Question

甲、乙兩球在光滑水平軌道上同向運(yùn)動(dòng),已知它們的動(dòng)量分別是p_甲=5 kg·m/s,p_乙=7 kg·m/s,甲追乙并發(fā)生碰撞,碰后乙球的動(dòng)量變?yōu)閜_乙′=10 kg·m/s.則兩球質(zhì)量m_甲與m_乙的關(guān)系可能是(    )

A.m_甲=m_乙           B.m_乙=2m_甲          C.m_乙=4m_甲          D.m_乙=6m_甲

Answer 1

C

解析:

由碰撞中動(dòng)量守恒可求得p_甲′=2 kg·m/s,要使甲追上乙,則必有:

v_甲>v_乙,即>                                                           ①

碰后p_甲′、p_乙′均大于零,表示同向運(yùn)動(dòng),則應(yīng)有:v_乙′≥v_甲′,

即≤                                                                ②

由于碰撞過程中,動(dòng)能不會(huì)增加,則

+≥+                                                  ③

代入數(shù)據(jù),由①②③式知,m_甲與m_乙的關(guān)系為≤m_乙≤5m_甲,可見選項(xiàng)C正確.