a).
(c) Since the speed of the antiprotons is about 0
.
78
c
but not over 0
.
79
c
, an antiproton will trigger C1.
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Momentum

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