µ±Ç°Î»ÖãºÊ×Ò³ > ÏÃÃÅÒ»ÖиßÈýÀí¿Æ×ÛºÏÄÜÁ¦²âÊÔÄ£ÄâÊÔÌâ¶þ
ÏÃÃÅÒ»ÖиßÈýÄ£ÄâÊÔÌâ
ÁªÁ¢ÇóµÃ£ºL?mv qB´úÈëÊý¾Ý£¬½âµÃ£º
L?5.8?10?2m £¨2·Ö£©
25£®£¨20·Ö£©
½â£º£¨1£©½â³ýËø¶¨ºóµ¯»É½«µ¯ÐÔÊÆÄÜÈ«²¿×ª»¯ÎªAÇòµÄ»úеÄÜ£¬Ôòµ¯»ÉµÄµ¯ÐÔÊÆÄÜΪ
Eµ¯?mgR £¨3·Ö£©
£¨2£©ABϵͳÓÉˮƽλÖû¬µ½Ô²¹ìµÀ×îµÍµãʱËÙ¶ÈΪv0£¬ÓÉ»úеÄÜÊØºã¶¨ÂÉ£¬ÓÐ
2mgR = 2mv02/2 £¨3·Ö£©
½â³ýµ¯»ÉËø¶¨ºó£¬µ¯»É»Ö¸´µ½Ô³¤Ê±£¬A¡¢BµÄËÙ¶È·Ö±ðΪvA¡¢vB £¬Óɶ¯Á¿ÊغãºÍÄÜÁ¿Êغ㣬ÔòÓÐ
2mv0 = mvA + mvB £¨3·Ö£© 2mv0/2?Eµ¯?mvA/2?mvB/2 £¨3·Ö£©
222
½âµÃ£ºvA?2gR?gR £¨²»·ûºÏÌâÒ⣬ÉáÈ¥£©
vA?2gR?gR £¨3·Ö£©
£¨3£©ÉèAÇòÏà¶Ô°ëÔ²²Û¿ÚÉÏÉý×î´ó¸ß¶ÈΪh£¬Ôò
mg?h?R??12mvA £¨3·Ö£© 2?1?½âµÃ£ºh???2?R?1.9R £¨2·Ö£©
?2?µÚ17Ò³£¬¹²13Ò³
¹²·ÖÏí92ƪÏà¹ØÎĵµ