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2019-2020 学年度第一学期期中学业水平检测 高二物理答案及评分标准 一、选择题:本大题共 12 小题,每小题 4 分,共 48 分。 二、实验:本大题共 2 小题,共 12 分。 13.(4 分)(1) 大于 (1 分) 等于 (1 分);( 2)AC (2 分)。 14.(8 分)(1)BC (1 分) DE (1 分); (2)2.07(2 分)2.06(2 分); (3)在误差允许的范围内,系统的动量守恒。(1 分) (4)无 (1 分) 三、解答题:本大题共 3 小题,共 30 分,解答时应写出必要的文字说明、证明过程 或演算步骤。 15.(7 分) 解: 假设鸡蛋到达地面行人时的速度为 v, 鸡蛋从 62.8m 高处自由落体 由运动学公式得 22gh v= ················································································ (2分) 对鸡蛋与安全帽撞击的过程,取向上为正方向 由动量定理: ( ) 0( )F mg t mv− ∆= −− ······························································ (3分) 解得 F = 2700.3 N ·························································································· (2分) 16.(9 分) 解: (1)由闭合电路欧姆定律可得: ()E IR r= + ·························································· (2分) 解得:I=1.5A ························································································ (1分) (2)对导体棒受力分析可得: 0= tan37F BIL mg=安 ································································································································· (2分) 题号 1 2 3 4 5 6 7 8 9 10 11 12 答案 B C B C D B C A BD AD ABC BC 高二物理答案 第 1 页(共 4 页) 解得:B=0.5T ······························································································· (1分) (3)对导体棒受力分析可得: 0= sin37BIL mg f+ ························································································· (1分) 0= cos37NF mg ······························································································· (1分) = NfFµ 联立解得: 3 0.1916 µ = ≈ ················································································ (1分) 17.( 11 分) 解 :( 1)设小物块的质量为 m,由于物块在做圆周运动,故: 2 2 2 4πsin = sinF mLT θθ ·················································································· (2分) 又: 1 =F mg ··································································································· (1分) 联立可得: 2 1 2 2 4π= FLg FT ····················································································· (2分) (2)在星球表面: =2 GMm mgR ··································································································· (2分) 解得: 22 1 2 2 4π= F LRM FT G ····················································································· (1分) (3)万有引力提供向心力,可得: 2 2) GMm mv Rh Rh ( ·························································································· (2分) 解得: 1 2 2π () FLRv T FRh   ······································································· (1分) 18.(13 分) 解 :( 1)小球 P 碰撞结束后获得的速度为 v0,在最低点由向心力公式可得: 2 0 0 PmvF mg l P ·························································································· (1分) 解得:vP= 4.5m/s 高二物理答案 第 2 页(共 4 页) 小球 p 摆到最低点的速度为: 2 00 1 2 Pm gl m v′= ···································································· (1分) 代入数据解得碰前 P 的速度为: 3m/sPv′ = ,方向水中向右 在小球 P 与 A 碰撞瞬间动量守恒,以水平向右为正方向, 则: ( )00PA Pm v Mv m v′ = +− ············································································ (1分) 代入数值可以得到物块 A 的速度: 3m/sAv = ,方向水中向右; ······························ (1分) (2)由题可知,木板 B 与挡板发生碰撞前,A 与 B 已经共速,设速度为 ABv 以 A、B 为系统,以水平向右为正方向,根据动量守恒: ( )A ABMMv mv= + ····················································································· (1分) 代入数据可以得到: 2.5m/sABv = ,方向水平向右 设第一碰撞前,A、B 相对位移 L ,则根据能量守恒可以得到: ( )2211 22A ABMMLv vMgmm = −+ ···································································· (2分) 代入数值可以得到: 0.75mL = ······································································· (1分) (3)由题可知木板 B 与挡板发生弹性碰撞,则碰后木板 B 以原速率反弹,而 A 仍以原速度前 进,即 A 减速前进,木板 B 先向右减速然后向左加速,当再次 A 共速后与挡板再次发生碰撞, 接着再次重复上述过程,直到最终二者动停止 由上面分析可知:第一次碰撞: ( )A ABvMM mv= + , 2.5m/sABv = 反弹之后再次共速: ( ) ' AB AB ABv mv MM mv−=+ ,则: ' 222.5m/s33AB ABvv= = × 第二次碰撞,反弹、共速: ( )' ' '' AB AB ABM v Mv m mv−=+ ,则: '' '1 222.5 /2 33AB ABv v ms= =×× 第三次碰撞,反弹、共速: ( )'' '' ''' AB AB ABv m M mvM v−=+ ,则 : ''' ''1 2222.5m/s2 333AB ABvv= =××× … 每次与挡板碰撞之后,木板 B 都是先向左做减速运动到零之后,再反向向右加速 只研究木板 B 向左减速过程,只有 A 对 B 的摩擦力做负功,设木板 B 向左减速运动的总路程为 s ,则根据动能定理可以得到: 为 高二物理答案 第 3 页(共 4 页) ( )2 '2 ''2 '''210 ...2 AB AB AB ABMgs mvvvvm − =− ++++ ·················································· (4分) 根据数学等比数列知识可以得到: 0.6ms = ······················································· (1分) 高二物理答案 第 4 页(共 4 页) 查看更多

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