Question

9£®ÈçͼËùʾ£¬³¤L=1m¡¢¹â»¬µÄ¸ËÇãб·ÅÖã¬ÓëˮƽÃæµÄ¼Ð½ÇΪ45¡ã£¬Ò»´øÕýµç¡¢ÖÊÁ¿Îªm=0.1kg¡¢µçºÉÁ¿Îªq=10^-4CµÄСÇòÌ×ÔÚ¸ËÉÏ£®ÒԸ˵Ä϶˵ãΪ×ø±êÔ­µã£¬ÔÚ¸ËËùÔÚµÄÊúֱƽÃæÄÚ½¨Á¢Æ½ÃæÖ±½Ç×ø±êϵxOy£¬xÖáˮƽ£¬Õû¸ö¿Õ¼ä´æÔÚƽÐÐÓÚxOyƽÃæµÄÔÈÇ¿µç³¡£®ÈÃСÇòÒÔijһ³õËÙ¶Èv´Ó¸ËµÄÉ϶˿ªÊ¼ÑظËÏ»¬£¬À뿪¸ËʱÆäËٶȴóСΪ2$\sqrt{2}$m/s£¬Í¨¹ýA£¨0.2£¬-0.6£©µãʱÆäËٶȴóСΪv_A=4m/s£¬Í¨¹ýB£¨-0.6£¬-3£©µãʱÆäËٶȴóСΪv_B=4$\sqrt{5}$m/s£¬È¡g=10m/s²£¬Çó£º
£¨1£©O¡¢AÁ½µãµÄµçÊƲîU_OA£¬O¡¢BÁ½µãµÄµçÊƲîU_OB£»
£¨2£©µç³¡Ç¿¶ÈµÄ´óСºÍ·½Ïò£»
£¨3£©Ð¡ÇòµÄ³õËÙ¶Èv£»
£¨4£©Ð¡ÇòÀ뿪Oµãºóͨ¹ýyÖáʱµÄ¶¯ÄÜ£®

Answer 1

·ÖÎö £¨1£©´ÓOµ½A£¬ºÍ´ÓOµ½BµÄ¹ý³ÌÖУ¬¸ù¾Ý¶¯Äܶ¨ÀíÁÐʽÇó½â£»
£¨2£©µç³¡Ïß·½ÏòÓëµÈÊÆÃæ´¹Ö±£¬¾Ý´Ë·ÖÎöµç³¡Ç¿¶È·½Ïò£¬¸ù¾ÝU=EdÇó½âµç³¡Ç¿¶È£»
£¨3£©¶ÔСÇòÔÚ¸ËÉÏÔ˶¯¹ý³ÌÊÜÁ¦·ÖÎö£¬µÃ³öСÇòµÄÔ˶¯Çé¿ö¼´¿ÉÇó½â£»
£¨4£©À뿪Oµãºó£¬Ð¡Çò×öÀàƽÅ×Ô˶¯£¬¸ù¾ÝƽÅ×Ô˶¯»ù±¾¹«Ê½½áºÏ¶¯Äܶ¨ÀíÇó½â£®

½â´ð ½â£º£¨1£©´ÓOµ½A£¬¸ù¾Ý¶¯Äܶ¨ÀíµÃ£º$mg{h}_{OA}+q{U}_{OA}=\frac{1}{2}m{{v}_{A}}^{2}-\frac{1}{2}m{{v}_{0}}^{2}$£¬
´úÈëÊý¾Ý½âµÃ£ºU_OA=-2000V£¬
´ÓOµ½B£¬¸ù¾Ý¶¯Äܶ¨ÀíµÃ£º$mg{h}_{OB}+q{U}_{OB}=\frac{1}{2}m{{v}_{B}}^{2}-\frac{1}{2}m{{v}_{0}}^{2}$£¬
´úÈëÊý¾Ý½âµÃ£ºU_OB=6000V£¬
£¨2£©Á¬½ÓAB£¬ÓëyÖá½»ÓÚCµã£¬Óɼ¸ºÎ¹Øϵ¿ÉÖª£¬OCΪµÈÊÆÃ棬Òò´Ëµç³¡·½ÏòÑØxÖḺ·½Ïò£¬Éèµç³¡Ç¿¶ÈΪE£¬Ôò$E=\frac{{U}_{OB}}{|{x}_{B}|}$
´úÈëÊý¾Ý½âµÃ£ºE=1¡Á10⁴V/m
£¨3£©¶ÔСÇòÔÚ¸ËÉÏÔ˶¯¹ý³ÌÊÜÁ¦·ÖÎö£¬ÈçͼËùʾ£¬¸ù¾ÝÌâÒ⣬G=F_µç£¬
¿É֪СÇòÔÚ¸Ë×ÓÔ˶¯Ê±£¬ºÏÁ¦ÎªÁ㣬СÇò×öÔÈËÙÔ˶¯£¬Òò´Ë£¬v=${v}_{0}=2\sqrt{2}m/s$£¬
£¨4£©À뿪Oµãºó£¬Ð¡Çò×öÀàƽÅ×Ô˶¯£¬$\overline{OD}cos45¡ã={v}_{0}t$£¬$\overline{OD}sin45¡ã=\frac{1}{2}a{t}^{2}$£¬a=$\sqrt{2}g$£¬
´úÈëÊý¾Ý½âµÃ£º$\overline{OD}=1.6m$
´ÓOµ½D£¬¸ù¾Ý¶¯Äܶ¨ÀíµÃ£º$mg{h}_{OD}={E}_{KD}-\frac{1}{2}m{{v}_{0}}^{2}$
´úÈëÊý¾Ý½âµÃ£ºE_KD=2.0J
´ð£º£¨1£©O¡¢AÁ½µãµÄµçÊƲîΪ-2000V£¬O¡¢BÁ½µãµÄµçÊƲîΪ6000V£»
£¨2£©µç³¡Ç¿¶ÈµÄ´óСΪ1¡Á10⁴V/m£¬·½ÏòÑØxÖḺ·½Ïò£»
£¨3£©Ð¡ÇòµÄ³õËÙ¶ÈvΪ$2\sqrt{2}m/s$£»
£¨4£©Ð¡ÇòÀ뿪Oµãºóͨ¹ýyÖáʱµÄ¶¯ÄÜΪ2.0J£®

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éÁ˶¯Äܶ¨Àí¡¢Æ½Å×Ô˶¯»ù±¾¹«Ê½ÒÔ¼°U=EdµÄÖ±½ÓÓ¦Óã¬ÒªÇóͬѧÃÇÄÜÕýÈ··ÖÎöСÇòµÄÊÜÁ¦Çé¿öºÍÔ˶¯Çé¿ö£¬ÄѶÈÊÊÖУ®