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动量守恒的条件性、矢量性和相对性受到大家重视,但动量守恒的同时性却易被忽视,不仅学生如此,连一些复习资料甚至教材都曾在这一问题上出现过疏忽。举两例来研究这一问题。例1,“双人滑”滑冰运动员甲、乙一起以速度 v_0沿一直线在光滑水平冰面上匀速滑行,他们的质量分别为 m_1、m_2。若甲将乙以相对速度 u 沿原滑行方向推出,求乙被推出后,甲的速度 v。原解:令原滑行方向为正,由题设可知,乙被推出后乙对冰面的速度为 v_0+u,则甲推乙前,系统总动量为(m_1+m_2)v_0;甲推出乙后,系统总动量为 m_1v+m_2·(v_0+u)。不计冰面摩擦,由动量守恒定律,有
The conditionality, vectority, and relativity of momentum conservation are valued by everyone. However, the simultaneity of conservation of momentum is easily overlooked. Not only students, but also some review materials and even textbooks have been neglected on this issue. Take two cases to study this issue. In Example 1, the “slippery” skaters A and B move along a straight line at a constant speed along a smooth horizontal ice surface. Their masses are m_1 and m_2, respectively. If A pushes B in the original taxiing direction at a relative speed u, find the speed v of A after B is pushed out. The original solution: the original taxiing direction is positive. It can be known from the title that when B is launched, the speed of B on the ice surface is v_0+u, then before A is pushed, the total momentum of the system is (m_1+m_2)v_0; After that, the total system momentum is m_1v+m_2·(v_0+u). Excluding friction on ice, from the law of conservation of momentum,