解析　(1)小球从曲面上滑下，只有重力做功，由机械能守恒定律知：

mgh＝mv                                                       ①

得v₀＝＝ m/s＝2 m/s.

(2)小球离开平台后做平抛运动，小球正好落在木板的末端，则

H＝gt²                                                                                                                                                      ②

＝v₁t                                                                                                                ③

联立②③两式得：v₁＝4 m/s

设释放小球的高度为h₁，则由mgh₁＝mv

得h₁＝＝0.8 m.

(3)由机械能守恒定律可得：mgh＝mv²

小球由离开平台后做平抛运动，可看做水平方向的匀速直线运动和竖直方向的自由落体运动，则：

y＝gt²                                                                                                                                                       ④

x＝vt                                                                                                                       ⑤

tan 37°＝                                                                                                          ⑥

v_y＝gt                                                                                                                      ⑦

v＝v²＋v                                                       ⑧

E_k＝mv                                                      ⑨

由④⑤⑥⑦⑧⑨式得：E_k＝32.5h                                                                       ⑩

考虑到当h>0.8 m时小球不会落到斜面上，其图象如图所示

答案　(1)2 m/s　(2)0.8 m　(3)E_k＝32.5h　图象见解析

解析　(1)如图所示，画出通过P₁、P₂的入射光线，交AC面于O，画出通过P₃、P₄的出射光线交AB面于O′.则光线OO′就是入射光线P₁P₂在三棱镜中的折射光线．

(2)在所画的图上注明入射角θ₁和折射角θ₂，并画出虚线部分，用量角器量出θ₁和θ₂(或用直尺测出线段EF、OE、GH、OG的长度)．

(3)n＝；(或因为sin θ₁＝，sin θ₂＝，

则n＝＝)．

答案　见解析