Question

（13分）如圖所示，質(zhì)量不計(jì)且足夠長的倒L型支架下端固定在質(zhì)量為2m的木板上，在其上端O處系一長為L的輕繩，繩的下端系一質(zhì)量為m的小球，小球可視為質(zhì)點(diǎn)。整個(gè)裝置在光滑的水平面上以速度v₀向右做勻速直線運(yùn)動(dòng)，水平面的右端為一矮墻。（重力加速度為g）

（1）若木板與墻壁相碰立即與墻壁粘合在一起，試求碰后小球上升至最高點(diǎn)時(shí)繩中的張力大小。（小球在運(yùn)動(dòng)過程中不與支架相碰）；

（2）若木板與墻壁相碰后以原速率反彈，要使繩的最大偏角不超過90°，則生產(chǎn)L應(yīng)滿足什么條件？

Answer 1

(13分)解:（1）當(dāng)時(shí),小球右擺,且設(shè)球上升到最高點(diǎn)時(shí)

     繩與豎直方向的夾角為

     對(duì)小球向上擺動(dòng)的過程有:

                                                                                                           

     小球在最高點(diǎn)時(shí)有:T=                                                           

     解得張力: T                                                                    

     （2）當(dāng)時(shí),繩松弛后小球做

     斜上拋運(yùn)動(dòng),當(dāng)小球上拋至

     最高點(diǎn)時(shí)繩仍松弛,故張力T=0.                                                          

 （3）當(dāng)時(shí),小球能做完整的圓周運(yùn)動(dòng),

設(shè)球運(yùn)動(dòng)到最高點(diǎn)時(shí)速度為,

對(duì)小球從最低點(diǎn)到最高點(diǎn)的過程有:

                                                            

小球在最高點(diǎn)時(shí)有:

                                                                            

解得張力為:T=                                                                 

(2)若繩長為L₀時(shí),碰后繩的最大偏角為,繩的偏角最大

時(shí),球與木板的共同速度為

據(jù)動(dòng)量守恒得:                                                       

據(jù)機(jī)械能守恒得:                                        

解得:  L₀=                                                                                   

所以當(dāng)L滿足 時(shí),繩的偏角不超過90⁰