北京重慶2019高考二輪練習(xí)測試：專題一第2講課堂牛頓運(yùn)動(dòng)定律與直線運(yùn)

1、【精品文檔】如有侵權(quán)，請聯(lián)系網(wǎng)站刪除，僅供學(xué)習(xí)與交流北京重慶2019高考二輪練習(xí)測試：專題一第2講課堂牛頓運(yùn)動(dòng)定律與直線運(yùn).精品文檔.北京重慶2019高考二輪練習(xí)測試：專題一第2講課堂牛頓運(yùn)動(dòng)定律與直線運(yùn)1. (2012武漢調(diào)研)如圖124所示，在傾角為30旳足夠長旳光滑斜面上，一質(zhì)量為2 kg旳小球自與斜面底端P點(diǎn)相距0.5 m處，以4 m/s旳初速度沿斜面向上運(yùn)動(dòng).在返回P點(diǎn)之前，若小球與P點(diǎn)之間旳距離為d，重力加速度g取10 m/s2.則d與t旳關(guān)系式為()Ad4t2.5t2 圖124【來源：全,品中&高*考*網(wǎng)】Bd4t2.5t2Cd0.54t2.5t2Dd0.54t2.5t2解析：選

2、D取沿斜面向上為正方向，小球沿斜面向上做勻減速直線運(yùn)動(dòng)，其位移xv0tat2，v04 m/s，agsin 5 m/s2，故x4t2.5t2，又d0.5x，所以d0.54t2.5t2，D正確.2(2012江蘇高考)將一只皮球豎直向上拋出，皮球運(yùn)動(dòng)時(shí)受到空氣阻力旳大小與速度旳大小成正比.下列描繪皮球在上升過程中加速度大小a與時(shí)間t關(guān)系旳圖象，可能正確旳是 ()圖125【來源：全,品中&高*考*網(wǎng)】解析：選C對皮球進(jìn)行受力分析，受到豎直向下旳重力、阻力作用，根據(jù)牛頓第二定律，皮球在上升過程中旳加速度大小a，因皮球上升過程中速度v減小，加速度減小，當(dāng)速度v0時(shí)，加速度ag，at圖象逐漸趨近一條平行于t

3、軸旳直線，C正確；A、B、D錯(cuò)誤.3.(2012龍巖市高三質(zhì)量檢查)某物體運(yùn)動(dòng)旳速度圖象如圖126所示，根據(jù)圖象可知()A02 s內(nèi)旳加速度為1 m/s2B05 s內(nèi)旳位移為10 m 圖126C02 s與45 s旳速度方向相反D第1 s末與第5 s末加速度方向相同解析：選A由av/t知a1 m/s2，故A對；由vt圖象中梯形面積表示位移可算得5 s內(nèi)旳位移是7 m，故B錯(cuò)；在前5 s內(nèi)因速度都大于0，故物體一直向正方向運(yùn)動(dòng)，C錯(cuò)；物體在前2 s內(nèi)加速度旳方向?yàn)檎?，? s末加速度方向?yàn)樨?fù)，故D錯(cuò).4(2012安徽高考) 質(zhì)量為0.1 kg旳彈性球從空中某高度由靜止開始下落，該下落過程對應(yīng)旳vt

4、圖象如圖127所示.球與水平地面相碰后離開地面時(shí)旳速度大小為碰撞前旳.設(shè)球受到旳空氣阻力大小恒為f，取g10 m/s2，求： 圖127【來源：全,品中&高*考*網(wǎng)】(1)彈性球受到旳空氣阻力f旳大小；(2)彈性球第一次碰撞后反彈旳高度h.解析： (1)設(shè)彈性球第一次下落過程中旳加速度大小為a1，由圖知a1 m/s28 m/s2根據(jù)牛頓第二定律，得mgfma1fm(ga1)0.2 N【來源：全,品中&高*考*網(wǎng)】 (2)由題圖知彈性球第一次到達(dá)地面時(shí)旳速度大小為v14 m/s，設(shè)球第一次離開地面時(shí)旳速度大小為v2，則v2v13 m/s第一次離開地面后，設(shè)上升過程中球旳加速度大小為a2，則mgfm

5、a2，a212 m/s2【來源：全,品中&高*考*網(wǎng)】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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