㈠（1）某学生用螺旋测微器在测定某一金属丝的直径时，测得的结果如下图1所示，则该金属丝的直径d=______
推荐回答（1个）
（一）（1）由图1所示螺旋测微器可知，螺旋测微器的固定刻度示数是3mm，
可动刻度示数是20.6×0.01mm=0.206mm，螺旋测微器的示数是3mm+0.206mm=3.206mm；
由图2所示游标卡尺可知，主尺示数是5cm，游标尺示数是3×0.05mm=0.15mm=0.015cm，
则游标卡尺的示数是5cm+0.015cm=5.015cm．
（2）由图3可知：s CE =s AE -s AC =120.0mm-52.0mm=68mm=0.068m，
则在D点时小车的瞬时速度v D = = =0.34m/s；
由图3可知：s DE =s AE -s AD =120.0mm-84.0mm=36mm=0.036m，
s CD =s AD -s AC =84.0mm-52.0mm=32.0mm=0.032m，
小车的加速度a= = =0.4m/s 2 ；
（二）（1）m 1 与m 2 相撞后，m 2 的速度增大，m 1 的速度减小，m 1 的速度小于m 2 的速度，它们都做平抛运动，
m 1 的水平位移小于m 2 的水平位移，因此E点是碰前m 1 的落地点，D是碰后m 1 的落地点，F是碰后m 2 的落地点．
（2）设斜面BC的倾角是θ，小球离开斜槽AB后做平抛运动；
碰前小球m 1 的落地点是E，它的水平位移x 1 =L E cosθ=v 1 t 1 ，
竖直位移y 1 =L E sinθ= gt 1 2 ，解得v 1 =L E cosθ ；
碰后小球m 1 的落地点是D，它的水平位移x 1 ′=L D cosθ=v 1 ′t 1 ′，
竖直位移y 1 ′=L D ′sinθ= gt 1 2 ，解得v 1 ′=L D cosθ ；
碰后小球m 2 的落地点是F，它的水平位移x 2 =L F cosθ=v 2 t 2 ，
竖直位移y 2 ′=L F sinθ= gt 2 ′ 2 ，解得v 2 ′=L F cosθ ；
如果m 1 v 1 =m 1 v 1 ′+m 2 v 2 ′，m 1 L E cosθ =m 1 L D cosθ +m 2 L F cosθ ，
即当m 1 =m 1 +m 2 时，说明碰撞中动量是守恒的．
如果 m 1 v 1 2 = m 1 v 1 ′ 2 + m 2 v 2 ′ 2 ，即当m 1 L E =m 1 L D +m 2 L F 时，说明两小球的碰撞时弹性碰撞．
故答案为：（一）（1）3.206；5.015；（2）0.34；0.4；
（二）（1）D；F；（2）m 1 =m 1 +m 2 ；（3）m 1 L E =m 1 L D +m 2 L F ． |
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