天津市部分区2021届高三上学期期末物理试题（扫描）

1110007 1110008 1110009 1110010 $$ 部分区期末练习高三物理答案 第 1 页（共 3 页） 天津市部分区 2020～2021 学年度第一学期期末练习 高三物理参考答案 第Ⅰ卷共 8 题，每题 5 分，共 40 分。 1．D 2．C 3．B 4．C 5．B 6．BC 7．AD 8．BC 第Ⅱ卷共 4 题，共 60 分。 9.（12 分） （1）①9.82（2 分） ②铝棒上墨线 A 点的速度不为零（2 分） （2）①A1：R1（各 2 分） ②0.83（0.80~0.87 范围内都给分）；1.42（1.40~1.44 范围内都给分）（各 2 分） 10.（14 分）解： （1）设地球的质量和半径分别为 M0 和 R0，火星的质量和半径分别为 M 和 R，由于忽略 星球的自转，对质量为 m 的物体，由万有引力定律得 地球上 0 02 0 M m G mg R  ·························································· ········（2 分） 火星上 2 Mm G mg R  ···································································（2 分） 代入数据解得 24m / sg  ·······························································（2 分） （2）对小球受力分析，由已知条件可知 37   在竖直方向上，由平衡条件得 cos mgF   ············································································（2 分） 代入数据解得 5NF  ····································································（2分） （3）对小球在水平方向上，由牛顿第二定律得 2π 2s n in 4 i s T F m L θ  ··············（2 分） 代入数据解得 π2 5 5 T  s ·······················································（2 分） 部分区期末练习高三物理答案 第 2 页（共 3 页） 11.（16 分）解： （1）由图像可知 0 6m / sv  ， 2 1 2m / sa  ··········································（2 分） 在沿杆方向上，对小球由牛顿第二定律得 1sin cos( ) ( )F mg F mg ma      ··········································（2 分） 代入数据解得 40NF  ·······························································（2 分） （2）2s 末小球速度 1 0 1  10m / sv v a t   ·········································· ··（1 分） 0—2s 小球的位移大小 11 2 0 1 2 x v t a t ············································（1 分） 代入数据解得 1 16mx  ····································································（2 分） 2s 后，对小球由牛顿第二定律得 2sin cosmg mg ma    ···························································（2 分） 代入数据解得 2 2 10m / sa  ····························································（1 分） 2s 后沿倾斜细杆向上运动的位移大小 2 1 2 22 5m v x a   ····························（1 分） 小球向上运动到最高点距出发点的距离为 1 2x x x  ·································（1 分） 代入数据解得 21mx  ··································