MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department. Problem Set 5 Solutions

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1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department Physics October, 003 Problem Set 5 Solutions Problem A Flying Brick, Resnick & Halliday, #, page 7. (a) The length contraction factor along the direction of motion is γ. Therefore, according to an observer moving with speed u relative to the brick, the volume of the brick is γ V 0, where γ = ( u ) / and u is in units of c. Note that the measured volume is independent of the orientation of the brick with respect to its direction of motion. To see this, imagine computing the volume of a brick in any orientation in its own rest frame S by breaking it up into a large number, N, of small cubes of volume l 3 that are aligned with the axes. The volume of the brick is then approximately Nl 3. Assuming as usual that S is S boosted in the x direction, each cube is contracted along one side and has volume γ l 3. This gives a total approximate volume of γ Nl 3. As N goes to infinity and l goes to zero the ratio tends to the rest volume, V 0. Therefore V = γ V 0. (b) The measured mass is simply γm 0. (c) The density of the brick is the quotient of its mass and its volume, which is ρ = γ m 0 /V 0 = γ ρ 0. In order for the density to be % larger than its rest value, we must have and thus u 0.. ρ ρ 0 = γ = 0.0, ρ 0 Problem A Small Change in Speed But a Large Change in Energy, Resnick & Halliday, #5, page 7. (a) The electron s kinetic energy, K is obtained by subtracting its rest mass, m 0 c from its total relativistic energy, γm 0 c, so that K = (γ )m 0 c. If β = 0.999, then γ.366, and so K kev 0.9 MeV. (b) If β is increased by 0.05% to a value of β , the kinetic energy increases by 43%. Problem 3 Limiting Rest Mass of a Neutrino In the limiting case that the neutrinos actually arrived at Earth 3 hours later than the light, their velocity must have been: u = 70, 000 c years 70, 000 years + 3 hours = c( 0 9 ) β = 0 9.

2 Using the high energy approximation found in Problem 8, page 8: m 0 c u/c, E we find: Δu m 0 c E = 634 ev. c Problem 4 The Fastest Particles, Resnick & Halliday, #3, page 8. (a) For a proton of energy 0 0 ev, we see that γ 0, given the proton rest energy of GeV. Therefore, β is of order γ 0 a very small number indeed. (b) The protons relativistic mass is 0 GeV/c 0 6 kg. (c) According to an observer moving with this highly relativistic proton, the contracted diameter of the Earth s orbit is γ km 3 m. Problem 5 Some Missing Algebra, Resnick & Halliday, #8, page 8. Begin with the following three relations: u M = m + m 0 ; M u = m u ; u =. + u /c Substituting M from the first equation into the second equation, we find m u = u. m + m 0 Now, substitute this expression for u in to the third equation: [ ] m u m u = u +. m + m 0 c m + m 0 If we cancel u and let q m 0 /m, this equation becomes [ ] u = +. + q c ( + q) After we multiply both sides by the term in square brackets, and then multiply by ( + q), we obtain ( + q) ( + q) + u = 0. c Finally, we expand ( + q) and collect terms to obtain u q = 0, c from which we derive the desired result that m = m 0 / u /c.

3 Problem 6 Solving the Equation of Motion, Resnick & Halliday, #33, page 0. (a) When F and a are parallel, the relativistic form of Newton s second law becomes du F u a =. dt m c Since, m = m 0 ( u /c ) /, we see that 3/ u F du = dt c m 0 (b) For the case where F does not depend on u or t, the equation above can be integrated quite easily, with the result / u F u = t + C, c m 0 where C is a constant of integration. If we set u = 0 at time t = 0, then C = 0. After multiplying both sides by ( u /c ) / and then squaring both sides, we obtain u F u = t, c m 0 c c from which follows the desired result after a bit of algebra: (F/m 0 )t u =. + (F/m 0 c) t (c) The equation for t(u) is obtained trivially from the first equation in part (b), with the desired result (m 0 /F )u t =. u /c (d) As u 0, we see that t(u) 0, and as t, we see that u(t) c. Problem 7 Not the Simplest Reference Frame, Resnick & Halliday, #40, page 0. (a) Let γ i = ( β i ) /, for i =,. For β = 0.8 and β = 0.6, we find that γ = 5/3 and γ = 5/4. The conserved relativistic momentum, is γ m 0 β c γ m 0 β c = (γ β γ β )m 0 c = 7m 0 c/. (b) The the value of the Newtonian momentum is obtain by setting γ i =, which yield the result (β β )m 0 c = m 0 c/5. (c) The conserved total relativistic energy is (γ + γ )m 0 c = 35m 0 c /. 3

