(1)当二者速度相等时,系统的电势能最大
由动量守恒得
4mv0-mv0=5mv v=0.6v0
由能量守恒得
0.5*mv0^2+0.5*4mv0^2=0.5*5mv^2+EP
EP=1.6MV0^2
(2)4mv0-mv0=m1v1+4mv2
0.5*mv0^2+0.5*4mv0^2=0.5*mv1^2+0.5*4mv2^2
v1=v2=v0
(1)当二者速度相等时,系统的电势能最大
由动量守恒得
4mv0-mv0=5mv v=0.6v0
由能量守恒得
0.5*mv0^2+0.5*4mv0^2=0.5*5mv^2+EP
EP=1.6MV0^2
(2)4mv0-mv0=m1v1+4mv2
0.5*mv0^2+0.5*4mv0^2=0.5*mv1^2+0.5*4mv2^2
v1=v2=v0