Question

7£®Èçͼ¼×Ëùʾ£¬ÔÚxOy×ø±êƽÃæµÄµÚÒ»ÏóÏÞ£¨°üÀ¨x¡¢yÖᣩÄÚ´æÔڴŸÐӦǿ¶È´óСΪB₀¡¢·½Ïò´¹Ö±ÓÚxOyƽÃæÇÒËæʱ¼ä×öÖÜÆÚÐԱ仯µÄÔÈÇ¿´Å³¡£¬ÈçͼÒÒËùʾ£¬¹æ¶¨´¹Ö±xOyƽÃæÏòÀïµÄ´Å³¡·½ÏòΪÕý£®ÔÚyÖá×ó²àÓÐÒ»¶ÔÊúÖ±·ÅÖõÄƽÐнðÊô°åM¡¢N£¬Á½°å¼äµÄµçÊƲîΪU₀£®Ò»ÖÊÁ¿Îªm¡¢µçÁ¿ÎªqµÄ´øÕýµçÁ£×Ó£¨ÖØÁ¦ºÍ¿ÕÆø×èÁ¦¾ùºöÂÔ²»¼Æ£©£¬´ÓÌù½üM°åµÄÖеãÎÞ³õËÙÊÍ·Å£¬Í¨¹ýN°åС¿×ºó´Ó×ø±êÔ­µãOÒÔijһËÙ¶ÈÑØxÖáÕý·½Ïò´¹Ö±ÉäÈë´Å³¡ÖУ¬¾­¹ýÒ»¸ö´Å³¡±ä»¯ÖÜÆÚT₀£¨T₀δ֪£©ºóµ½´ïµÚÒ»ÏóÏÞÄÚµÄijµãP£¬´ËʱÁ£×ÓµÄËٶȷ½ÏòÇ¡ºÃÑØxÖáÕý·½Ïò£®

£¨1£©ÇóÁ£×Ó½øÈë´Å³¡×÷ÔÈËÙÔ²ÖÜÔ˶¯Ê±µÄÔ˶¯°ë¾¶£»
£¨2£©ÈôÁ£×ÓÔÚt=0ʱ¿Ì´ÓOµãÉäÈë´Å³¡ÖУ¬ÇóÁ£×ÓÔÚPµã×Ý×ø±êµÄ×î´óÖµy_m¼°ÏàÓ¦µÄ´Å³¡±ä»¯ÖÜÆÚT₀µÄÖµ£»
£¨3£©ÈôÔÚÉÏÊö£¨2£©ÖУ¬µÚÒ»ÏóÏÞÄÚy=y_m´¦Æ½ÐÐxÖá·ÅÖÃÓÐÒ»ÆÁÄ»£¬Èçͼ¼×£¬´Å³¡±ä»¯ÖÜÆÚΪÉÏÊö£¨2£©ÖÐT₀£¬µ«M¡¢NÁ½°å¼äµÄµçÊƲîU¿ÉÒÔÔÚU₀£¼U£¼9U£¬·¶Î§Äڱ仯£¬Á£×ÓÈÔÔÚt=0ʱ¿Ì´ÓOµãÉäÈë´Å³¡ÖУ¬ÇóÁ£×Ó¿ÉÄÜ»÷ÖеÄÆÁÄ»·¶Î§£®

Answer 1

·ÖÎö £¨1£©Óɶ¯Äܶ¨ÀíÇó³öÁ£×Ó½øÈë´Å³¡Ê±µÄËٶȣ¬È»ºóÓÖÂåÂ××ÈÁ¦ÌṩÏòÐÄÁ¦Çó³öÁ£×ÓÔ˶¯µÄ°ë¾¶£»
£¨2£©¸ù¾Ý¼¸ºÎ֪ʶÇó³öPµãºá×ø±êºÍ×Ý×ø±êÓëÁ£×ÓÔ²ÖÜÔ˶¯°ë¾¶µÄ¹Øϵ£®¸ù¾ÝÁ£×ÓÔÚµÚÒ»ÏóÏÞÔ˶¯µÄÌõ¼þÇó½âPµãµÄ×Ý×ø±êµÄ×î´óÖµ¼°ÏàÓ¦µÄ´Å³¡±ä»¯ÖÜÆÚT₀µÄÖµ£®
£¨3£©½áºÏÇ°Á½ÎʵĹ«Ê½£¬Ð´³öÁ£×ӵİ뾶µÄ·¶Î§£¬È»ºó»­³öÁ£×ÓÔ˶¯µÄ¹ì¼£Í¼£¬×îºó½áºÏ¼¸ºÎ¹ØϵµÃ³öÁ£×Ó¿ÉÄܵķ¶Î§£®

