Question

一質(zhì)量為m的小球以v₀的速度與靜止在光滑水平面上的質(zhì)量為M的小球發(fā)生對(duì)心碰撞，碰撞時(shí)無(wú)機(jī)械能損失.求碰后兩球的速度.

Answer 1

解析：設(shè)碰后m的速度為v₁,M的速度為v₂,由動(dòng)量守恒定律得

mv₀=mv₁+Mv₂                                                                                                ①

因碰撞時(shí)無(wú)機(jī)械能損失，總動(dòng)能守恒

mv₀²=mv₁²+Mv₂²                                                                                   ②

聯(lián)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?yōu)閙(v₀-v₁)=Mv₂                                                                                        ③

②變?yōu)?IMG align="middle" height=41 src="http://thumb.zyjl.cn/pic1/1898/img/02//73//12/189802731210003712/4.gif" width=16 v:shapes="_x0000_i1308">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(shí)，v₁為正；m＜M時(shí)，v₁為負(fù)，

即被反向彈回；m=M時(shí)，v₁=0,v₂=v₀，即碰后兩球交換速度.

（2）當(dāng)m＞＞M時(shí)，v₁≈v₀,v₂≈2v₀;

當(dāng)m＜＜M時(shí)，v₁≈-v₀,v₂≈0.