Question

如图三个大小相同的小球A、B、C置于光滑水平面上，三球的质量分别为m_A=2kg、m_B=4kg、m_C=2kg，取水平向右方向为动量的正方向，某时刻A球的动量P_A=20kgm/s，B球此刻的动量大小和方向未知，C球的动量为零。A球与B球先碰，随后B球与C球碰，碰撞均在同一直线上，且A球与B球以及B球与C球之间分别只相互碰撞一次，最终所有小球都以各自碰后的速度一直匀速运动。所有的相互作用结束后，ΔP_C=10kgm/s、ΔP_B=4kgm/s，最终B球以5m/s的速度水平向右运动。求：

   Ⅰ．A球对B球的冲量大小与C球对B球的冲量大小之比；

   Ⅱ．整个过程系统由于碰撞产生多少热量？

Answer 1

解：Ⅰ.由A、B、C组成系统动量守恒ΔP_A+ΔP_B+ΔP_C=0··································· （1分）

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

 由A、B相碰时对A用动量定理可得：I_BA=ΔP_A  I_AB=－I_BA=14kgm/s········ （1分）

 由B、C相碰时对C用动量定理可得：I_BC=ΔP_C  I_CB=－I_BC=－10kgm/s····· （1分）

 则I_AB：I_CB=7:5（1分）

Ⅱ.设A、B碰前A的动量为P_A，B的动量为P_B，C的动量为P_C，所有的作用结束后A的动量为P_A^/，B的动量为P_A^/，B的动量为P_C^/，由A、B、C组成系统动量守恒得：

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）

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