Unformatted text preview: e largest. 33. REASONING The total energy E and the magnitude p of the relativistic momentum are
related according to Equation 28.7: E 2 = p 2c 2 + m 2c 4 or p2 = E 2 − m 2c 4
c2 (28.7) We are given a value for the total energy, but do not have a value for the mass m. However,
we recognize that the rest energy is E0 = mc2 (Equation 28.5). With this substitution,
Equation 28.7 becomes
2
2
E 2 − m 2c 4 E − E0
2
p=
=
c2
c2
We can obtain a value for the rest energy, because the total energy is the sum of the kinetic
energy and the rest energy or E = KE + E0. In other words, the rest energy is E0 = E − KE, 1498 SPECIAL RELATIVITY and we have values for both E and KE. Using this substitution for the rest energy, our
expression for p2 becomes
p=
2 2
E 2 − E0 c2 = E 2 − ( E − KE ) 2 (1) c2 SOLUTION Using Equation (1), we find that p= E 2 − ( E − KE )
c2 2 = ( 5.0 ×1015 J )2 − ( 5.0 ×1015 J ) − ( 2.0 ×1015 J ) 2
8
( 3.0 ×10 m/s ) 2 = 1.3 ×107 kg ⋅ m/s 34. REASONING Let’s define the following relative velocities:
vAB = velocity of galaxy A relative to galaxy B
vAE = velocity of galaxy A relative to Earth
vEB = velocity of Earth relative to galaxy B
These three velocities are related by the velocityaddition formula, Equation 28.8: vAB = vAE + vEB
vv
1 + AE 2 EB
c Let’s assume that galaxy A is moving to the right, along the +x direction, so its velocity
relative to earth is vAE = + 0.75c . Galaxy B is moving along the –x axis with a speed of
0.55c relative to earth, so its velocity is vBE = –0.55c. According to the velocityaddition
formula, we need to know vEB, not vBE. However, they are related by vEB = – vBE, so that
vEB = – (–0.55c) = +0.55c. SOLUTION The velocity of galaxy A relative to galaxy B is vAB = vAE + vEB
+0.75c + 0.55c
=
= + 0.920c
vAE vEB
( +0.75c )( +0.55c )
1+
1+
2
c2
c The speed of galaxy A relative to galaxy B is the magnitude of this result, or 0.920c .
______________________________________________________________________________ Chapter 28 Problems 35. 1499 SSM REASONING Let’s define the following relative velocities, assuming that the
spaceship and exploration vehicle are moving in the positive direction.
vES = velocity of Exploration vehicle relative to the Spaceship.
vEO = velocity of Exploration vehicle relative to an Observer on earth = +0.70c
vSO = velocity of Spaceship relative to an Observer on earth = +0.50c
The velocity vES can be determined from the velocityaddition formula, Equation 28.8: vES = vEO + vOS
vv
1 + EO 2 OS
c The velocity vOS of the observer on earth relative to the spaceship is not given. However,
we know that vOS is the negative of vSO, so vOS = – vSO = – (+0.50c) = –0.50c.
SOLUTION The velocity of the exploration vehicle relative to the spaceship is vES = vEO + vOS
+0.70c + ( −0.50c )
=
= + 0.31c
( +0.70c )( −0.50c )
vEO vOS
1+
1+
c2
c2 The speed of the exploration vehicle relative to the spaceship is the magnitude of this result
or 0.31 c .
______________________________________________________________________________
36. REASONING If the speeds of your car and the truck are much less than the speed of light,
the relative speed at which the truck approaches you is the same in parts (a) and (b). Let’s
suppose that you are traveling due east, which is taken to be the positive direction, and the
truck is traveling due west, which is the negative direction. The relative velocities are:
vTC = velocity of the Truck relative to the Car
vTG = velocity of the Truck relative to the Ground
vCG = velocity of the Car relative to the Ground (Note that the velocity vGC of the
Ground relative to the Car is vGC = – vCG.)
The velocity vTC of the truck with respect to the car is equal to the velocity vTG of the truck
with respect to the ground plus the velocity vGC of the ground with respect to the car:
vTC = vTG + vGC, as discussed in Section 3.4.
When vTG = –35 m/s and vGC = –25 m/s, the velocity of the truck relative to the car is
vTC = –60 m/s, where the minus sign indicates that the relative velocity is westward. 1500 SPECIAL RELATIVITY When vTG = –55 m/s and vCG = +5.0 m/s, the relative velocity of the truck with respect to
the car is still vTC = vTG + vGC = –55 m/s – 5 m/s = –60 m/s. In either case, the speed is the
magnitude of vTC, or 60 m/s.
However, the relative velocities and, hence, the relative speeds would not be the same in
parts (a) and (b) if the speeds were comparable to the speed of light. According to special
relativity, the correct relation is the velocityaddition formula, Equation 28.8: vTC = vTG + vGC
vv
1 + TG 2GC
c
2 Because of the presence of the term vTGvGC/c in the denominator, different results are
obtained when vTG = –35 m/s and vGC = –25 m/s than when vTG = –55 m/s and
vGC = –5.0 m/s.
SOLUTION
a. When vTG = –35 m/s and vGC = –25 m/s, the velocity of the truck relative to the car is vTC = vTG + vGC
−35 m/s − 25 m/s
=
= − 49.7 m/s
vTG vGC
( −35 m/s )( −25 m/s )
1+
1+
c2
( 65 m/s )2 The speed of the truck relative to the car is the magnitude of this result, or 49.7 m/s .
b. Whe...
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 Spring '13
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 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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