1 B的质量越大获得动量越大,因为动量守恒MV1=MV1'+KMV2',能量守恒0.5MV1^2=0.5MV1'^2+0.5KMV2'^2,解得V2'=2V1/(K+1)得P2=KMV2'=2MV1K/(K+1)显然K越大,P2越大,故选A2 对椭圆AX^2+BY^2=1求导得2AX+2BY'Y=0解出Y'即得切线斜率,再带点得直线方程.或者用现成的结论对二次曲线AX^2+BY^2+CX+DY+EXY+F=0过曲线上一点(X0,Y0)的切线为AX0X+BY0Y+C(X+X0)/2+D(Y+Y0)/2+E(X0Y+Y0X)/2+F=0实在不行就老老实实用判别式求....
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