Question

Èçͼ¼×Ëùʾ£¬Á½Æ½ÐнðÊô°å¼ä¾àΪ2l£¬¼«°å³¤¶ÈΪ4l£¬Á½¼«°å¼ä¼ÓÉÏÈçͼÒÒËùʾµÄ½»±äµçѹ£¨t=0ʱÉϼ«°å´øÕýµç£©£®ÒÔ¼«°å¼äµÄÖÐÐÄÏßOO₁ΪxÖὨÁ¢×ø±êϵ£¬ÏÖÔÚƽÐаå×ó²àÈë¿ÚÕýÖв¿Óпí¶ÈΪlµÄµç×ÓÊøÒÔƽÐÐÓÚxÖáµÄ³õËÙ¶Èv₀´Ót=0ʱ²»Í£µØÉäÈëÁ½°å¼ä£®ÒÑÖªµç×Ó¶¼ÄÜ´ÓÓÒ²àÁ½°å¼äÉä³ö£¬Éä³ö·½Ïò¶¼ÓëxÖáƽÐУ¬ÇÒÓеç×ÓÉä³öµÄÇøÓò¿í¶ÈΪ2l£®µç×ÓÖÊÁ¿Îªm£¬µçºÉÁ¿Îªe£¬ºöÂÔµç×ÓÖ®¼äµÄÏ໥×÷ÓÃÁ¦£®
£¨1£©Çó½»±äµçѹµÄÖÜÆÚTºÍµçѹU₀µÄ´óС£»
£¨2£©Ôڵ糡ÇøÓòÍâ¼Ó´¹Ö±Ö½ÃæµÄÓнçÔÈÇ¿´Å³¡£¬¿ÉʹËùÓеç×Ó¾­¹ýÓнçÔÈÇ¿´Å³¡¾ùÄÜ»á¾ÛÓÚ£¨6l£¬0£©µã£¬ÇóËù¼Ó´Å³¡´Å¸ÐӦǿ¶ÈBµÄ×î´óÖµºÍ×îСֵ£»
£¨3£©Çó´ÓOµãÉäÈëµÄµç×Ó¸Õ³ö¼«°åʱµÄ²àÏòλÒÆ£®

Answer 1

£¨1£©µç×ÓÔڵ糡ÖÐˮƽ·½Ïò×öÔÈËÙÖ±ÏßÔ˶¯£¬
Ôò£º4l=v₀nT£¬½âµÃ£ºT=

4l

nv0

£¨n=1£¬2£¬3¡­£©£¬
µç×ÓÔڵ糡ÖÐÔ˶¯×î´ó²àÏòλÒÆ£º

l

2

=2n?

1

2

a(

T

2

)2£¬ÓÉÅ£¶ÙµÚ¶þ¶¨Âɵãºa=

eU0

2lm

£¬
½âµÃ£ºU0=

nm v 20

4e

£¨n=1£¬2£¬3¡­£©£»
£¨2£©Á£×ÓÔ˶¯¹ì¼£ÈçͼËùʾ£º

ÓÉͼʾ¿ÉÖª£¬×î´óÇøÓòÔ²°ë¾¶Âú×㣺rm2=(2l)2+(rm-l)2£¬½âµÃ£ºr_m=2.5l£¬
¶ÔÓÚ´øµçÁ£×Óµ±¹ì¼£°ë¾¶µÈÓڴų¡ÇøÓò°ë¾¶Ê±£¬´øµçÁ£×Ó½«»ã¾ÛÓÚÒ»µã£¬
ÓÉÅ£¶ÙµÚ¶þ¶¨Âɵãºqv0Bmin=

