Question

2£®ÈçͼaËùʾ£¬ÔÈÇ¿´Å³¡´¹Ö±ÓÚxOyƽÃ棬´Å¸ÐӦǿ¶ÈB₁°´Í¼bËùʾ¹æÂɱ仯£¨´¹Ö±ÓÚÖ½ÃæÏòÍâΪÕý£©£®t=0ʱ£¬Ò»±ÈºÉΪ$\frac{q}{m}$=1¡Á10⁵C/kgµÄ´øÕýµçÁ£×Ó´ÓÔ­µãÑØyÖáÕý·½ÏòÉäÈ룬ËٶȴóСv=5¡Á10⁴m/s£¬²»¼ÆÁ£×ÓÖØÁ¦£®
£¨1£©Çó´øµçÁ£×ÓÔÚÔÈÇ¿´Å³¡ÖÐÔ˶¯µÄ¹ìµÀ°ë¾¶£®
£¨2£©Çót=$\frac{¦Ð}{2}$¡Á10^-4sʱ´øµçÁ£×ÓµÄ×ø±ê£®
£¨3£©±£³ÖbÖдų¡²»±ä£¬ÔÙ¼ÓÒ»´¹Ö±ÓÚxOyƽÃæÏòÍâµÄºã¶¨ÔÈÇ¿´Å³¡B₂£¬Æä´Å¸ÐӦǿ¶ÈΪ0.3T£¬ÔÚt=0ʱ£¬Á£×ÓÈÔÒÔÔ­À´µÄËٶȴÓÔ­µãÉäÈ룬ÇóÁ£×ӻص½×ø±êÔ­µãµÄʱ¿Ì£®

Answer 1

·ÖÎö £¨1£©¸ù¾ÝÂåÂ××ÈÁ¦ÌṩÏòÐÄÁ¦Áз½³Ì¼´¿ÉÇó³ö´øµçÁ£×ÓÔÚÔÈÇ¿´Å³¡ÖÐÔ˶¯µÄ¹ìµÀ°ë¾¶£»
£¨2£©ÏÈÇó³ö´øµçÁ£×ÓÔڴų¡ÖÐÔ˶¯µÄÖÜÆÚ£¬ÔÙ·Ö±ðÇó³öÔÚ0¡«$\frac{¦Ð}{4}$¡Á10^-4sºÍ$\frac{¦Ð}{4}$¡Á10^-4s¡«$\frac{¦Ð}{2}$¡Á10^-4s¹ý³ÌÖУ¬Á£×ÓÔ˶¯Á˵ÄÖÜÆÚºÍÔ²»¡¶ÔÓ¦µÄÔ²ÐĽǣ¬»­³öÁ£×ÓµÄÔ˶¯¹ì¼£Í¼£¬ÀûÓü¸ºÎ¹ØϵÇó³ö´øµçÁ£×ÓµÄ×ø±ê£»
£¨3£©»­³öÊ©¼ÓB₂=0.3TµÄÔÈÇ¿´Å³¡ÓëÔ­´Å³¡µþ¼Óºó¹æÂɱ仯ͼ£¬·Ö±ðÇó³öµ±nT¡Üt¡ÜnT+$\frac{T}{2}$ºÍnT+$\frac{T}{2}$¡Üt¡Ü£¨n+1£©T£¨n=0£¬1£¬2£¬¡­£©Ê±Á£×ÓÔ˶¯Á˵ÄÖÜÆÚ£¬»­³öÁ£×ÓÔ˶¯¹ì¼£Í¼¼´¿ÉÇó³öÁ£×ӻص½×ø±êÔ­µãµÄʱ¿Ì£®

