Question

如圖所示，質(zhì)量M="3.5" kg的小車靜止于光滑水平面上靠近桌子處，其上表面與水平桌面相平，小車長L="1.2" m，其左端放有一質(zhì)量為m₂="0.5" kg的滑塊Q。水平放置的輕彈簧左端固定，質(zhì)量為m₁="1" kg的小物塊P置于桌面上的A點(diǎn)并與彈簧的右端接觸。此時(shí)彈簧處于原長，現(xiàn)用水平向左的推力將P緩慢推至B點(diǎn)(彈簧仍在彈性限度內(nèi))時(shí)，推力做的功為W_F，撤去推力后，P沿桌面滑動(dòng)到達(dá)C點(diǎn)時(shí)的速度為2 m/s，并與小車上的Q相碰，最后Q停在小車的右端，P停在距小車左端S="0.5" m處。已知AB間距L₁="5" cm，A點(diǎn)離桌子邊沿C點(diǎn)距離L₂="90" cm，P與桌面間動(dòng)摩擦因數(shù)μ₁=0.4，P、Q與小車表面間動(dòng)摩擦因數(shù)μ₂=0.1。(g="10" m/s。)求：
(1)推力做的功WF；
(2)P與Q碰撞后瞬間Q的速度大小和小車最后速度v.

Answer 1

(1) W_F="6J                                                                        "
(2) 0.4m/s

(1)對P由A→B→C應(yīng)用動(dòng)能定理，得
                                                                
解得                                                                                            
(2)設(shè)P、Q碰后速度分別為v₁、v₂，小車最后速度為v，由動(dòng)量守恒定律得
m₁vc=m₁v₁+-m₂v₂                                                                                 
m₁v_c=(m₁+m₂+M)v                                                                                
由能量守恒得
                       
解得，v₂=2m/s
v₂′=
v=0.4m/s                                                                                                 
　當(dāng)v2′=時(shí)，v1=>v₂′不合題意，舍去。                            
即P與Q碰撞后瞬間Q的速度大小為v₂=2m/s
小車最后速度為0.4m/s