???????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)
5
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