(20分)如图所示,物块A的质量为M,物块B、C的质量都是m,都可以看作质点,且m<M<2m。A与B、B与C用不可身长的轻线通过轻滑轮相连,A与地面用劲度系数为
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(20分)如图所示,物块A的质量为M,物块B、C的质量都是m,都可以看作质点,且m<M<2m。A与B、B与C用不可身长的轻线通过轻滑轮相连,A与地面用劲度系数为k的轻弹簧连接,物块B与物块C的距离和物块C到地面的距离相等,假设C物块落地后不反弹。若物块A距滑轮足够远,且不计一切阻力。则:
(1)若将B与C间的轻线剪断,求A下降多大距离时速度最大; (2)若B与C间的轻线不剪断,将物块A下方的轻弹簧剪断后,要使物块B不与物块C相碰,则M与m应满足什么关系?(不计物块B、C的厚度) |
答案
(1)A下降的距离为x=x1+x2=时速度最大 (2)当M>m时,B物块将不会与C相碰。 |
解析
(1)因为m<M<2m,所以开始时弹簧处于伸长状态,其伸长量x1,则 (2m-M)g=kx1····················································································································· (2分) 得x1=g······················································································································ (1分) 若将B与C间的轻线剪断,A将下降B将上升,当它们的加速度为零时A的速度最大,此时弹簧处于压缩状态,其压缩量x2,则 (M-m)g=kx2······················································································································· (2分) 得x2=g······················································································································· (1分) 所以,A下降的距离为x=x1+x2=时速度最大··································································· (2分) (2)A、B、C三物块组成的系统机械能守恒,设B与C、C与地面的距离均为L,A上升L时,A的速度达到最大,设为v,则 2mgL-MgL=(M+2m)v2 ·································································································· (4分) 当C着地后,A、B两物块系统机械能守恒。 若B恰能与C相碰,即B物块再下降L时速度为零,此时A物块速度也为零,则 MgL-mgL=(M+m)v2········································································································ (4分) 解得:M=m···················································································································· (3分) 由题意可知,当M>m时,B物块将不会与C相碰。············································(1分) |
举一反三
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