Question

如圖三個大小相同的小球A、B、C置于光滑水平面上，三球的質(zhì)量分別為m_A=2kg、m_B=4kg、m_C=2kg，取水平向右方向為動量的正方向，某時刻A球的動量P_A=20kgm/s，B球此刻的動量大小和方向未知，C球的動量為零。A球與B球先碰，隨后B球與C球碰，碰撞均在同一直線上，且A球與B球以及B球與C球之間分別只相互碰撞一次，最終所有小球都以各自碰后的速度一直勻速運動。所有的相互作用結(jié)束后，ΔP_C=10kgm/s、ΔP_B=4kgm/s，最終B球以5m/s的速度水平向右運動。求：

   Ⅰ．A球?qū)?i>B球的沖量大小與C球?qū)?i>B球的沖量大小之比；

   Ⅱ．整個過程系統(tǒng)由于碰撞產(chǎn)生多少熱量？

Answer 1

解：Ⅰ.由A、B、C組成系統(tǒng)動量守恒ΔP_A+ΔP_B+ΔP_C=0··································· （1分）

 解得：ΔP_A=－14kgm/s·········································································· （1分）

 由A、B相碰時對A用動量定理可得：I_BA=ΔP_A  I_AB=－I_BA=14kgm/s········ （1分）

 由B、C相碰時對C用動量定理可得：I_BC=ΔP_C  I_CB=－I_BC=－10kgm/s····· （1分）

 則I_AB：I_CB=7:5（1分）

Ⅱ.設(shè)A、B碰前A的動量為P_A，B的動量為P_B，C的動量為P_C，所有的作用結(jié)束后A的動量為P_A^/，B的動量為P_A^/，B的動量為P_C^/，由A、B、C組成系統(tǒng)動量守恒得：

P_A+P_B+P_C= P_A^/+ P_B^/+ P_C^/········································································ （1分）

P_A^/₌ P_A+ΔP_A）························································································ （1分）

P_C^/₌ P_C+ΔP_C）························································································ （1分）

······································································ （1分）

聯(lián)立解得：Q=48J··················································································· （1分）