Question

如圖所示，質(zhì)量為2m的木板停在光滑的水平面上，其左端有質(zhì)量為m、可視為質(zhì)點(diǎn)的遙控電動(dòng)賽車，由靜止出發(fā)，經(jīng)過(guò)時(shí)間t后關(guān)閉電動(dòng)機(jī)，此時(shí)賽車速度為v₁，賽車在木板上滑行一段距離后，恰好停在木板的右端。若通電后賽車以恒定功率P行駛，賽車在運(yùn)動(dòng)過(guò)程中受到木板的摩擦阻力恒為f，不計(jì)空氣阻力，求

（1）賽車由靜止出發(fā)經(jīng)過(guò)時(shí)間t后木板速度v₂的大小和方向；

（2）賽車由靜止出發(fā)在t時(shí)刻與木板左端之間的距離L₁；

（2）木板長(zhǎng)度L 。

Answer 1

【標(biāo)準(zhǔn)解答】（1）對(duì)賽車和木板組成的系統(tǒng)，由動(dòng)量守恒定律：

mv₁ – 2mv₂ = 0···································································································· ①（3分）

解得：v₂＝v₁·································································································· ②（1分）

速度v₂的方向水平向左

（2）對(duì)賽車和木板組成的系統(tǒng)，由能量守恒定律：

Pt – fL₁ = mv₁² +  · 2mv₂² ··········································································· ③（4分）

L₁ = (Pt – mυ₁²)························································································· ④（1分）

（3）設(shè)賽車恰好停在木板的右端時(shí)，賽車和木板的共同速度為v，則以賽車和木板組成的系統(tǒng)為研究對(duì)象，對(duì)全過(guò)程，由動(dòng)量守恒定律：

3mv = 0·············································································································· ⑤（3分）

得v = 0·············································································································· ⑥（1分）

由能量守恒定律：

Pt – fL = 3mv² – 0··························································································· ⑦（4分）

得L =  ······································································································· ⑧（1分）

【思維點(diǎn)拔】本題的關(guān)鍵在于判斷在關(guān)閉電動(dòng)機(jī)之前賽車的牽引力是變力，導(dǎo)致賽車或木板的合外力均為變力，不能用動(dòng)力學(xué)觀點(diǎn)來(lái)求解，結(jié)合動(dòng)量和能量觀點(diǎn)來(lái)解答；在關(guān)閉電動(dòng)機(jī)之后，賽車或木板的合外力均為恒力，可以用動(dòng)力學(xué)觀點(diǎn)來(lái)求解。在處理問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。在本題中要注意電動(dòng)機(jī)對(duì)賽車要做功的同時(shí)，對(duì)木板也要做功。拓展：若通電后賽車以恒定加速度a行駛，其額定功率為P，當(dāng)維持勻加速直線運(yùn)動(dòng)的速度達(dá)到最大時(shí)，關(guān)閉電動(dòng)機(jī)，則情況又如何？