Physics Solution Manual for 1100 and 2101

Using this expression in equation 298 we find that h

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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 more-massive 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 break-up and the system would have a non-zero momentum. According to the principle of conservation of momentum, the total momentum after the break-up must be zero, just as it was before the break-up. 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 . ______________________________________________________________________________ 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 X-ray photon has a much greater frequency than does a microwave photon (see Section 24.2). The X-ray 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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