Question

如图所示，质量为2m的木板停在光滑的水平面上，其左端有质量为m、可视为质点的遥控电动赛车，由静止出发，经过时间t后关闭电动机，此时赛车速度为v₁，赛车在木板上滑行一段距离后，恰好停在木板的右端。若通电后赛车以恒定功率P行驶，赛车在运动过程中受到木板的摩擦阻力恒为f，不计空气阻力，求

（1）赛车由静止出发经过时间t后木板速度v₂的大小和方向；

（2）赛车由静止出发在t时刻与木板左端之间的距离L₁；

（2）木板长度L 。

Answer 1

【标准解答】（1）对赛车和木板组成的系统，由动量守恒定律：

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

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

速度v₂的方向水平向左

（2）对赛车和木板组成的系统，由能量守恒定律：

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

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

（3）设赛车恰好停在木板的右端时，赛车和木板的共同速度为v，则以赛车和木板组成的系统为研究对象，对全过程，由动量守恒定律：

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）

【思维点拔】本题的关键在于判断在关闭电动机之前赛车的牵引力是变力，导致赛车或木板的合外力均为变力，不能用动力学观点来求解，结合动量和能量观点来解答；在关闭电动机之后，赛车或木板的合外力均为恒力，可以用动力学观点来求解。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。在本题中要注意电动机对赛车要做功的同时，对木板也要做功。拓展：若通电后赛车以恒定加速度a行驶，其额定功率为P，当维持匀加速直线运动的速度达到最大时，关闭电动机，则情况又如何？