A.mv02 B.
mv02 C.
mv02 D.
mv02
A.mv02 B.
mv02 C.
mv02 D.
mv02
解析:碰撞过程中动量守恒,所以mv0=mv1+3mv2,又mv12=(1-75%)×
mv02,解得v1=±0.5v0.代入动量守恒方程求得v2=
v0(舍去)或v2=0.5v0.求得B的动能为EkB=
mv02.
答案:D