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