高二期中物理答案.pdf

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）