½â´ð ½â£º£¨1£©ÉèÁ£×Ó±»µç³¡¼ÓËÙ»ñµÃµÄËÙ¶ÈÊÇv₀£¬Óɶ¯Äܶ¨ÀíµÃ£º$q{U}_{0}=\frac{1}{2}m{v}_{0}^{2}$
½âµÃ£º${v}_{0}=\sqrt{\frac{2q{U}_{0}}{m}}$
Á£×Ó½øÈë´Å³¡ºó×öÔ²ÖÜÔ˶¯µÄ¹ìµÀ°ë¾¶Îªr£¬¸ù¾ÝÅ£¶ÙµÚ¶þ¶¨ÂɵÃ$q{v}_{0}{B}_{0}=\frac{m{v}_{0}^{2}}{r}$
ËùÒÔ£º$r=\frac{m{v}_{0}}{q{B}_{0}}=\frac{1}{{B}_{0}}•\sqrt{\frac{2m{U}_{0}}{q}}$
£¨2£©ÉèÁ½¶ÎÔ²»¡µÄÔ²ÐÄO₁O₂µÄÁ¬ÏßÓëyÖá¼Ð½ÇΪ¦È£¬PµãµÄ×Ý×ø±êΪy£¬Ô²ÐÄO₂µ½yÖáÖ®¼äµÄ¾àÀëΪx£¬ÔòÓɼ¸ºÎ¹Øϵ£¬µÃ
  $\overline{O{O}_{1}}=\overline{P{O}_{2}}=r$£¬$\overline{{O}_{1}{O}_{2}}=2r$
   y=2r+2rcos¦È
   sin¦È=$\frac{x}{2r}$
±£Ö¤Á£×ÓÔÚµÚÒ»ÏóÏÞÄÚÔ˶¯£¬x¡Ýr
Óɼ¸ºÎ¹Øϵ¿ÉÖªµ±¦È=30¡ãʱ£¬yÈ¡×î´ó£¬${y}_{m}=2r+2rcos30¡ã=£¨2+\sqrt{3}£©r=£¨2+\sqrt{3}£©•$$\frac{1}{{B}_{0}}•\sqrt{\frac{2m{U}_{0}}{q}}$
ÉèÁ£×ÓÔڴų¡ÖÐÔ˶¯µÄÖÜÆÚΪT£¬Ôò$T=\frac{2¦Ðr}{{v}_{0}}=\frac{2¦Ðm}{q{B}_{0}}$Óɼ¸ºÎ¹Øϵ֪Á£×ÓÔڴų¡ÖÐת¹ý£¨180¡ã-30¡ã£©=150¡ãʱ´Å³¡¿ªÊ¼¸Ä±ä·½Ïò£¬¼´Ôڴų¡±ä»¯µÄ°ë¸öÖÜÆÚÄÚÁ£×Óת¹ý150¡ã£¬Ôò£º$\frac{{T}_{0}}{2}=\frac{150¡ã}{360¡ã}•T=\frac{5}{12}T$
½âµÃ£º${T}_{0}=\frac{5}{6}T=\frac{5¦Ðm}{3q{B}_{0}}$

£¨3£©ÓÉU₀£¼U£¼9U£¬ºÍ${v}_{0}=\sqrt{\frac{2q{U}_{0}}{m}}$µÃ£ºv₀£¼v£¼3v₀£®
Ôڴų¡ÖÐÔ˶¯µÄ°ë¾¶£ºr£¼R£¼3r£¬Ô˶¯¹ì¼£Èçͼ£¬ÔòÓɼ¸ºÎ¹Øϵ¿ÉµÃ£¬´òÔÚÆÁÄ»Éϵķ¶Î§µÄ×î×ó¶ËÈçͼÖТÚËùʾ£¬¸ÃµãµÄºá×ø±ê£ºx₁=0£¬
´òÔÚ×îÓҶ˵Ĺ켣ÈçͼÖТÛËùʾ£¬Óɼ¸ºÎ¹ØϵµÃ£º$£¨{y}_{m}-{R}_{m}£©^{2}+{x}_{2}^{2}={R}_{m}^{2}$
ÁªÁ¢½âµÃ£º${x}_{2}=\sqrt{5+2\sqrt{3}}r$=$\sqrt{\frac{2£¨5+3\sqrt{3}£©m{U}_{0}}{q{B}_{0}^{2}}}$
Á£×Ó¿ÉÄÜ»÷ÖеÄÆÁÄ»·¶Î§ÊÇ£º$0¡Üx¡Ü\sqrt{\frac{2£¨5+3\sqrt{3}£©m{U}_{0}}{q{B}_{0}^{2}}}$
´ð£º£¨1£©Á£×Ó½øÈë´Å³¡×÷ÔÈËÙÔ²ÖÜÔ˶¯Ê±µÄÔ˶¯°ë¾¶ÊÇ$\frac{1}{{B}_{0}}•\sqrt{\frac{2m{U}_{0}}{q}}$£»
£¨2£©ÈôÁ£×ÓÔÚt=0ʱ¿Ì´ÓOµãÉäÈë´Å³¡ÖУ¬Á£×ÓÔÚPµã×Ý×ø±êµÄ×î´óÖµÊÇ$\frac{2+\sqrt{3}}{{B}_{0}}•\sqrt{\frac{2m{U}_{0}}{q}}$£¬ÏàÓ¦µÄ´Å³¡±ä»¯ÖÜÆÚT₀µÄÖµÊÇ$\frac{5¦Ðm}{3q{B}_{0}}$£»
£¨3£©Á£×Ó¿ÉÄÜ»÷ÖеÄÆÁÄ»·¶Î§ÊÇ$0¡Üx¡Ü\sqrt{\frac{2£¨5+3\sqrt{3}£©m{U}_{0}}{q{B}_{0}^{2}}}$£®

µãÆÀ ±¾ÌâÊÇ´øµçÁ£×ÓÔÚ½»±ä´Å³¡ÖÐÔ˶¯µÄÎÊÌ⣬»­³öÁ£×ÓÔ˶¯µÄ¹ì¼££¬¸ù¾Ý¼¸ºÎ֪ʶÇó½â°ë¾¶Êǹؼü£®