2.如图所示,质量为m的A球以速度v0在光滑水平面上运动,-|||-与原来静止的质量为4m的B球碰撞,碰撞后A球以 =(v)_(0)(v)_(0)-|||-(alpha lt 1) 的速率弹回,并与挡板P发生完全弹性碰撞,若要使-|||-A球能追上B球再相碰,则α的取值范围为-|||-P A/v0 B-|||-777-|||-A. dfrac (1)(5)lt alpha lt dfrac (1)(3) B. dfrac (1)(3)lt alpha lt dfrac (2)(3)-|||-C. dfrac (1)(3)lt alpha lt dfrac (2)(5) D. dfrac (1)(3)lt alpha leqslant dfrac (3)(5)-|||-[解题思路]由题意可知,A、B两小球在碰撞过程中动量守-|||-恒,以A球初速度v的方向为正方向,设碰后B球的速度为-|||-vB,则由动量守恒定律可得 (v)_(0)=-ma(v)_(0)+4m(v)_(B) ,A与挡板P-|||-碰撞后能追上B发生再次碰撞的条件是 _(U)(U)_(0)gt (U)_(B) ,两式联立-|||-可解得 alpha gt dfrac (1)(3) ;碰撞前后两小球的机械能应满足 dfrac (1)(2)m({v)_(0)}^2geqslant -|||-dfrac (1)(2)m((-{{V)_(0)}_(0))}^2+dfrac (1)(2)times 4m({V)_(B)}^2 ,解得 alpha leqslant dfrac (3)(5) ,综合可得 dfrac (1)(3)lt -|||-alpha leqslant dfrac (3)(5) o