动能定理和动量守恒方程联立后怎么解?

2024-11-01 09:55:44
推荐回答(2个)
回答1:

mv0=mv1+5mv2 (1)
(1/2)m(v0)^2=1/2m(v1)^2+ (1/2) 5m(v2)^2 (2)
Sub (2) into (1)
(v1+5v2)^2 = (v1)^2 + 5(v2)^2
10v1v2 + 25(v2)^2 = 5(v2)^2
v2(2v2+v1)=0
v2 = 0
or v2 = -(1/2)v1
when v2=0
from (1)
v0= v1
when v2= -(1/2)v1
v0=v1-5/2v1
v1= -(2/3)v0

(v1,v2) = ( v0,0) or (-(2/3)v0, (2/3)v0)

回答2:

好像缺个条件·····