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Unformatted text preview: ruiser at a velocity
U = +1.000c – 0.800c = +0.200c
d. The aliens see the ions moving away from the cruiser at a velocity
U′ = +0.994c – 0.800c = +0.194c
______________________________________________________________________________ Chapter 28 Problems 1507 47. REASONING The total linear momentum of the system is conserved, since no net external
force acts on the system. Therefore, the final total momentum p1 + p 2 of the two fragments
must equal the initial total momentum, which is zero since the particle is initially at rest. As
a result, p1 = − p 2 , where Equation 28.3 must be used for the magnitudes of the momenta
p1 and p2. Thus, we find
m1v1
2
1 − (v1 / c 2 ) = −m2v2
2
1 − (v2 / c 2 ) SOLUTION Letting fragment 2 be the moremassive fragment, we have that v1 = +0.800c ,
m 1 = 1.67 × 10 –27 kg , and m 2 = 5.01 × 10 –27 kg . Squaring both sides of the above
equation, rearranging terms, substituting the known values for v1, m1, and m2, we find that
2 22 v2
2 2 1 − (v2 / c ) = m1 v1 2
2
m2 1 − (v1 / c 2 ) = (1.67 × 10
(5.01×10 –27 –27 2 2 kg) (+0.800c)
2
= 0.1975c
2
2
kg) 1 – (+0.800c / c) Therefore,
2
2
2
v2 = 0.1975c 2 1 − (v2 / c 2 ) = 0.1975c 2 – 0.1975v2 Solving for v2 gives
2 0.1975c
v2 = ±
= ±0.406c
1.1975
We reject the positive root, since then both fragments would be moving in the same
direction after the breakup and the system would have a nonzero momentum. According
to the principle of conservation of momentum, the total momentum after the breakup must
be zero, just as it was before the breakup. The momentum of the system will be zero only
if the velocity v2 is opposite to the velocity v1. Hence, we chose the negative root and v2 = –0.406c .
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48. REASONING AND SOLUTION Any change in the energy of a system is equivalent to a
change in the mass of the system, according to ∆ E 0 = ( ∆ m) c 2 (see Section 28.6).
Therefore, the change in mass of a system when its energy changes by an amount ∆ E 0 is
given by
∆E
∆m = 20
(1)
c 1508 SPECIAL RELATIVITY We want to know how close two stationary electrons have to be positioned so that their total
mass is twice what it is when the electrons are very far apart. Let m total = 2 m e represent
the total mass of the system when the electrons are very far apart. Then, when the electrons
have been brought together so that their total mass becomes 2 m total , we have
mtotal + ∆m = 2mtotal (2) When the two electrons are very far apart, their electric potential energy is zero. When they
are brought together so that their separation is r, their electric potential energy is EPE = ke
r 2 where e is the charge on the electron (see Equations 19.3 and 19.6). Therefore, the change
in energy of the system is
ke
∆E0 = EPE – 0 =
r 2 (3) Combining Equations (1) and (2), we have
mtotal + ∆E0
c 2 = 2mtotal Substituting the right hand side of Equation (3) for ∆ E 0 , we have 1 ke2 mtotal + 2 = 2mtotal cr Further simplification and solving for r leads to
1 ke2 = mtotal c2 r r= ke 2 = ke 2 = 9 2 2 (8.99 ×10 N ⋅ m /C )(1.60 ×10 –19 C) 2 = 1.40 ×10 –15 m mtotalc
2me c
2(9.11×10 kg)(3.00 ×10 m/s)
______________________________________________________________________________
2 2 –31 8 2 Chapter 28 Problems 49. 1509 SSM WWW REASONING Only the sides of the rectangle that lie in the direction of
motion will experience length contraction. In order to make the rectangle look like a square,
each side must have a length of L = 2.0 m. Thus, we move along the long side, taking the
proper length to be L0 = 3.0 m. We can solve for the speed using Equation 28.2. Then, with
this speed, we can use the relation for length contraction to find L for the short side as we
move along it.
SOLUTION From Equation 28.2, L = L0 1 − (v 2 / c 2 ) , we find that
2 2
L 2.0 m v = c 1− = c 1− = 0.75 c
L 3.0 m 0 Moving at this speed along the short side, we take L0 = 2.0 m and find L:
2 L = L0 2 v 0.75 c 1 − = (2.0 m) 1 − = 1.3 m
c
c The observed dimensions of the rectangle are, therefore, 3.0 m × 1.3 m , since the long side is not contracted due to motion along the short side.
______________________________________________________________________________ CHAPTER 29 PARTICLES AND WAVES
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS
1. (a) At the higher temperature, the intensity per unit wavelength is greater, and the aximum
occurs at a shorter wavelength (see Section 29.2). 2. (b) An Xray photon has a much greater frequency than does a microwave photon (see
Section 24.2). The Xray photon also has a greater energy E, because E = hf (Equation
29.2), where h is Planck’s constant and f is the frequency. The wavelength λ and frequency
of a photon are related by λ = c/f (Equation 16.1), where c is the speed of light in a vacuum.
Since the microwave photon has the smaller frequency, it has the greater wavelength. 3. (c) A green photon has a greater frequency than does a red photon (see Section 24.2).
Therefore, the green p...
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