Unformatted text preview: ˆ†t1 (4) Substituting Equations (1) into Equation (4) yields
A2 âˆ’ A1 = âˆ†t02 âˆ’ âˆ†t01 âˆ’ âˆ†t2 + âˆ†t1 = âˆ†t2 1 âˆ’ 2
v2 c2 âˆ’ âˆ†t1 1 âˆ’ 2
v1 c2 âˆ’ âˆ†t2 + âˆ†t1 v2
v2 1 âˆ’ 2 âˆ’ 1 âˆ’ âˆ†t 1 âˆ’ 1 âˆ’ 1 = âˆ†t 2 1
c2
c2 (5) Substituting Equations (3) into Equation (5), we obtain d d v2
v2
v2
v2
A2 âˆ’ A1 = âˆ†t2 1 âˆ’ 2 âˆ’ 1 âˆ’ âˆ†t1 1 âˆ’ 1 âˆ’ 1 = 1 âˆ’ 2 âˆ’ 1 âˆ’ 1 âˆ’ 1 âˆ’ 1 c2
c2
c2
c2 v2 v1 (6) In applying Equation (6), we make use of the fact that the speed c of light in a vacuum is
equal to 1 lightyear per year: c = 1 lightyear/year: 1488 SPECIAL RELATIVITY ( 0.500c )2
12.0 lightyears A2 âˆ’ A1 =
1âˆ’
âˆ’ 1 0.500 lightyears/year c2 ( 0.900c )2
12.0 lightyears âˆ’
1âˆ’
âˆ’ 1 = 4.3 years 0.900 lightyears/year c2 b. Since A2 â€“ A1 is positive, A2 is greater than A1. Thus, when the twins meet again at the
earliest possible time, the second twin (traveling at 0.500 c) is older . 18. REASONING The relativistic momentum p of an object of mass m is given by
mv
p=
(Equation 28.3), where v is the speed of the object and c is the speed of light in
v2
1âˆ’ 2
c
a vacuum. For part (a), we will solve Equation 28.3 to determine the mass of the ion. In part
(b), we will use the mass determined in part (a) in Equation 28.3 to calculate the relativistic
momentum of the ion at its final speed.
SOLUTION
a. Solving Equation 28.3 for m and using the initial speed of v = 0.460c, we obtain 2 ( ( 0.460 c )
1âˆ’ ) v
5.08 Ã—10âˆ’17 kg â‹… m/s
2
c2
c=
v
0.460 3.00 Ã—108 m/s p 1âˆ’
m= ( ) 2 = 3.27 Ã—10âˆ’25 kg b. Substituting the value for m found in part (a) and the final speed of v = 0.920c into
Equation 28.3 yields
p= mv
1âˆ’ v2
c2 (3.27 Ã—10âˆ’25 kg ) ( 0.920) (3.00 Ã—108 m/s) = 2.30 Ã—10âˆ’16 kg â‹… m/s
=
( 0.920 c )
1âˆ’ 2 c2
______________________________________________________________________________
19. REASONING
The relativistic momentum has a magnitude p that is given by
Equation 28.3. This expression is valid for any speed v, no matter if it is small compared to
the speed c of light in a vacuum or if it is an appreciable fraction of the speed of light. Thus,
we will use it for both parts of this problem. Chapter 28 Problems 1489 SOLUTION
a. Using Equation 28.3, we obtain
mv p= 1âˆ’ v2
c2 (1.2 Ã—105 kg ) (140 m/s ) = 1.7 Ã—107 kg â‹… m/s
=
1âˆ’ (140 m/s )2 ( 3.00 Ã— 108 m/s )2 b. Using Equation 28.3 and the hypothetical value of c = 170 m/s, we obtain
mv p= 1âˆ’ v2
c2 = (1.2 Ã—105 kg ) (140 m/s ) = 3.0 Ã—107 kg â‹… m/s
(
)2
1 âˆ’ 140 m/s 2
(170 m/s ) 20. REASONING In special relativity the momentum of a particle is given by Equation 28.3 as ( 2 p = mv / 1 âˆ’ v / c 2 ) . Because of the ( 2 1âˆ’ v /c 2 ) term in the denominator, doubling the particleâ€™s speed more than doubles its momentum.
An examination of Equation 28.3 shows that the relativistic momentum is directly
proportional to the mass m. Thus, halving the particleâ€™s mass also halves its momentum.
mv The relation p = v2
1âˆ’ 2
c
particles a, b, and c are:
p=
a mv
v2
1âˆ’ 2
c indicates that the magnitudes of the relativistic momenta for ,p =
b ( 1 m ) ( 2v ) =
2
1âˆ’ ( 2v )
c 2 2 mv
4v 2
1âˆ’ 2
c ,p=
c ( 1 m ) ( 4v ) =
4
1âˆ’ ( 4v )
c2 2 mv
16v 2
1âˆ’ 2
c . These results show that particle c has the greatest momentum magnitude, followed by
particle b and then by particle a.
SOLUTION The momenta of the three particles are: Particle a p= mv
1âˆ’ v2
c2 (1.20 Ã— 10âˆ’8 kg ) ( 0.200 ) (3.00 Ã— 108 m/s ) = 0.735 kg â‹… m/s
=
( 0.200c )2
1âˆ’
c2 1490 SPECIAL RELATIVITY Particle b p= mv
v2
c2 1âˆ’ Particle c p= mv
1âˆ’ v2
c2 ( 12 Ã—1.20 Ã— 10âˆ’8 kg ) ( 2 Ã— 0.200 ) (3.00 Ã— 108 m/s ) = 0.786 kg â‹… m/s
=
1âˆ’ ( 2 Ã— 0.200c )2
c2 ( 1 Ã— 1.20 Ã— 10âˆ’8 kg ) ( 4 Ã— 0.200) (3.00 Ã— 108 m/s ) = 1.20 kg â‹… m/s
4
=
( 4 Ã— 0.200c )2
1âˆ’
c2 As expected, the ranking of the momenta (largest first) is c, b, a.
______________________________________________________________________________
21. SSM REASONING The height of the woman as measured by the observer is given by Equation 28.2 as h = h0 1 âˆ’ (v / c) 2 , where h0 is her proper height. In order to use this
equation, we must determine the speed v of the woman relative to the observer. We are
given the magnitude of her relativistic momentum, so we can determine v from p.
2
2
2
2
SOLUTION According to Equation 28.3 p = mv / 1 âˆ’ v / c , so mv = p 1 âˆ’ v / c
Squaring both sides, we have m2v 2 = p 2 (1 â€“ v 2 / c 2 ) or p2 v 2 m2 + 2 = p 2 c m2 v 2 + p 2 or v2
= p2
2
c p2 v2 = m2 + p2
c2 Solving for v and substituting values, we have
v= p
m2 + 2 p
c2 2.0 Ã— 1010 kg â‹… m/s = 2.0 Ã—1010 kg â‹… m/s (55 kg)2 + 8 3.00 Ã— 10 m/s = 2.3 Ã— 108 m/s 2 Then, the height that the observer measures for the woman is
2 2.3 Ã— 108 m/s v
h = h0 1 âˆ’ = (1.6 m) 1 âˆ’ = 1.0 m
8 c 3.0 Ã— 10 m/s ______________________________________________________________________________
2 Chapter 28 Problems 1491 22. REASONING The magnitude prel of the relativistic momentum of the spacecraft is given
mv
by prel =
(Equation 28.3), where m is the mass of the spacecraft, v is its speed, and
v2
1âˆ’ 2
c
c is...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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