在“验证动量守恒定律”实验中（1）实验中，在确定小球落地点

问题问答题

在“验证动量守恒定律”实验中

（1）实验中，在确定小球落地点的平均位置时通常采用的做法是用圆规画一个尽可能小的圆把所有的落点圈在里面，圆心即平均位置，其目的是减小实验中的______（选填系统误差、偶然误差）．

（2）图中O点是小球抛出点在地面上的垂直投影．实验时，先让入射球m_l多次从斜轨上S位置静止释放，找到其平均落地点的位置P，测量平抛射程OP．然后，把被碰小球m₂静置于轨道的水平部分，再将入射球m_l从斜轨上S位置静止释放，与小球m₂相碰，并多次重复．

接下来要完成的必要步骤是______．（填选项前的符号）

A．用天平测量两个小球的质量m_l、m₂；B．测量小球m₁开始释放高度h；C．测量抛出点距地面的高度H；D．分别找到m₁、m₂相碰后平均落地点的位置M、N；E．测量平抛射程OM，ON．

（3）若两球相碰前后的动量守恒，其表达式可表示为______（用②中测量的量表示）；若碰撞是弹性碰撞，那么还应满足的表达式为______ （用②中测量的量表示）．

（4）经测定，m₁=45.0g，m₂=15.0g，小球落地点的平均位置距O点的距离分别为OP=44.00cm、ON=60.00cm、OM=23.00cm，则碰撞前、后总动量的比值为______．（保留三位有效数字）

答案

（1）由于落点比较密集，又较多，每次测量距离很难，确定落点平均位置的方法是最小圆法，即用尽可能最小的圆把各个落点圈住，这个圆的圆心位置代表落点的平均位置以减小偶然误差；

（2）实验时，先让入射球m_l多次从斜轨上S位置静止释放，找到其平均落地点的位置P，测量平抛射程OP．然后，把被碰小球m₂静置于轨道的水平部分，再将入射球m_l从斜轨上S位置静止释放，与小球m₂相碰，并多次重复．测量平均落点的位置，找到平抛运动的水平位移，因此步骤中D、E是必须的，而且D要在E之前．至于用天平秤质量先后均可以．所以答案是ADE

（3）设落地时间为t，则v₁=

，v′1=

，v′2=

而动量守恒的表达式是m₁v₁=m₁v₁′+m₂v₂′

所以若两球相碰前后的动量守恒，则m₁?OM+m₂?ON=m₁?OP 成立

若碰撞是弹性碰撞，动能是守恒的，则有

m₁v₁²=

m₁v₁′²+

m₂v₂′^{2
即m1?OM2+m2?ON2=m1?OP2成立
（4）碰撞前后总动量之比：
P′1
P′2
=m1OP
m1OM+m2ON
=45×44
45×23+15×60
=1980
1935
=1.02
故答案为：（1）偶然误差；（2）ADE；（3）m₁?OM+m₂?ON=m₁?OP；m₁?OM₂+m₂?ON₂=m₁?OP₂；（4）1.02}

完形填空。
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