Question

如圖所示,水平桌面離地面高h(yuǎn)=1.25 m.小物塊A靜止在桌面上,距桌右邊緣l=1.00 m,小物塊B從桌面的左邊緣向A運(yùn)動(dòng),并與之發(fā)生正碰(碰撞時(shí)間極短).碰后A從桌面的右邊緣以垂直邊緣的速度飛出,其落地點(diǎn)距桌面右邊緣的水平距離s=0.75 m,B剛好到達(dá)桌面右邊緣.A和B質(zhì)量均為m=0.1 kg,與桌面間的動(dòng)摩擦因數(shù)均為μ=0.2.重力加速度為g=10 m/s².試求:

(1)小物塊A從桌面邊緣飛出時(shí)的速度大小;

(2)小物塊B與A碰撞過(guò)程中B與A組成的系統(tǒng)損失的機(jī)械能.

Answer 1

解:(1)物塊A做平拋運(yùn)動(dòng)的時(shí)間t==0.5 s                                   ①

從桌面邊緣飛出時(shí)的速度v==1.5 m/s                                           ②

(2)碰撞前A的速度為v₀,碰撞后,A的速度為v₁,B的速度為v₂,

據(jù)動(dòng)能定理,有-μmgl=mv²-mv₁²                                              ③

-μmgl=0-mv₂²                                                               ④

代入數(shù)據(jù),解得v₁=2.5 m/s                                                        ⑤

v₂=2 m/s                                                                      ⑥

據(jù)動(dòng)量守恒定律,有mv₀=mv₁+mv₂                                                 ⑦

解得v₀=4.5 m/s                                                                 ⑧

碰撞損失的機(jī)械能E_損=mv₀²-mv₁²-mv₂²=0.5 J                               ⑨