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问题缘由:2013年上海高考物理卷第20题涉及双拖驳船问题,原题和参考解答有几处问题值得探讨.摘录原题如下.图1图1为在平静海面上,两艘拖船A、B拖着驳船C运动的示意图.A、B的速度分别沿着缆绳CA、CB方向,A、B、C不在一条直线上.由于缆绳不可伸长,因此C的速度在CA、CB方向的投影分别与A、B的速度相等,由此可知C的(A)速度大小可以介于A、B的速度大小之间.(B)速度大小一定不小于A、B的速度大小.
The cause of the problem: The 20th issue of the 2013 Shanghai Physical Education Examination Paper involves the issue of double-tug barges. There are several problems with the original questions and the reference answers. The excerpts from the original questions are as follows. Figure 1 Figure 1 shows two tugs on a calm sea. B. Diagram of the motion of the barge C. The speeds of A and B are along the directions of cables CA and CB respectively. A, B and C are not on a straight line. Because the cable cannot be extended, the speed of C is projected in the direction of CA and CB. The speeds of A and B are equal to each other. It can be seen that the speed of (A) of C can be between the speeds of A and B. (B) The speed is not less than the speed of A and B.