sorry but it's very inconvenient for me to type Chinese on the bbs.. So I choose to type in English. ※ 引述《zi98btcc (幼斤)》之銘言： : : : 上例稍改一下就類似火箭排氣推進，把5m/s拿掉，4m/s當作排氣的速度， : : : 請問要如何用f=dm/dt*v+m*dv/dt來解(求f)？ : : : |-------------| : : : | v=3m/s --> | : : : f ---> | a=1m/s/s--> | 2kg/s : : : | m=6kg | -----------> : : : | | 4m/s : : : |-------------| : : : _________________________________________________ : 那是因為f=(2kg/s)*(4m/s-3m/s)+(6kg)*(1m/s/s)這樣算 : 是根據系統原則：系統總外力=(系統總動量)'=(m1v1)'+(m2v2)'+(m3v3)'+... : : 上述火箭這題，系統分2部分，第1部分為排氣，第2部分就是火箭本身， : 系統總外力就是f， : 所以得到 f=排氣動量變化率+火箭動量變化率 : 因為 排氣動量變化率=(2kg/s)*(4m/s-3m/s) : 火箭動量變化率=(6kg)*(1m/s/s) : f=(2kg/s)*(4m/s-3m/s)+(6kg)*(1m/s/s) : 再次強調不是根據 f=dm/dt*v+m*dv/dt To me, your approach is the same as f=dm/dt*v+m*dv/dt : : 牛頓第二運動定律最多寫成 f=P'=(mv)'=ma : : 不宜再寫成 f=dm/dt*v+m*dv/dt. I don't see the difference... Alright...I know your point...then let me follow your logic and use what you think is qualified to be f=dm/dt*v+m*dv/dt to solve this problem. Let the system be composed of the rocket and the exhaust. M : mass of the rocket ( including the fuel which has not be burned and ejected as exhaust ) v : velocity of the rocket ( including........) m : mass of the exhaust u : velocity of the exhaust f = M dv/dt + v dM/dt + m du/dt+ u dm/dt M = 6 (kg) dv/dt = 1 (m/s^2) du/dt = 0. ( The exhaust is not accelerated ) v = 3 (m/s) u = 4 (m/s) dm/dt = 2 (kg/s) dM/dt = -dm/dt = -2 (kg/s) ( conservation of mass ) => Answer.... --

※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 128.227.146.228 ※ 編輯: chungweitw 來自: 128.227.146.228 (03/23 04:02)