[題目] 高中分子動力論
題目
容器A面能使1/3的入射分子黏住 其餘2/3反跳
然後B面能使1/4的入射分子穿過 1/4黏住 剩下反跳
求AB面壓力比
想法一:
P(壓力)和△p(動量變化量=N(m△V))成正比
A =>PA(壓力)=△p1+△p2(動量變化量)
|<--V
| (1/3N) =>PA=△p1+△p2= 1/3Nm(V-0)+2/3Nm(V-(-V))=5/3NmV
|-->0
|
|<--V
| (2/3N)
|-->V
B =>PB(壓力)=△p1+△p2(動量變化量)
|<--V (1/4N)
| =>PB=△p1+△p2= 1/4Nm(V-V)+1/4Nm(V-0)+1/2Nm(V-(-V))
|<--V (1/4N) = 5/4NmV
|-->0
|
|<--V (1/2N)
|-->V PA/PB = 4/3.....END
想法二:
P= F/A= (N*△p) / (△t*A)
其中△t= 兩次碰撞的長度所走的長/速度
PA= N*(p'-p) / (2l/v')*A (因動量守恆 mv=(2/3m)*(v') )
= N* 2mv * v' / 2l *A =>v'= 2/3v
= (3/2) *N*m*(v^2)/ l*A
PB= N*(p'-p) / (2l/v')*A (因1/4穿過所以後來質量是為原本的3/4)
= N*(2mv)*v' / 2l*A (因動量守恆 (3/4m)*v = (1/2m)*(v') )
= (3/2)(3/4)*N*m*(v^2)/ l*A =>v'=2/3v
PA/PB = 4/3.....END
兩個想法有很大的衝突....
一個是說動量守恆 一個沒有動量守恆
可是答案一樣.....
請問哪個想法比較正確?
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