1.5 m/s.

评分标准：①②③式各2分，④⑤⑥⑦式各1分.

物块滑到小车上时的初速度为v₀，则mgh=mv²₀，v₀="5" m/s                   ①

设物块与挡板相碰，则碰后瞬间小车速度最大，此时物块、小车速度分别为v₁、v₂则

mv₀=mv₁+Mv₂                                                                ②

μmgL=mv²₀-mv₁²-Mv₂²                                                 ③

所以2v₂²-5v₂+3=0

所以v₂="1" m/s或v₂="1.5" m/s                                                     ④

讨论：①当v₂="1" m/s时，v₁="2" m/s>v₂，此为碰前瞬间的速度                        ⑤

②当v₂="1.5" m/s时，v₁="0.5" m/s<v₂，此为碰后瞬间的速度                           ⑥

即小车的最大速度为1.5 m/s                                                    ⑦