4 Problem 8 The Great Neutron Shootout, Resnick & Halliday, #43, page 0. Sam s frame is riding on the neutrons that Sue is firing. (a) For u = 0.60c, we find γ =.5. Then Δx = 0.8 Δx = 8 km. (b) β = 0.6. Therefore, let the speed of Jim s neutrons in Sam s frame be: 0.6c ( 0.6c) u u v = = = 0.88 c. uv/c ( 0.36) (c) In lecture we set up the relativistic energy and momentum so that, in fact, they would be conserved quantities. Thus, Jim and Sam agree on the fact that relativistic energy and momentum are conserved though they each measure different values for these quantities. (d) We are told that one of the neutrons is deflected in a collision by an angle of 30 o in Jim s frame. We can find the angle in Sam s frame by using ordinary velocity transforms to obtain the following, after the collision. Consider one of Jim s electrons moving from right to left with speed v, and then undergoing a collision in which the direction is changed by 30, but the speed is unchanged. After the collision, the velocity components are then: u x u y = v cos 30 = v sin 30. These are easily transformed into Sam s frame (moving to the right with speed v): v cos 30 v u x = ( v ) cos 30 /c v sin 30 u y =. γ[ ( v ) cos 30 /c ] In turn, these can be used to find the angle θ in Sam s frame: u y v sin 30 β sin 30 tan θ = = = = 0.44, u x vγ(cos 30 + ) cos 30 + and therefore θ =.. (e) This is a simple exercise in time dilation. 0,000 neutrons emerge in second in the lab frame. In Sam s frame this takes Δt = γδt =.5 seconds. This means that in Sam s frame 0, 000/.5 = 8, 000 neutrons come out every second; rate of fire R = 8, 000 s. Problem 9 Cosmic Ray Protons in the Galactic Magnetic Field, Resnick & Halliday, #50, page. Because this is an extremely energetic particle we can assume that its momentum is simply p = E/c. The relativistic radius of curvature is given by p E 0 0 ev( J/eV) r = = = qb qcb ( C)(3 0 8 m/s)(0 9 T) = m ly 4

5 This value of curvature is approximate to the radius of the luminal galaxy. Problem 0 The Ultimate Terrestrial Accelerator, Resnick & Halliday, #5, page. Using the equation from Problem 9 again, we see that at that radius and magnetic field: pc = rqcb = ( m)( C)(3 0 8 m/s)(5.0t )/( J/eV) = MeV E. (b) The rest energy of the proton is 836 times larger than that of an electron, which is 0.5 MeV. Therefore, m mc E MeV 0 7 = =. m 0 m 0 c m 0 c 938. MeV Problem The Decay of a Pion, Resnick & Halliday, #56, page. To solve this problem with conservation of momentum and energy we must assign the momentum p ν = E ν /c to the (approximately) massless neutrino. We can calculate the momentum of the muon from: p µ c = K + m 0µ c K = 4. + (05.7)4. MeV = 9.7 MeV. Assuming the two momenta are equal and opposite, we get E ν = 9.7 MeV. The total energy is then E = E µ + E ν = K µ + m 0µ c + E ν = 39.5 MeV Problem Water is Heavier Than Ice, Resnick & Halliday, #66, page 3. Since the energy required to melt the ice is acquired by the liquid water, the rest massenergy of liquid water is slightly larger than that of the ice. If the mass of the ice is M 0, then energy required to melt the ice is M J kg ; division by M 0 c gives the fractional increase in mass energy, Problem 3 A Ton of Sunlight, Resnick & Halliday, #69, page 3. The Sun gives off energy at a rate of R = W. The fraction of this that hits the Earth in one day is: E = ( day)r πre ( m) = s( W) 4( m) =.57 0 J, 4πR ES where R ES is the distance from the Earth to the Sun, and R E is the radius of the Earth. This energy is the mass equivalent of E/c = kg. Since 000 kg equals one metric ton ( 00 lbs), the total mass received is approximately m = 70 tons per day. 5

