③          (2分)

小物块从A到B做自由落体运动，根据运动学公式有           ①    (2分)

代入数据解得v_B=4m/s，方向竖直向下                                                                (2分)

（2）设小物块沿轨道切线方向的分速度为v_Bx，因OB连线与竖直方向的夹角为60°，故v_Bx=v_Bsin60°

                                                                ②      (2分)

 从B到C，只有重力做功，根据机械能守恒定律有

  小车最后速度为0.4m/s

24导与练上有

35(20分) 解：（1）由几何关系可知，AB间的距离为R (1分)

　当v2′=时，v1=>v₂′不合题意，舍去。                                 (2分)

  即P与Q碰撞后瞬间Q的速度大小为v₂=2m/s

  v=0.4m/s                                                                                                       (3分)

  v₂′=

μ₂m1gS+μ₂m2gL=                                   (3分)

  解得，v₂=2m/s

    W_F-μ₁m₁g(2L₁+L₂)=                                                                       (4分)

  解得W_F=6J                                                                                                         (3分)

  (2)设P、Q碰后速度分别为v₁、v₂，小车最后速度为v，由动量守恒定律得

  m₁vc=m₁v₁+-m₂v₂                                                                                        (2分)

  m₁v_c=(m₁+m₂+M)v                                                                                      (2分)

  由能量守恒得

34．解：(1)对P由A→B→C应用动能定理，得

（4=3+1分）