Question

一质量为m的小球以v₀的速度与静止在光滑水平面上的质量为M的小球发生对心碰撞，碰撞时无机械能损失.求碰后两球的速度.

Answer 1

解析：设碰后m的速度为v₁,M的速度为v₂,由动量守恒定律得

mv₀=mv₁+Mv₂                                                                                                ①

因碰撞时无机械能损失，总动能守恒

mv₀²=mv₁²+Mv₂²                                                                                   ②

联立求解方程①、②便可得到v₁、v₂，但该二元二次方程组用代入法解很麻烦，需变形后再解

①变为m(v₀-v₁)=Mv₂                                                                                        ③

②变为m(v₀²-v₁²)= Mv₂²                                                                             ④

④除以③得

v₀+v₁=v₂                                                                                                        ⑤

解①⑤组成的方程组可得v₁=v₀                                                               ⑥

v₂=v₀.                                                                ⑦

讨论:（1）m＞M时，v₁为正；m＜M时，v₁为负，

即被反向弹回；m=M时，v₁=0,v₂=v₀，即碰后两球交换速度.

（2）当m＞＞M时，v₁≈v₀,v₂≈2v₀;

当m＜＜M时，v₁≈-v₀,v₂≈0.