Physics Solution Manual for 1100 and 2101

A according to table 123 the latent heat of fusion

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Unformatted text preview: frame. 8. (c) The observers will always disagree about the time interval and the length, as indicated by the time-dilation and length-contraction equations. However, each will measure the same relative speed for the other’s motion. 9. 22.7 m 10. (c) The expression p = mv for the magnitude of the relativistic momentum applies 1 − v2 / c2 at any speed v. When it is used, the conservation of linear momentum is valid for an isolated system no matter what the speeds of the various parts of the system are. 11. (d) Both of the expressions can be used provided that v << c. Expression B differs from expression A only by a negligible amount in this limit of small speeds. 12. 0.315 kg·m/s 1476 SPECIAL RELATIVITY 13. (b) The mass m of an object is proportional to the object’s rest energy E0, according to E0 = mc 2 . The rest energy includes all forms of energy except kinetic energy, which plays no role here, because the glass is not moving. To freeze half the liquid into ice, energy in the form of heat must be removed from the liquid, so the water in possibility B has more mass than the water in possibility A. To freeze the remaining liquid into ice, more heat must be removed, so the water in possibility A has more mass than the water in possibility C. Thus, the ranking in descending order (largest first) is B, A, C. 14. 1.88 × 10−5 kg 15. (b) The total energy is the sum of the rest energy and the kinetic energy. The rest energy includes all forms of energy (including potential energy) except kinetic energy. 16. 2.60 × 108 m/s E 2 − m2c 4 , c where E is the total energy. The total energy is E = E0 + KE . Since the kinetic energy is 17. (a) According to Equation 28.7, the magnitude of the momentum is p = equal to the rest energy, the total energy is E = 2 E0 . Substituting this result into the expression for p and using the fact that mc = E0 give p = 2 18. 2.18 × 108 m/s 2 2 4 E0 − E0 c = 3mc 2 . c Chapter 28 Problems 1477 CHAPTER 28 SPECIAL RELATIVITY PROBLEMS ______________________________________________________________________________ 1. SSM REASONING Since the "police car" is moving relative to the earth observer, the earth observer measures a greater time interval ∆t between flashes. Since both the proper time ∆t0 (as observed by the officer) and the dilated time ∆ t (as observed by the person on earth) are known, the speed of the "police car" relative to the observer can be determined from the time dilation relation, Equation 28.1. SOLUTION According to Equation 28.1, the dilated time interval between flashes is ∆t = ∆t0 / 1 − (v 2 / c 2 ) , where ∆t0 is the proper time. Solving for the speed v, we find 2 ∆t 1.5 s 8 v = c 1 − 0 = (3.0 ×108 m/s) 1 − = 2.4 ×10 m/s ∆t 2.5 s ______________________________________________________________________________ 2. 2 REASONING a. The two events in this problem are the creation of the pion and its subsequent decay (or breaking apart). Imagine a reference frame attached to the pion, so the pion is stationary relative to this reference frame. To a hypothetical person who is at rest with respect to this reference frame, these two events occur at the same place, namely, at the place where the pion is located. Thus, this hypothetical person measures the proper time interval ∆t0 for the decay of the pion. On the other hand, the person standing in the laboratory sees the two events occurring at different locations, since the pion is moving relative to that person. The laboratory person, therefore, measures a dilated time interval ∆t. The relation between these two time intervals is given by ∆t = ∆t0 / 1 − v 2 / c 2 (Equation 28.1). b. According to the hypothetical person who is at rest in the reference frame attached to the moving pion, the distance x that the laboratory travels before the pion breaks apart is equal to the speed v of the laboratory relative to the pion times the proper time interval ∆t0 , or x = v ∆t0 . The speed of the laboratory relative to the pion is the same as the speed of the pion relative to the laboratory, namely, 0.990c. SOLUTION a. The proper time interval is 1478 SPECIAL RELATIVITY ( 0.990 c ) = 4.9 × 10 –9 s v2 ∆t0 = ∆t 1 – 2 = ( 3.5 × 10 –8 s ) 1 – c c2 2 (28.1) b. The distance x that the laboratory travels before the pion breaks apart, as measured by the hypothetical person, is x = v ∆t0 = ( 0.990 ) (4 ×108 m/s ) ( 4.9 × 10−9 s ) = 1.5 m 3.00 1444 24444 3 0.990c ______________________________________________________________________________ 3. SSM REASONING The expression for time dilation is, according to Equation 28.1, ∆t = ∆t0 1 − v2 / c2 For a given event, it relates the proper time interval ∆t0 to the time interval ∆ t that would be measured by an observer moving at a speed v relative to the frame of reference in which the event takes place. We must consider two situations; in the first situation, the Klingon spacecraft has a speed of 0.75c with respect to the earth. In the second situation, the craft has a speed of 0.94c relative to th...
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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.

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