m v 20

rm

£¬½âµÃ£ºBmin=

2mv0

5el

£¬
×îСÇøÓòÔ²°ë¾¶Îªr_n=0.5l£¬
ÓÉÅ£¶ÙµÚ¶þ¶¨Âɵãºqv0Bmax=

m v 20

rn

£¬½âµÃ£ºBmax=

2mv0

el

£»
£¨3£©Éèʱ¼äΪ¦Ó£¬

T

2

£¾¦Ó£¾0£¬Èôt=kT+¦ÓÇÒ(

T

2

£¾¦Ó£¾0)ʱµç×Ó½øÈëµç³¡£¬
Ôò£º

y1=n[ 1 2 a( T 2 -¦Ó)2?2- 1 2 a¦Ó2?2]=n[ 1 4 aT2-aT¦Ó]= (4k+1)l 2 - nv0t 2

£¬ÆäÖУ¨n=1£¬2£¬3¡­£¬k=0£¬1£¬2£¬3¡­£©£¬
Èôt=(k+

1

2

)T+¦ÓÇÒ(

T

2

£¾¦Ó£¾0)½øÈëµç³¡
Ôò£ºy2=-n[

1

4

aT2-aT¦Ó]=

nv0

2

(t-kT-

1

2

T)-

l

2

=

nv0t

2

-

(4k+3)

2

l£¬ÆäÖУ¨n=1£¬2£¬3¡­£¬k=0£¬1£¬2£¬3¡­£©£»
»ò£ºÈôµç×ÓÔÚt=kT+¦ÓÇÒ(T£¾¦Ó£¾

T

2

)½øÈëµç³¡Ê±£¬³öµç³¡µÄ×ܲàÒÆΪ£º

y2=n[- 1 2 a(T-¦Ó)2?2+ 1 2 a(¦Ó- T 2 )2] =n[- 3 4 aT2+aT¦Ó]=- (4k+3) 2 l+ nv0t 2

£¬ÆäÖУ¨n=1£¬2£¬3¡­£¬k=0£¬1£¬2£¬3¡­£©£»
ÆäËû½â·¨£ºÈôkT£¼t£¼kT+

T

2

£¬Ôò
µç×ÓÑØ+y·½ÏòµÚÒ»´Î¼ÓËÙµÄʱ¼äΪ

T

2

-(t-kT)
µç×ÓÑØ-y·½ÏòµÚÒ»´Î¼ÓËÙµÄʱ¼äΪt-kTy={

1

2

a[

T

2

-(t-kT)]2-

1

2

a(t-kT)2}?2n
½âµÃ£ºy=

4k+1

4

naT2-naTt£¬ÆäÖÐaT2=

2l

n

£¬aT=

1

2

v0£¬
¡ày=

4k+1

2

l-

1

2

nv0t£¨n=1£¬2£¬3¡­£¬k=0£¬1£¬2£¬3¡­£©
ÈôkT+

T

2

£¼t£¼kT+T£¬Ôò
µç×ÓÑØ-y·½ÏòµÚÒ»´Î¼ÓËÙµÄʱ¼äΪT-£¨t-kT£©
µç×ÓÑØ+y·½ÏòµÚÒ»´Î¼ÓËÙµÄʱ¼äΪt-kT-

T

2

y={-

1

2

a[T-(t-kT)]2+

1

2

a(t-kT-

T

2

)2}?2n
½âµÃ£ºy=

4k+1

4

naT2-naTt£¬ÆäÖÐaT2=

2l

n

£¬aT=

1

2

v0£¬¡ày=

1

2

nv0t-

4k+3

2

l£¨n=1£¬2£¬3¡­£¬k=0£¬1£¬2£¬3¡­£©£»
´ð£º£¨1£©½»±äµçѹµÄÖÜÆÚT=

4l

nv0

£¨n=1£¬2£¬3¡­£©£¬µçѹU0=

nm v 20

4e

£¨n=1£¬2£¬3¡­£©£»
£¨2£©Ëù¼Ó´Å³¡´Å¸ÐӦǿ¶ÈBµÄ×î´óÖµBmax=

2mv0

el

£»×îСֵBmin=

2mv0

5el

£»
£¨3£©´ÓOµãÉäÈëµÄµç×Ó¸Õ³ö¼«°åʱµÄ²àÏòλÒÆΪ

(4k+1)l

2

-

nv0t

2

ÆäÖУ¨n=1£¬2£¬3¡­£¬k=0£¬1£¬2£¬3¡­£©£¬»ò

nv0t

2

-

(4k+3)

2

l£¬ÆäÖУ¨n=1£¬2£¬3¡­£¬k=0£¬1£¬2£¬3¡­£©£®