中南大学大学物理双语版答案Problem 1-22

From Eq.(3), we have v0?v?

Eq.(4) + Eq.(5) 2v0??l? ??1ML2? (5) 3ml1ML2? 3ml6mlv0?0 (Answer)

ML2?3ml2Inserting ? into Eq.(4), obtains

ML2?3ml2v0 (Answer) v??ML2?3ml2 Solution: (b)

121I??Mgh, I?ML2 23mlvI?26L2h???(202)2

2MggML?3ml

6. Solution: (a) angular momentum is

conserved

1 mvid?(mR2?MR2)?

22mvid???0 (Answer) 22 Fig. 7-6 MR?2mR

(b) Mechanical energy is not conserved

because the clay and the solid cylinder undergo an inelastic collision. In this process, some kinetic energy must be lost.:

m2vi2d212 Ef?I?? 22(M?2m)R12Ef2I?d21??2?1， Ef?E0 E01mv2R1?M/2mi2

Problem 9. Answers: 1. (a) 3.07 MeV（Or: 6mc2, 4.92?10?13 J）, (b) 0.986c. Solution: (a)(??1)mc2?5mc2, ??6,

For an electron, mc2?0.511MeV Total energy: ?mc2?3.07MeV

Solution: (b)

11?v/c22?6, v?0.986c

2. 2mc2 ? 2?0.511=1.02 MeV. (1.64?10?13 J ) 3. c.

Solution:

u?v?0.75c?0.75cu???

1?u?v/c21?0.752c2/c2??0.96c

4. d

Solution: K?(11?v/c22)mc2,

v?c,K??.

5. 1.63?103 MeV/c.

Solution: ?mc2?2mc2, ??2, v?3c 2 p??mv?3mc?3mc2/c,

1.67?10?27?9?1016?939MeV For a proton, mc?1.6?10?192 p?3mc2/c?1.63?103MeV/c 6. (a) 5.37?10?11 J = 335 MeV (5.36?10?11 J) (b) 1.33?10?9 J = 8.31 GeV (1.33?10?9 J) Solution: W??K

?K?(?2?1)mc2?(?1?1)mc2?(?2??1)mc2 (a) W?(?2??1)mc2?939(11?0.7511?0.99522?11?0.5?12)?335MeV;

(b) W?(?2??1)mc2?939(1?0.52)?8.31GeV

7. Solution: E2?p2c2?(mc2)2, ?E??mc2 ??2m2c4?p2c2, ?m2c?

?1222424?(pc?mc)/mc, 221?v/c1m2c4?p2c2?m2c4, 221?v/c

v2m2c4 1?2?22,

cpc?m2c4v2m2c4p2 2?1?22 ?22422cpc?mcp?mcv?

c1?(mc/p)2

Problem 10. Answers: 1. 1.2?1022 for 37?C. Solution: pV??RT,

pV(761/760)1.01?105?500?10?6????0.02mol

RT8.31?(273?37) N??NA?0.0? 21026.?0223?10?1.22

2. 385 K, 7.97?10-21J, molar mass of the gas.

pV1.6?106?7.0?10?3??385K Solution: pV??RT, T??R3.5?8.31 K? v?33kT??1.38?10?23?385?7.97?10?21J 228RT, M is the molar mass of the gas. ?M 3. b. ??1kT ?222?dn2?dp 4. a. vp?2RT M12v0a?v0a?1, a?; 23v0 5. (a) 2/3v0, (b) N/3, (c) 1.22v0, (d) 1.31v0. Solution: (a)

2.0v0??0f(v)dv?1,

(b) N1?? (c) v??21.5v0Nf(v)dv?N(2v0?1.5v0)a?Nv0v02N?? 23v032v00vf(v)dv??2v0202v0av11vdv??vadv?v0?1.2v0

v0v09 (d) v??0vf(v)dv??vf(v)dv??02v0avdv??v2adv

v0v0v022v0v0v2f(v)dv

??v20v0433v0?v0)aa(8v0 ? ?4v031142 ?(?)v0

69 vrms?1.31v0 (Answer)

6. Solution:

p10525??1.8?10 (a) p?nkT,n?. ?23kT1.38?10?400(b) vp?2kT2kT2?1.38?10?23?400?26,m?2??5.31?10kg, 2mvp456