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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1-2-4所示,在倾角为θ=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( )
A.d=4t+2.5t2 图1-2-4【来源:全,品…中&高*考*网】
B.d=4t-2.5t2
C.d=0.5+4t+2.5t2
D.d=0.5+4t-2.5t2
解析:选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x=v0t-at2,v0=4 m/s,a=gsin θ=5 m/s2,故x=4t-2.5t2,又d=0.5+x,所以d=0.5+4t-2.5t2,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1-2-5【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v=0时,加速度a=g,a-t图象逐渐趋近一条平行于t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1-2-6所示,根据图象可知( )
A.0~2 s内旳加速度为1 m/s2
B.0~5 s内旳位移为10 m 图1-2-6
C.0~2 s与4~5 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a=Δv/Δt知a=1 m/s2,故A对;由v-t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v-t图象如图1-2-7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g=10 m/s2,求: 图1-2-7【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/s2=8 m/s2
根据牛顿第二定律,得
mg-f=ma1
f=m(g-a1)=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v1=4 m/s,设球第一次离开地面时旳速度大小为v2,则
v2=v1=3 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mg+f=ma2,a2=12 m/s2【来源:全,品…中&高*考*网】
于是,有0-v22=-2a2h
解得h= m.
答案:(1)0.2 N (2) m
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