根據(jù)動(dòng)量守恒,
mv=(m+2m+...+nm)V
即v=(1+2+...+n)V=n(n+1)/2 V
V=2/(n(n+1)) v
動(dòng)能損失即
0.5mv^2-0.5(m+2m+...+nm)V^2
=0.5m(v^2-n(n+1)/2 V^2)
=0.5m(v^2-n(n+1)/2 *(2/(n(n+1)) v)^2)
=0.5mv^2(1-2/n(n+1))
n足夠大的時(shí)候貌似損失的不錯(cuò)呀.