动能定理和动量守恒方程联立后怎么解?
例如这两个方程
mv0=mv1+5mv2
1/2m(v0)2=1/2m(v1)2+1/2*5m(v2)2
第二个方程是这样的
1/2m(v0)平方=1/2m(v1)平方+1/2*5m(v2)平方
人气:246 ℃ 时间:2020-05-04 09:37:37
解答
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)
推荐
猜你喜欢
- The farmer soon came back to get his shoes,but when he put his foot into one of his shoes and felt something hard,he fou
- 求y=lnx在点M(e,1)的切线方程和法线方程.
- It feel like three
- 骨骼不含钙有没有无机盐
- 硫酸铵和硫酸铁铵的ph值大小
- A.refresh B.renew C.stimulate D.encourage
- 如图所示,在△ABC中,AB=AC,点MN分别在BC所在的直线上,且AM=AN,BM与CN相等吗?两种解答方法.
- 巳知电压为380v功率45千瓦,用多大的空气开关,线的大少,是怎么算的!请高手多指教谢谢了.