Question

如圖所示，勻強(qiáng)電場方向豎直向上，勻強(qiáng)磁場方向水平且垂直紙面向里，有兩個(gè)帶電小球a和b，a恰能在垂直于磁場方向的豎直平面內(nèi)做半徑r=0.8 m的勻速圓周運(yùn)動(dòng)，b恰能以v=2 m/s的水平速度在垂直于磁場方向的豎直平面內(nèi)向右做勻速直線運(yùn)動(dòng)．小球a、b質(zhì)量m_a=10 g，m_b=40 g，電荷量q_a=1×10^－2C，q_b= 2×10^－2C，g=10m/s²。求：

（1）小球a和b分別帶什么電？電場強(qiáng)度E與磁感應(yīng)強(qiáng)度B的大�。�

（2）小球a做勻速周周運(yùn)動(dòng)繞行方向是順時(shí)針還是逆時(shí)針？速度大小v_a是多大？

（3）設(shè)小球b的運(yùn)動(dòng)軌跡與小球a的運(yùn)動(dòng)軌跡的最低點(diǎn)相切，當(dāng)小球a運(yùn)動(dòng)到最低點(diǎn)即切點(diǎn)時(shí)小球b也同時(shí)運(yùn)動(dòng)到切點(diǎn)，a、b相碰后合為一體，設(shè)為c，在相碰結(jié)束的瞬間，c的加速度a_c=？

 

Answer 1

【答案】

（1）小球a在互相垂直的勻強(qiáng)電場和勻強(qiáng)磁場中做勻速圓周運(yùn)動(dòng)，電場力和重力的合力為零，電場力方向向上，所以小球a帶正電，且有

m_ag=q_aE··································································①

E=10 N / C································································②

小球b做勻速直線運(yùn)動(dòng)，合力為零，帶正電，且有

m_bg=q_bvB+q_bE·····························································③

B=5T·····································································④

（2）小球a做勻速圓周運(yùn)動(dòng)繞行方向是逆時(shí)針方向。

由

得······························································⑤

v_a=4 m / s·································································⑥

（3）a、b相碰前速度方向相同，設(shè)碰后的共同速度為v_c，則

m_a v_a+m_b v=(m_a+m_b) v_c······················································⑦

v_c=2.4 m / s································································⑧

由牛頓第二定律得

m_ca_c=q_cE+q_c v_cB – m_eg······················································⑨

m_c=m_a+m_b

q_c=(q_a+q_b)

解得a_c=3.2 m / s²··························································⑩

評(píng)分標(biāo)準(zhǔn)：本題滿分18分。其中判斷a、b帶電性質(zhì)各1分，a繞行方向1分，①式2分，②式1分，③式2分，④式1分，⑤式2分，⑥式1分，⑦式2分，⑧式1分，⑨式2分，⑩式1分。

【解析】略