This preview shows page 1. Sign up to view the full content.
Unformatted text preview: the speed of light in a vacuum. The numerator of Equation 28.3 is the magnitude of the
nonrelativistic momentum p = mv (Equation 7.2), so we have that mv prel = 1− v2 p = 1− c2 (1) v2
c2 We are told that the pilot measures the proper time interval ∆t0 between two events to be
half the dilated time interval ∆t: ∆t0 = 1 ∆t . Thus, the dilated time interval is
2 ∆t = 2∆t0 (2) In general, the proper time interval ∆t0 and the dilated time interval ∆t are related by
∆t = ∆t0 (28.1) v2
1− 2
c SOLUTION Substituting Equation (2) into Equation 28.1, and solving for the expression 1− v2
that also appears in Equation (1), we obtain
c2 ∆t0 ∆t = 2 ∆t0 = v2
1− 2
c or 2= 1
v2
1− 2
c or v2 1
1− 2 =
2
c (3) Substituting Equation (3) into Equation (1), we find that prel = p
1− v2 = p
= 2 p = 2 (1.3 ×1013 kg ⋅ m/s ) = 2.6 ×1013 kg ⋅ m/s
1 2 c2
______________________________________________________________________________ 1492 SPECIAL RELATIVITY 23. SSM REASONING AND SOLUTION The total momentum of the man/woman system is
conserved, since no net external force acts on the system. Therefore, the final total
momentum pm + pw must equal the initial total momentum, which is zero. As a result,
pm = – pw where Equation 28.3 must be used for the momenta pm and pw. Thus, we find mm vm
1 – ( vm / c ) 2 =– mw vw
1 – ( vw / c ) 2 We know that mm = 88 kg, mw = 54 kg, and vw = +2.5 m/s. Remember that c has the
hypothetical value of 3.0 m/s. Solving for vm, we find vm = –2.0 m/s . ______________________________________________________________________________
24. REASONING AND SOLUTION The total energy for each particle is given by E= mc 2 v
1– c 2 (9.11 × 10
= –31 )( kg 3.00 × 108 m/s 0.20c 1– c ) 2 2 = 8.4 × 10 –14 J (28.4) The annihilation energy is twice this value, so E ′ = 2E = 1.7 × 10−13 J .
______________________________________________________________________________
25. SSM REASONING AND SOLUTION
a. In Section 28.6 it is shown that when the speed of a particle is 0.01c (or less), the
relativistic kinetic energy becomes nearly equal to the nonrelativistic kinetic energy. Since
the speed of the particle here is 0.001c , the ratio of the relativistic kinetic energy to the nonrelativistic kinetic energy is 1.0 .
b. Taking the ratio of the relativistic kinetic energy, Equation 28.6, to the nonrelativistic
1 kinetic energy, 2 mv2, we find that 1 –1 mc 2 2
2
2 1 1 – (v / c ) = 2 c –1 1 mv 2 v 1 – (v 2 / c 2 ) 2
1 –1 = 6.6 1 – (0.970c)2 / c 2 ______________________________________________________________________________
c
= 2 0.970c 2 Chapter 28 Problems 1493 26. REASONING Compressing the spring stores elastic potential energy, which increases the
total energy of the spring. Because the spring is not in motion, it is the spring’s rest energy
E0 that increases, as well as its mass m, according to E0 = mc 2 (Equation 28.5). We will use
the hypothetical value of c = 3.00×102 m/s for the speed of light in a vacuum in Equation
28.5. The increase of ∆m = 0.010 g = 0.010×10−3 kg in the mass of the spring, therefore,
corresponds to an increase ∆E0 in the rest energy of the spring, where ∆E0 = ( ∆m ) c 2 (1) The increase of the rest energy is equal to the spring’s elastic potential energy
PEelastic = 1 kx2 (Equation 10.13), where k is the spring constant of the spring, and x is the
2
distance by which the spring is compressed from its equilibrium length. Therefore, we have
that
∆E0 = PEelastic = 1 kx2
(2)
2 SOLUTION Setting the right sides of Equations (1) and (2) equal to one another and
solving for x2 yields ∆E0 = 1 kx 2 = ( ∆m ) c 2
2 x2 = or 2 ( ∆m ) c 2
k (3) Taking the square root of Equation (3), we obtain ( −3 ) 2 0.010 ×10 kg
2 ( ∆m ) c2
2 ( ∆m )
x=
=c
= 3.00 ×102 m/s
= 0.046 m
k
k
850 N/m
______________________________________________________________________________ ( ) 27. REASONING The mass m of the aspirin is related to its rest energy E0 by Equation 28.5,
2 8 E0 = mc . Since it requires 1.1 × 10 J to operate the car for twenty miles, we can calculate
the number of miles that the car can go on the energy that is equivalent to the mass of one
tablet.
SOLUTION We begin by converting the mass m from milligrams (mg) to kilograms (kg): ( 325 mg ) 1000 mg 1000 g = 325 ×10−6 1g 1 kg The number N of miles the car can go on one aspirin tablet is kg 1494 SPECIAL RELATIVITY N= E0 mc 2 = (1.1 × 108 J ) / ( 20.0 mi ) (1.1 × 108 J ) / ( 20.0 mi ) (325 × 10−6 kg )(3.0 × 108 m/s )
= 2 (1.1 × 10 J ) / ( 20.0 mi )
8 = 5.3 × 106 mi ______________________________________________________________________________
28. REASONING In order to change a certain mass of ice at 0 °C into liquid water at 0 °C heat
must be added, and heat is a form of energy. Therefore, the energy of the liquid water is
greater than that of the ice. According to special relativity, energy and mass are equivalent.
Since the liquid water has the greater energy, it also has the greater...
View
Full
Document
This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

Click to edit the document details