Question

如图所示，质量为3m的足够长木板 C 静止在光滑水平面上，质量均为 m 的两个物体 A、B 放在 C 的左端，A、B 间相距 s₀，现同时对A、B施加水平向右的瞬间冲量而使之分别获得初速度v₀和2v₀，若A、B与C之间的动摩擦因数分别为 μ 和 2 μ ，则：

（1）最终A、B、C的共同速度为多大

（2）求A达到最小速度时，系统产生的热量Q。

Answer 1

（1）0.6v₀；（2）1.5mv₀²

解析:

（1）对A、B、C组成的系统，最终相对静止，设共同速度为v_共，由动量守恒定律：

mv₀＋2mv₀＝（m＋m＋3m）v_共······································································· ①（4分）

解得：v_共＝0.6v₀ ······················································································· ②（1分）

（2）A、B相对C向右匀减速运动阶段，C做匀加速运动，设A的加速度为a_A，B的加速度为a_B，C的加速度为a_C ，设t时刻A、C达到共同速度v_A，B的速度为v_B

由牛顿第二定律：

a_A＝gμ ··································································································· ③（1分）

a_B＝2gμ ································································································· ④（1分）

a_C＝＝gμ ·························································································· ⑤（1分）

t时间内，对A，由速度公式：v_A＝v₀－a_At ························································ ⑥（1分）

对B：v_B＝2v₀－a_Bt ···················································································· ⑦（1分）

对C：v_A＝a_C t ·························································································· ⑧（1分）

解得：v_A＝0.5v₀     ·················································································· ⑨（1分）

v_B＝v₀ ···································································································· ⑩（1分）

此后A、C一起匀加速运动，B继续以加速度a_B减速运动，故运动过程中A的最小速度为v_A＝0.5v₀

············································································································ ⑾（1分）

根据系统的能量守恒定律，系统产生的热量

Q＝mv₀²＋m(2v₀)²－mv??_A²－×3mv??_A²－mv??_B² ·················································· ⑿（4分）

解得 Q＝1.5mv₀²························································································ ⒀（2分）

【评析】本题的关键在于平时对滑块与滑板叠加问题的积淀，能快速准确分析相对运动的临界条件、运动过程中出现速度相同的物体间摩擦力的转型、快速准确运用牛顿运动定律及运动学公式计算各个物体的加速度、速度、位移及物体间相对位移。在处理问题的方法与技巧上要特别重视动量及能量守恒往往能快刀斩乱麻地找到初末状态的速度，很好地回避中间过程，这也是两大守恒的抽象思维能力的体现，在运用时要清醒的认识守恒的对象、过程与条件。另外本题后面的运动过程未设置问题，实际上A与C相对静止后，将一起加速运动到0.6v₀，整个过程中B的运动最简单，可考虑A与B最终相距多远、整个过程中A、B与木板C因摩擦所产生的热量之比、运动时间等问题。特别提醒：运动学公式、动量及能量中的矢量均以相对地面的物理量代入公式计算，相对位移用对地的位移差或功能关系解决，相对路程问题（具有重复划痕的往返相对运动）要么分段、要么用功能关系解决。