求快招

[f50509123 10]

Pitcher and catcher on boats: a pitcher stands in a boat, throws a baseball of

mass m with speed v(relative to the pitcher)to the catcher on another boat; the

catcher sends the ball back with a slower speed of v/2(relative to the catcher)

to the pitcher. Assume the drag forcrs applying to the boats and the baseball

are negligibly small, and the mass of the picher together with his boat is the

same as the mass of the catcher and his boat(denoted by M).If they start this

playing from rest, what is the speed of the pitcher after he caught the ball

back? 9[15points]

簡單敘述

一個投手和一個捕手各站在一艘船上，投手將球以相對投手速度Ｖ投出，捕手接到後再以相對捕手速度Ｖ／２丟回，投手接到後，請問投手速度多少？小球質量ｍ，人＋船的質量Ｍ

請問這題有比較快的解法嗎?

我用了快一張A4紙 才把答案算出來

[如夢幻夜™ 10]

照定義去算還真不是普通的難算XD

大學很多題目都是這樣如此吃力不討好

好不容易算出一個可能是答案的答案

卻連自己有沒有算對都不敢肯定w

我不確定自己算的是不是對的，我只知道我算完之後變得好無力XDD

但還是把算法PO上來看看好了

因為動量守恆，已知拋出後投手與球相對速度為V

假設投出後投手速度為a，球速b

Ma+mb=0

b-a=V

a=-m/(M+m)V

b=M/(M+m)V

捕手接到球後，又動量守恆

mb=(m+M)v’

v‘=Mm/(M+m)^2 V(捕手船速)

從中計算得捕手投出球後之速度為

(M+m)v’=mx+My

y-x=V/2

x=v’-MV/2(M+m)

最後投手接到球，維持動量守恆

設末速V’

Ma+mx=(M+m)V’

V’=(m-3m^2-6Mm-3M^2)MmV/2(M+m)^4

[f50509123 10]

謝謝 不過我跟你的答案有點不一樣耶~~~~~

你V'的因次怪怪的 ~~~~

[[email protected] 10]

答案是多少??

[=冰河= 16]

照定義去算還真不是普通的難算XD

大學很多題目都是這樣如此吃力不討好

好不容易算出一個可能是答案的答案

卻連自己有沒有算對都不敢肯定w

我不確定自己算的是不是對的，我只知道我算完之後變得好無力XDD

但還是把算法PO上來看看好了

因為動量守恆，已知拋出後投手與球相對速度為V

假設投出後投手速度為a，球速b

Ma+mb=0

b-a=V

a=-m/(M+m)V

b=M/(M+m)V

捕手接到球後，又動量守恆

mb=(m+M)v’

v‘=Mm/(M+m)^2 V(捕手船速)

從中計算得捕手投出球後之速度為

(M+m)v’=mx+My

y-x=V/2

x=v’-MV/2(M+m)

最後投手接到球，維持動量守恆

設末速V’

Ma+mx=(M+m)V’

V’=(m-3m^2-6Mm-3M^2)MmV/2(M+m)^4

這行我有疑惑XDDD

可否請TM大大解釋

[如夢幻夜™ 10]

這行是由

(M+m)v’=mx+My

y-x=V/2

求得的x

x代表捕手投出球後之球速

雖然最後算錯了但我到現在還沒找到一個空閒的時間把正解求出來(汗