Environmental Engineering Reference
In-Depth Information
(x
ct) where x
=
x B
x A etc. Putting the numbers in gives γ
=
2 and hence v/c
0 . 866. Note that you could get this directly from the
length contraction formula since the events occur at the same time in one
of the frames. We can make use of “ t =
=
vx/c 2 ) ” to determine the
γ(t
corresponding time interval, i.e.
2 0
10 8
1000
t =
10 6 s .
0 . 866
×
=−
5 . 77
×
3
×
The minus sign means that the event at larger x is observed first in the primed
frame.
6.14 The emission of the two pulses corresponds to two events with x
=
4km
5 µ s. Given t =
0 we need to use t =
vx/c 2 ) to
and t
=
γ(t
0 . 375.
6.15 Key here is to identify the relevant events. Let event A be the impact of
the comet and event B be the event in the party that is simultaneous with
event A according to an observer whizzing past the Earth. Then t =
determine the speed, i.e. v/c
=
ct/x
=
0
by definition (primes indicate the arbitrary inertial frame) and we want
to know the corresponding time interval in the Earth frame ( t ). Since
we know t
and x the relevant equation is t =
vx/c 2 ) .
γ(t
The extreme values of t occur when v
c whence t
x/c . Thus
10 11 /( 3
10 8 ) s
89 minutes. The
impact of the comet occurs (in the Earth frame) at the midpoint of the
party whilst the party ends with the students observing the impact using
their telescopes.
6.16 A diagram will help to clarify the way in which you must use the velocity
addition formula. The speed is ( 0 . 70
the party must last a time 2
×
8
×
×
=
0 . 97 c .
6.17 As in the last question, a diagram will help. Required speed is ( 0 . 5
+
0 . 85 )c/( 1
+
0 . 7
×
0 . 85 )
=
0 . 8 )c/( 1
0 . 5
×
0 . 8 )
=−
0 . 5 c . The minus sign implies the rocket moves
towards the Earth.
6.18 Velocity addition gives V
=
(v
+
c/n)/( 1
+
v/(nc)) .Noteif n
=
1 then
V
=
c . Fizeau would have worked with v
nc hence the denominator can
v/(nc)) 1
be simplified using ( 1
+
1
v/(nc) . This yields the result after
neglecting terms suppressed in v/c .
6.19 Work in A's rest frame and determine the velocity of B in that frame, then
use tan 30 =
v By /v Bx to determine u . In more detail: use velocity addition
to obtain v Bx =
= 3 which implies
u and v By =
u/γ (u) and hence γ(u)
= 2 / 3.
u/c
PROBLEMS 7
10 27
10 8 ) 2 /( 1 . 6022
10 13 ) MeV
7.1 Use E
=
1 . 673
×
×
( 2 . 9979
×
×
=
938
mc 2 . Hence m
938 MeV /c 2 .
MeV
=
=
7.2 (a) γ
1
=
1 hence v
=
0 . 866 c .(b) γ
1
=
5 hence v
=
0 . 986 c .
c 2 d m/ d t . Hence rate of mass loss is 4.2 million tonnes per
7.3 Power output
=
second.
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