Question

如圖所示，勻強(qiáng)電場(chǎng)方向豎直向上，勻強(qiáng)磁場(chǎng)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(chǎng)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(chǎng)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(chǎng)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

（1）小球a在互相垂直的勻強(qiáng)電場(chǎng)和勻強(qiáng)磁場(chǎng)中做勻速圓周運(yùn)動(dòng)，電場(chǎng)力和重力的合力為零，電場(chǎng)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?lè)较颉?br />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。

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