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

（1）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.

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.