9.ABD【解析】带电粒子进入偏转电场时,根据动能定理得qU=bmv,进入偏转电场后偏转的位移为有y=u-1y-号,与q、m无关,当y相同时,x相同,可知两个粒子运动轨迹重合,打到感光板上的位置相同,故A项正确;设粒子在加速电场中加速时间为t1,加速位移为d,在偏转电场中运动时间为h,偏转位移为y有d=是m,可得=VqU,由于d减小,粒子在加速电场中加速时间td减小,由y=2m知粒子在偏转电场中运动时间a不变,因此质子和a粒子打到感光板的时间将减小,故B项正确;从开始运动到打到感光板上,根据动能E定理有q+9Ey=E.-0,故E=2=2,C项错误;设质子在O2处的速度为v,a粒子在O2处的速度为v,根据动能定理得qU=bmv2,可得v20,质子的比荷里=1,a粒子的比荷坐一1,解得弘=2,故D项正确。

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