6 Problem 4 An Enormous Source of Energy, Resnick & Halliday, #70, page 3. If the power output of the quasar, L, is due to the conversion of mass into energy, then L = M c, where M is the rate of mass consumption. Letting L = 0 4 W, we find that Ṁ 0 4 kg s 7 M yr. Problem 5 Nuclear and TNT Explosions Compared, Resnick & Halliday, #7, page 3. (a) The energy released in the nuclear explosion is kg c = J. (b) The mass of TNT required to yield the same explosion energy as the nuclear bomb is M = J 0.7 kg mol kg J mol (c) For the fission, the mass energy conversion efficiency is 0 3, while for TNT, the efficiency is, using the numbers already calculated, J/( kg c ) 0 0. Therefore, the fission bomb is 0 7 more efficient at converting mass to energy than TNT. Problem 6 The energy liberated in each of the following reactions is shown in parentheses: H + H He 3 + γ (5.49 MeV) He 3 + He 3 He 4 + H (.86 MeV) Li 7 + H He 4 (7.36 MeV) N 5 + H C + He 4 (4.96 MeV) C 3 + H N 4 + γ (7.55 MeV). Optional Problem A The power generated per unit mass of the Sun is 0 4 W kg, while that of a person is W kg. The energy generation in a person is via chemical reactions, which is many orders of magnitude less efficient than nuclear energy which powers the Sun. This disparity in power per gram (seemingly in the wrong sense) suggests that a typical atom in a person is reacting much more quickly than is the average nucleus in the Sun. Thus, even though the Sun generates less power per gram than a person, its total energy output per gram over its lifetime (of 0 0 ) years is millions of times greater than that of a person. Optional Problem B If the luminosity of the Sun, L, is produced at the expense of gravitational potential energy specifically, due to the Sun s contraction then d GM GM dr L =. dt R R dt In order to obtain a rough estimate of the required contraction timescale, T, simply set dr = 0.5R and dt = T. Therefore, L 0.5GM /R T, so that T 0 7 yr. 6

7 Optional Problem C The total light travel time is Also, the angles θ i and θ t are related by y 0 T = (sec θ i + n sec θ t ). c x 0 = y 0 (tan θ i + tan θ t ). The minimum travel time must correspond to dt /dθ i = 0 (θ t could also have been chosen as the independent variable), so that dt = 0 = y 0 (sin θ i sec θ i + n sin θ t sec θ t dθ t ). dθ i c dθ i The derivative dθ t /dθ i is obtained by differentiating the constraint relation: 0 = sec θ i + sec θ t dθ t. dθ i After substituting this result, we see that Snell s law immediately follows: sin θ i n sin θ t = 0. Optional Problem D Lagrange s Equations of Motion (a) Identify the independent variable as z. Therefore the only kinetic term is T = Mż There are two potentials to consider, however; that of the spring and that of gravity. These can be represented as: V = kz + M gz. Here I have defined increasing z as the positive direction. Performing the required derivatives for Euler s Equation we find: 0 = kz M g Mz Mz = kz M g, which is what we expect from the sum of forces equal to the mass times the acceleration. (b) This is a slightly trickier problem to set up. To start, let us define two Cartesian coordinates: x = l sin φ and z = l cos φ. This way the kinetic energy is: T = M ( ẋ + ż ) = M l (φ cos φ + φ sin φ) = M l φ 7.

8 This result is not surprising as we are told this can be a one dimensional problem. Define the potential as: V = Mgz = Mgl( cos φ), where V = 0 when the mass is hanging vertically. Thus, L = Ml φ Mgl( cos φ). Plugging these values into the appropriate Euler equation we find: L d L = 0, φ dt φ g or 0 = Ml φ g Mgl sin φ φ = sin φ φ, l l where the approximation is only valid for small angles. We see the solution to this differential equation is sinusoidal with an oscillation frequency of ω = g/l. 8

Physics 202. Professor P. Q. Hung. 311B, Physics Building. Physics 202 p. 1/2

Physics 202. Professor P. Q. Hung. 311B, Physics Building. Physics 202 p. 1/2 Physics 202 p. 1/2 Physics 202 Professor P. Q. Hung 311B, Physics Building Physics 202 p. 2/2 Momentum in Special Classically, the momentum is defined as p = m v = m r t. We also learned that momentum