合肥工业大学大学物理试题答案

1. S: 2kv dt
dv
a -==
2kv dx
dv
v dt dx dx dv -==
k d x v dv
x
x v
v -=⎰⎰
)(ln
00
x x k v v
--= )(00x x k e v v --= (answer)
2. S: j t i t dt r
d v )3cos 15()3sin 15(+-== j
t i t dt
v d a )3sin 45()3cos 45(-+-==
()()j t i t j t i t v r
)3cos 15()3sin 15()3sin 5()3cos 5(+-⋅+=⋅
j j t t i i t t
⋅⋅+⋅⋅-=)3c o s 3s i n 75()3sin 3cos 75( 0= (proved c)
3. S: dt
dv v m k m f a =-==
dt m
k
v dv t t v v -=⎰⎰
0)(0
t m
k
v t v -=0)(ln t m k
e v t v -=0)( (answer) D: t m k e v dt
dx
v -==0
dt e v dx t m k t
t x -⎰⎰
=0
0)
(0
k
mv x e k
mv e
k
mv t x t m k t t m
k 0
max 0
0),1()(=
-=-=--
4. S: )()32(j y d i dx j i x r d f dw
+⋅+=⋅=
dy xdx dw w f
i
32+==⎰⎰
dy xdx 323
3
4
2
⎰
⎰
--+
=
= -6 J (answer)
5. S: 23230.60.4)0.30.4(t t t t t dt
d
dt d +-=+-==θω, t t t dt
d
dt d 60.6)30.60.4(2+-=+-==
ωα 0.40300.60.4)0(2=⨯+⨯-=ω (answer of a)
0.28)0.4(30.40.60.4)0.4(2=⨯+⨯-=ω rad/s (answer of a ) 60.266)0.2(=⨯+-=α rad/s 2 (answer of b )
t t 60.6)(+-=α is time varying not a constant (answer of c) 6. S: ω200
3
1
222ML L v m L mv +⋅= ML
mv ML L mv 4343020=
=
ω (answer a)
)c o s 1(2
)31(21m a x 22θω-=L
Mg ML ]1631[cos 2
2
21
max
gL
M v m -=-θ (answer b) 7. G: m =1.0g, M =0.50kg, L =0.60m, I rod =0.0602m kg ⋅,
s rod /5.4=ω
R:
I sys , v 0
S: I sys =I rod +(M+m)L
2
=0.060+(0.50+0.0010)×0.602= 0.24 2m kg ⋅(answer)
the system ’s angular momentum about rotating axis is conservative in the collision.
sys
I L mv ω=0
s m mL I v sys
/108.160
.00010.024
.05.430⨯=⨯⨯=
=
ω (answer )
D: The bullet momentum 0v m p
=(before impact), its angular momentum
about rotating axis can be expressed as L mv 0(a scalar) 8. S:
γ==
00.800
x x
t v c -∆=
=
08
1
180
0.600 3.0010
t t γ
∆=
∆=⨯⨯ 9. S: 2022
02)(mc E cp E E γγ==+=
222c p m c m c m c =
10. S: 0i n t =-=∆n e t n e t W Q E n e t n e t W Q = 1
(3010)(4.0 1.0)2
=-- J 30= (answer)
11. S: from nRT PV =and K T A 300= we can get:
K
T K T C B 100300== (answer of a)
Change of internal energy between A and B:
0)(2
3
int =-=∆A B T T k n E (answer of b)
The net work of the cycle:))(100300()13(2
1
21m N AC BC W ⋅-⋅-=⋅=
J 200= (answer of c) From the first law : W E Q +∆=int we can derive:
the net heat of the whole cycle is J W Q 200== (answer)
12. S: 13
1)(3
20
==
=
⎰
⎰
∞
F v Av dv Av dv v p F
33
F
v
A =
(answer of a ) F F v a v g v Av dv vAv v F
4
3
41420
==
=
⎰
13. G: T 1=T 2=T , m 1, p 1, v rms,1, m 2, p 2=2p 1, v avg,2 = 2v rms,1 R: m 1 / m 2 S: v avg,2 =1.602
m kT
v rms,1 = 1.73
1
m kT
v avg,2 = 2v rms,1
67.4)60
.173.12(2
21=⨯=m m (answer) 14. S: dE int =dQ – dW
d Q = dE int + dW = n C v dT+pdV V
dV
nR T dT nC dV T p T dT nC T dQ dS v v +=+==
i
f i f v V
V v T T V V nR T T nC V dV
nR T dT nC ds S f i f
i
ln
ln +=+=
=∆⎰⎰
⎰ 15. S: dA E q θεcos 0⎰=212100)0.60100(1085.8⨯-⨯⨯=- C 61054.3-⨯= 16. S: 2
041
)(r Q
r E πε=
(R < r <∞) dr r
Q dr r E udV dU 2
02
2
208421πεπε=⋅== R
Q r dr Q udV U R
02
20288πεπε=
==⎰
⎰∞
(answer) R
Q r dr Q U r r R
επεεπε02
202
*
88=
=⎰
∞
(answer ) 18. S: in the shell of r – r + dr
dr r R r dV r dq 204)/1()(πρρ-==
)34(31)/(4)(43
03
2
00
r R
r dr R r r dq r q r
-=-==⎰
⎰πρπρfrom the shell theorems , within the spherical symmetry distribution )34(12)(41)(200
2
0r Rr R
r r q r E -==ερπε (answer of b)
R r r R R
dr dE 3
2
0)64(12*00
=
⇒
=-=ερ 0
0200*max 9])32(3324[12)(ερερR
R R R R r E E =-⨯=
= 19. S: j y
V i x V V gradV y x E
∂∂-∂∂-=-∇=-=),( )0.20.2(y x x V
E x +-=∂∂-
= x y
V E y 0.2-=∂∂-= )/(480.2)0.20.2()0.2,0.2(m V j i j x i y x E
--=-+-=
20. S: Q in = - q , Q out = q (answer ) 1
010
2412
41)0(R q
q V q πεπε=
=
1
04)0(R q
V in πε-=
2
04)0(R q V o u t πε=
)0()0()0()0(out in q V V V V ++= )1
1(
42
10R R q +=
πε
21. S: from the planar symmetry and superposition principle, E
must in normal direction of the plates and 1σ,2σ,3σ,4σ must be const. From
charge conservation
A Q S =+)(21σσ ⇒ S
Q A
=+21σσ (1) B Q S =+)(43σσ ⇒ S
Q B
=
+43σσ (2) Apply Gauss ’ law in the closed surface shown in Fig. 032=+σσ (3)
within the metal, 0=p E
which leads to
00222243210
4
030201=-++⇒=-++σσσσεσεσεσεσ
From(1), (2), (3), (4) yield:
⎪⎪⎩
⎪⎪⎨
⎧
-=-=+==S Q Q S
Q Q B A
B A 223241σσσσ (answer of a) (6 points) 0
04030201122222ε
εσεσεσεσS Q Q E B
A p -=
--+= (1 point) 0
04030201222222εεσεσεσεσS Q Q E B
A p +=+++=
(1 point) (answer of b) d S Q Q d E d E V B
A p A
B 0
12ε-=
=⋅= (2 points) (answer of c)。