Question

如圖，一個(gè)質(zhì)量為0.6kg                                        的小球以某一初速度從P點(diǎn)水平拋出，恰好從光滑圓弧ABC的A點(diǎn)的切線方向進(jìn)入圓�。ú挥�(jì)空氣阻力，進(jìn)入圓弧時(shí)無機(jī)械能損失）．已知圓弧的半徑R=0.3m，θ=60°，小球到達(dá)A點(diǎn)時(shí)的速度                                                                             v=4m/s．取g=10m/s²，求：                          

（1）小球做平拋運(yùn)動(dòng)的初速度v₀；                                                                                 

（2）P點(diǎn)與A點(diǎn)的高度差；                                                                                            

（3）小球到達(dá)圓弧最高點(diǎn)C時(shí)對(duì)軌道的壓力．                                                                

                                                                                                 

                                                                                                                                        

Answer 1

（1）小球恰好從光滑圓弧ABC的A點(diǎn)的切線方向進(jìn)入圓弧，則小球到A點(diǎn)的速度與水平方向的夾角為θ，所以：

  v₀=v_x=v_Acosθ=4×0.5m/s=2m/s

（2）v_y=v_Asinθ=4×m/s=2m/s

由平拋運(yùn)動(dòng)的規(guī)律得：v_y²=2gh

帶入數(shù)據(jù)，解得：h=0.6m．

（3）從A到C的運(yùn)動(dòng)過程中，運(yùn)用動(dòng)能定理得：

﹣=﹣mgR（1+cosθ）

帶入數(shù)據(jù)解之得：v_C=m/s．

由圓周運(yùn)動(dòng)向心力公式得：N_C+mg=m

代入數(shù)據(jù)解之得：N_C=8N

由牛頓第三定律，得：小球?qū)�軌道的壓力大�?N，方向豎直向上．

答：（1）小球做平拋運(yùn)動(dòng)的初速度v₀為2m/s；（2）P點(diǎn)與A點(diǎn)的高度差為0.6m；（3）小球到達(dá)圓弧最高點(diǎn)C時(shí)對(duì)軌道的壓力大小為8N，方向豎直向上．