Question

如圖所示，質(zhì)量為3m的足夠長(zhǎng)木板 C 靜止在光滑水平面上，質(zhì)量均為 m 的兩個(gè)物體 A、B 放在 C 的左端，A、B 間相距 s₀，現(xiàn)同時(shí)對(duì)A、B施加水平向右的瞬間沖量而使之分別獲得初速度v₀和2v₀，若A、B與C之間的動(dòng)摩擦因數(shù)分別為 μ 和 2 μ ，則：

（1）最終A、B、C的共同速度為多大

（2）求A達(dá)到最小速度時(shí)，系統(tǒng)產(chǎn)生的熱量Q。

Answer 1

（1）0.6v₀；（2）1.5mv₀²

解析:

（1）對(duì)A、B、C組成的系統(tǒng)，最終相對(duì)靜止，設(shè)共同速度為v_共，由動(dòng)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相對(duì)C向右勻減速運(yùn)動(dòng)階段，C做勻加速運(yùn)動(dòng)，設(shè)A的加速度為a_A，B的加速度為a_B，C的加速度為a_C ，設(shè)t時(shí)刻A、C達(dá)到共同速度v_A，B的速度為v_B

由牛頓第二定律：

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

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

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

t時(shí)間內(nèi)，對(duì)A，由速度公式：v_A＝v₀－a_At ························································ ⑥（1分）

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

對(duì)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一起勻加速運(yùn)動(dòng)，B繼續(xù)以加速度a_B減速運(yùn)動(dòng)，故運(yùn)動(dòng)過(guò)程中A的最小速度為v_A＝0.5v₀

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

根據(jù)系統(tǒng)的能量守恒定律，系統(tǒng)產(chǎn)生的熱量

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

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

【評(píng)析】本題的關(guān)鍵在于平時(shí)對(duì)滑塊與滑板疊加問(wèn)題的積淀，能快速準(zhǔn)確分析相對(duì)運(yùn)動(dòng)的臨界條件、運(yùn)動(dòng)過(guò)程中出現(xiàn)速度相同的物體間摩擦力的轉(zhuǎn)型、快速準(zhǔn)確運(yùn)用牛頓運(yùn)動(dòng)定律及運(yùn)動(dòng)學(xué)公式計(jì)算各個(gè)物體的加速度、速度、位移及物體間相對(duì)位移。在處理問(wèn)題的方法與技巧上要特別重視動(dòng)量及能量守恒往往能快刀斬亂麻地找到初末狀態(tài)的速度，很好地回避中間過(guò)程，這也是兩大守恒的抽象思維能力的體現(xiàn)，在運(yùn)用時(shí)要清醒的認(rèn)識(shí)守恒的對(duì)象、過(guò)程與條件。另外本題后面的運(yùn)動(dòng)過(guò)程未設(shè)置問(wèn)題，實(shí)際上A與C相對(duì)靜止后，將一起加速運(yùn)動(dòng)到0.6v₀，整個(gè)過(guò)程中B的運(yùn)動(dòng)最簡(jiǎn)單，可考慮A與B最終相距多遠(yuǎn)、整個(gè)過(guò)程中A、B與木板C因摩擦所產(chǎn)生的熱量之比、運(yùn)動(dòng)時(shí)間等問(wèn)題。特別提醒：運(yùn)動(dòng)學(xué)公式、動(dòng)量及能量中的矢量均以相對(duì)地面的物理量代入公式計(jì)算，相對(duì)位移用對(duì)地的位移差或功能關(guān)系解決，相對(duì)路程問(wèn)題（具有重復(fù)劃痕的往返相對(duì)運(yùn)動(dòng)）要么分段、要么用功能關(guān)系解決。