中南大学大学物理双语版答案Problem 1-22

???????Problem 1. Answers: 1. v?2i?16j; a?8j; 7.13?.(a?v?avcos?) 2. t?(3vf/k)1/3 3. a-e, b-d, c-f. 4. [d]: x2?y2?L2, xdxdy?y?0 dtdtdxdyx?v, ?vB, xv?yvB?0, vB?v?vcot? dtdty??t3? 5. (a)r?i?(10t?2t2)j, (Answer)

3???????r(3)?r(0)??3i?4j, (Answer) (b) r?9i?12j, vavg?3????v(3)?v(0)?aavg??3i?4j (Answer)

3?vy2??(c) v?9i?2j tan????, ???12.?5 (Answer)

vx96. Solution: From the definition of acceleration for a straight line motion

dv a?,

dtand the given condition a??kv, we have dv. dtApply chain rule to dv/dt, the equation can be rewritten as

dvdxdv?v ?kv?dxdtdxSeparating the variables gives

?kv? xvdv??kdTake definite integration for both sides of the equation with initial conditions,

we have

?0v0, xvdv???kd0x3/2or x?2v03k (Answer)

Problem 2. Answers:

1. 13.0 m/s2, 5.7m/s, 7.5 m/s2, 2.

gR, (2?1)R.

Solution: At initial moment

when the ball is just kicked out:

ar?v2i??g, ??v. g2i

In order the ball not to hit the rock,

vi2???R, vi?gR; (Answer) gFor the vertical motion:

12gt?R, t?2R/g, 2vit?R?(2?1)R (Answer)

3. d. 4. d.

5. Solution: at any moment the speed of the projectile is

2v?v0?(gt)2 dvg2t?Tangential acceleration:at?,

222dtv0?gtRadial acceleration:ar?g2?at2?gv0v?gt2022,

From ar?v2?20,

we have the radius of curvature is given by

??(v?gt)v?argv02223/2 (Answer)

6. ??86.7?

?????Solution: vsd?vsg?vgd?vsg?vdg

?? tanvdgvsg35?10/3600??17. 360.8? ??86.?7?86? 42

v2Problem 3. Answers: 1. Solution: T1cos??T2cos??m

R T1sin??T2sin??mg

mv2mg??108N, (a)T1?2Rcos?2sin? (b) T2?T1?mg?56N sin? 2. (a) Mg, (b) 3. d. 4. e

Mgr/m

Fig. 3-1

5. Solution: from the given condition v?vie?ct and with vi?10.0m/s, at 20 s,

v?5.0m/s, we have (a)0.0347s?1,

With 40.0 s, we can get (b) 2.50 m/s,

Solution: (c) From v?vie?ct, we have

dv??cvie?ct dtTherefore a??cv, (Answer)

which means the acceleration of the boat is proportional to the speed at any time.

6.

Solution: From Newton’s second law, we have

2?m a ?kmv By the definition of acceleration, this equation can be rewritten as

dv?kmv2?m

dt Separating the variables obtains

dv ?2?kdt

v Take the definition integral with the initial conditions, we hve

vtdv ??2??kdt

v00v

11??kt vv0Then v?v0/(1?ktv0). (Answer)

Problem 4. Answers: 1.(a) 21J, (b) cos??17, ? =19.4? 513???23 2. F?(7?9xy)i?3xj. 3. c. 4. d

??0AB 5. (a) U??F?dr??(?Ax?Bx2)dx=x2?x3,

xx235A19B? (b) ?U? 2319B5A?(c) ?U??K?0, ?K???U? 32

6. Solution: (a) In the process of the pendulum swinging down, only the

gravity does work , mechanic energy is conserved:

1mgL(1?cos?)?mv2,

2When the sphere is released from a certain height, in order to make the ball will return to this height after the string strikes the peg:

12mv?mg(L?d), 20mgL(1?cos?)?mg(L?d)

Lcos??d. (Answer)

Solution: (b) If the pendulum is to swing in a complete circle centered on the peg, at the top of the path, the tension T on the cord must not be zero.

v2T?mg?m,

L?dBecause T?0 So v2?g(L?d)

From the conservation of mechanical energy, we have

12?mv?2m(g?L) mgL d2Inserting v2?g(L?d)into above equation, obtains

mgL?1mg(L?d)?2mg(L?d) 23Ld? (Answer)

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