Question

一質量為m的小球以v₀的速度與靜止在光滑水平面上的質量為M的小球發生對心碰撞，碰撞時無機械能損失.求碰后兩球的速度.

Answer 1

（1）v₁=v₀                                                               ⑥
v₂=v₀.                                                                ⑦
討論:（1）m>M時，v₁為正；m<M時，v₁為負，
即被反向彈回；m=M時，v₁=0,v₂=v₀，即碰后兩球交換速度.
（2）當m>>M時，v₁≈v₀,v₂≈2v₀;
當m<<M時，v₁≈-v₀,v₂≈0.

設碰后m的速度為v₁,M的速度為v₂,由動量守恒定律得
mv₀=mv₁+Mv₂                                                                                                ①
因碰撞時無機械能損失，總動能守恒
mv₀²=mv₁²+Mv₂²                                                                                   ②
聯立求解方程①、②便可得到v₁、v₂，但該二元二次方程組用代入法解很麻煩，需變形后再解①變為m(v₀-v₁)=Mv₂                                                                                     ③
②變為m(v₀²-v₁²)= Mv₂²                                                                             ④
④除以③得
v₀+v₁=v₂                                                                                                       ⑤
解①⑤組成的方程組可得v₁=v₀                                                               ⑥
v₂=v₀.                                                                ⑦
討論:（1）m>M時，v₁為正；m<M時，v₁為負，
即被反向彈回；m=M時，v₁=0,v₂=v₀，即碰后兩球交換速度.
（2）當m>>M時，v₁≈v₀,v₂≈2v₀;
當m<<M時，v₁≈-v₀,v₂≈0.