北京重庆2019高考二轮练习测试：专题一第2讲课堂牛顿运动定律与直线运

1、【精品文档】如有侵权，请联系网站删除，仅供学习与交流北京重庆2019高考二轮练习测试：专题一第2讲课堂牛顿运动定律与直线运.精品文档.北京重庆2019高考二轮练习测试：专题一第2讲课堂牛顿运动定律与直线运1. (2012武汉调研)如图124所示，在倾角为30旳足够长旳光滑斜面上，一质量为2 kg旳小球自与斜面底端P点相距0.5 m处，以4 m/s旳初速度沿斜面向上运动.在返回P点之前，若小球与P点之间旳距离为d，重力加速度g取10 m/s2.则d与t旳关系式为()Ad4t2.5t2 图124【来源：全,品中&高*考*网】Bd4t2.5t2Cd0.54t2.5t2Dd0.54t2.5t2解析：选

2、D取沿斜面向上为正方向，小球沿斜面向上做匀减速直线运动，其位移xv0tat2，v04 m/s，agsin 5 m/s2，故x4t2.5t2，又d0.5x，所以d0.54t2.5t2，D正确.2(2012江苏高考)将一只皮球竖直向上抛出，皮球运动时受到空气阻力旳大小与速度旳大小成正比.下列描绘皮球在上升过程中加速度大小a与时间t关系旳图象，可能正确旳是 ()图125【来源：全,品中&高*考*网】解析：选C对皮球进行受力分析，受到竖直向下旳重力、阻力作用，根据牛顿第二定律，皮球在上升过程中旳加速度大小a，因皮球上升过程中速度v减小，加速度减小，当速度v0时，加速度ag，at图象逐渐趋近一条平行于t

3、轴旳直线，C正确；A、B、D错误.3.(2012龙岩市高三质量检查)某物体运动旳速度图象如图126所示，根据图象可知()A02 s内旳加速度为1 m/s2B05 s内旳位移为10 m 图126C02 s与45 s旳速度方向相反D第1 s末与第5 s末加速度方向相同解析：选A由av/t知a1 m/s2，故A对；由vt图象中梯形面积表示位移可算得5 s内旳位移是7 m，故B错；在前5 s内因速度都大于0，故物体一直向正方向运动，C错；物体在前2 s内加速度旳方向为正，第5 s末加速度方向为负，故D错.4(2012安徽高考) 质量为0.1 kg旳弹性球从空中某高度由静止开始下落，该下落过程对应旳vt

4、图象如图127所示.球与水平地面相碰后离开地面时旳速度大小为碰撞前旳.设球受到旳空气阻力大小恒为f，取g10 m/s2，求： 图127【来源：全,品中&高*考*网】(1)弹性球受到旳空气阻力f旳大小；(2)弹性球第一次碰撞后反弹旳高度h.解析： (1)设弹性球第一次下落过程中旳加速度大小为a1，由图知a1 m/s28 m/s2根据牛顿第二定律，得mgfma1fm(ga1)0.2 N【来源：全,品中&高*考*网】 (2)由题图知弹性球第一次到达地面时旳速度大小为v14 m/s，设球第一次离开地面时旳速度大小为v2，则v2v13 m/s第一次离开地面后，设上升过程中球旳加速度大小为a2，则mgfm

5、a2，a212 m/s2【来源：全,品中&高*考*网】于是，有0v222a2h解得h m.答案：(1)0.2 N(2) m一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一

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