½â´ð ½â£º£¨1£©´øµçÁ£×ÓÔÚÔÈÇ¿´Å³¡ÖÐÔ˶¯£¬ÂåÂØ×ÈÁ¦ÌṩÏòÐÄÁ¦£¬$qv{B_1}=m\frac{v^2}{r}$
´úÈëÊý¾Ý½âµÃ£º
r=1m                        
£¨2£©´øµçÁ£×ÓÔڴų¡ÖÐÔ˶¯µÄÖÜÆÚ£¬${T_0}=\frac{2¦Ðr}{v}=\frac{2¦Ð}{5}¡Á{10^{-4}}$s      
ÔÚ0¡«$\frac{¦Ð}{4}¡Á{10^{-4}}$s¹ý³ÌÖУ¬Á£×ÓÔ˶¯ÁË$\frac{{5{T_0}}}{8}$£¬
Ô²»¡¶ÔÓ¦µÄÔ²ÐĽǣ¬${¦È_1}=\frac{5¦Ð}{4}$
ÔÚ$\frac{¦Ð}{4}¡Á{10^{-4}}$s¡«$\frac{¦Ð}{2}¡Á{10^{-4}}$s¹ý³ÌÖУ¬Á£×ÓÓÖÔ˶¯ÁË$\frac{{5{T_0}}}{8}$£¬
Ô²»¡¶ÔÓ¦µÄÔ²ÐĽǣ¬${¦È_2}=\frac{5¦Ð}{4}$
¹ì¼£ÈçͼaËùʾ£¬¸ù¾Ý¼¸ºÎ¹Øϵ¿ÉÖª£¬
ºá×ø±ê£º$x=2r+2rsin\frac{¦Ð}{4}=£¨2+\sqrt{2}£©m¡Ö3.41$m   
×Ý×ø±ê£º$y=-2rcos\frac{¦Ð}{4}=-\sqrt{2}m¡Ö-1.41$m        
´øµçÁ£×ÓµÄ×ø±êΪ£¨3.41m£¬-1.41m£©
£¨3£©Ê©¼ÓB₂=0.3TµÄÔÈÇ¿´Å³¡ÓëÔ­´Å³¡µþ¼Óºó£¬ÈçͼbËùʾ£¬
¢Ùµ±$nT¡Üt£¼nT+\frac{T}{2}$£¨n=0£¬1£¬2£¬¡­£©Ê±£¬${T_1}=\frac{2¦Ðm}{{q£¨{{B_1}+{B_2}}£©}}=\frac{¦Ð}{4}¡Á{10^{-4}}$s                 
¢Úµ±$nT+\frac{T}{2}¡Üt£¼£¨n+1£©T$£¨n=0£¬1£¬2£¬¡­£©Ê±£¬${T_2}=\frac{2¦Ðm}{{q£¨{{B_1}-{B_2}}£©}}=¦Ð¡Á{10^{-4}}s$
Á£×ÓÔ˶¯¹ì¼£ÈçͼcËùʾ£¬ÔòÁ£×ӻص½Ô­µãµÄʱ¿ÌΪ£¬${t_1}=£¨\frac{¦Ð}{4}+2n¦Ð£©¡Á{10^{-4}}s$
${t_2}=2£¨n+1£©¦Ð¡Á{10^{-4}}s$£¨n=0£¬1£¬2£¬¡­£©     
´ð£º£¨1£©Çó´øµçÁ£×ÓÔÚÔÈÇ¿´Å³¡ÖÐÔ˶¯µÄ¹ìµÀ°ë¾¶Îª1m£»
£¨2£©t=$\frac{¦Ð}{2}$¡Á10^-4sʱ´øµçÁ£×ÓµÄ×ø±êΪ[3.41m£¬-1.41m]£»
£¨3£©Á£×ӻص½×ø±êÔ­µãµÄʱ¿ÌΪt₁=£¨$\frac{¦Ð}{4}$+2n¦Ð£©¡Á10^-4s£¬t₂=2£¨n+1£©¦Ð¡Á10^-4s £¨n=0£¬1£¬2£¬¡­£©£®

µãÆÀ ´ËÌâÊÇ´øµçÁ£×ÓÔÚÔÈÇ¿´Å³¡ÖеÄÔ˶¯ÎÊÌ⣬½âÌâʱҪÈÏÕæ·ÖÎöÁ£×ÓµÄÔ˶¯¹ì¼££¬»­³öÁ£×ÓµÄÔ˶¯¹ì¼£Í¼²¢½áºÏ¼¸ºÎ֪ʶ½øÐÐÇó½â£¬Í¬Ê±Òª×¢ÒâÁ£×ÓÔ˶¯µÄÖÜÆÚÐÔ£¬¹ý³Ì±È½Ï¸´ÔÓ£¬ÒâÔÚ¿¼²éѧÉú×ۺϷÖÎöÎÊÌâµÄÄÜÁ¦£®