、完全非单性碰撞 碰撞后速度相同v1=v2=ν 动量守恒m1v1+m 2120=(m1+m +nk = tm2 动能损失 △E 2 110 +- n-(m1+m2)y lI1 2(m,+m 10 20 2三、完全非弹性碰撞 碰撞后速度相同v1=v2 =v 动量守恒 m v m v (m m )v 1 10 + 2 20 = 1 + 2 1 2 1 10 2 20 m m m v m v v + + = 动能损失 ( ) ( ) ( ) 2 1 0 2 0 1 2 1 1 2 1 2 2 2 2 0 2 1 1 0 2 2 1 2 1 2 1 v v m m m m E m v m v m m v − + = − + = +