质量为m木的小物块以水平速度v0滑上原来静止在光滑水平面上质量为m木的小车上,

(1)整体分析,摩擦力为内力抵消,所以动量守恒
mV0=(m+M)v
v=mV0/(M+m)
(2)动量定理
因为一直是受恒定摩擦力umg
ft=mΔv
umg*t=m*(v0-v)
t=MV0/[ug(M+m)]
(3)热量=能量损失
=(1/2)mV0^2-(1/2)mv^2-(1/2)Mv^2
=(1/2)mV0^2-(1/2)(M+m)[mV0/(M+m)]^2
=(1/2)MmV0^2/(M+m)
小车滑行距离,小车一直加速,加速度为a=umg/M
时间为t=MV0/[ug(M+m)]
s=(1/2)at^2=(1/2)umg/M[MV0/[ug(M+m)]]^2
=(1/2)MmV0^2/[ug(M+m)^2]