一质量为5Kg的长木板，静止在光滑水平桌面上．一质量为2Kg的小滑块以水平速度V0=4m/S2从长木板的一端开始在木板上滑动，直到离开木板．滑块刚离开木板时的速度为2m/s．若将以木块固定在

Question

一质量为5Kg的长木板，静止在光滑水平桌面上．一质量为2Kg的小滑块以水平速度V₀=4m/S²从长木板的一端开始在木板上滑动，直到离开木板．滑块刚离开木板时的速度为2m/s．若将以木块固定在水平桌面上，其他条件相同，求滑块离开木板时的速度大小．₀²

Answer 1

设第一次滑块离开时木板速度为v，由系统的动量守恒，有
mv₀0=mv′+Mv
解得v=(m(v0−v′))/(M)＝(2×(4−2))/(5)＝0.8m/s
设滑块与木板间摩擦力为f，木板长L，滑行距离s，如图，

由动能定理对木板：fs=(1)/(2)Mv²　　　
对滑块：-f（s+L）=(1)/(2)mv′²-(1)/(2)mv02
即  fL=(1)/(2)mv02-(1)/(2)mv′²-(1)/(2)Mv²=(1)/(2)×2×16−(1)/(2)×2×4−(1)/(2)×5×0.64=10.4J
当板固定时　 fL=(1)/(2)mv02-(1)/(2)mv″²
解得：v″=√(((1)/(2)×2×16−10.4)/((1)/(2)×2))=2.37m/s
答：滑块离开木板时的速度大小2.37m/s (m(v0−v′))/(M)m(v0−v′)m(v0−v′)m(v0−v′)0−v′)MMM＝(2×(4−2))/(5)2×(4−2)2×(4−2)2×(4−2)555＝0.8m/s
设滑块与木板间摩擦力为f，木板长L，滑行距离s，如图，

由动能定理对木板：fs=(1)/(2)Mv²　　　
对滑块：-f（s+L）=(1)/(2)mv′²-(1)/(2)mv02
即  fL=(1)/(2)mv02-(1)/(2)mv′²-(1)/(2)Mv²=(1)/(2)×2×16−(1)/(2)×2×4−(1)/(2)×5×0.64=10.4J
当板固定时　 fL=(1)/(2)mv02-(1)/(2)mv″²
解得：v″=√(((1)/(2)×2×16−10.4)/((1)/(2)×2))=2.37m/s
答：滑块离开木板时的速度大小2.37m/s (1)/(2)111222Mv²2　　　
对滑块：-f（s+L）=(1)/(2)mv′²-(1)/(2)mv02
即  fL=(1)/(2)mv02-(1)/(2)mv′²-(1)/(2)Mv²=(1)/(2)×2×16−(1)/(2)×2×4−(1)/(2)×5×0.64=10.4J
当板固定时　 fL=(1)/(2)mv02-(1)/(2)mv″²
解得：v″=√(((1)/(2)×2×16−10.4)/((1)/(2)×2))=2.37m/s
答：滑块离开木板时的速度大小2.37m/s (1)/(2)111222mv′²2-(1)/(2)mv02
即  fL=(1)/(2)mv02-(1)/(2)mv′²-(1)/(2)Mv²=(1)/(2)×2×16−(1)/(2)×2×4−(1)/(2)×5×0.64=10.4J
当板固定时　 fL=(1)/(2)mv02-(1)/(2)mv″²
解得：v″=√(((1)/(2)×2×16−10.4)/((1)/(2)×2))=2.37m/s
答：滑块离开木板时的速度大小2.37m/s (1)/(2)111222mv02
即  fL=(1)/(2)mv02-(1)/(2)mv′²-(1)/(2)Mv²=(1)/(2)×2×16−(1)/(2)×2×4−(1)/(2)×5×0.64=10.4J
当板固定时　 fL=(1)/(2)mv02-(1)/(2)mv″²
解得：v″=√(((1)/(2)×2×16−10.4)/((1)/(2)×2))=2.37m/s
答：滑块离开木板时的速度大小2.37m/s 02
即  fL=(1)/(2)mv02-(1)/(2)mv′²-(1)/(2)Mv²=(1)/(2)×2×16−(1)/(2)×2×4−(1)/(2)×5×0.64=10.4J
当板固定时　 fL=(1)/(2)mv02-(1)/(2)mv″²
解得：v″=√(((1)/(2)×2×16−10.4)/((1)/(2)×2))=2.37m/s
答：滑块离开木板时的速度大小2.37m/s 2
即  fL=(1)/(2)mv02-(1)/(2)mv′²-(1)/(2)Mv²=(1)/(2)×2×16−(1)/(2)×2×4−(1)/(2)×5×0.64=10.4J
当板固定时　 fL=(1)/(2)mv02-(1)/(2)mv″²
解得：v″=√(((1)/(2)×2×16−10.4)/((1)/(2)×2))=2.37m/s
答：滑块离开木板时的速度大小2.37m/s (1)/(2)111222mv02-(1)/(2)mv′²-(1)/(2)Mv²=(1)/(2)×2×16−(1)/(2)×2×4−(1)/(2)×5×0.64=10.4J
当板固定时　 fL=(1)/(2)mv02-(1)/(2)mv″²
解得：v″=√(((1)/(2)×2×16−10.4)/((1)/(2)×2))=2.37m/s
答：滑块离开木板时的速度大小2.37m/s 02-(1)/(2)mv′²-(1)/(2)Mv²=(1)/(2)×2×16−(1)/(2)×2×4−(1)/(2)×5×0.64=10.4J
当板固定时　 fL=(1)/(2)mv02-(1)/(2)mv″²
解得：v″=√(((1)/(2)×2×16−10.4)/((1)/(2)×2))=2.37m/s
答：滑块离开木板时的速度大小2.37m/s 2-(1)/(2)mv′²-(1)/(2)Mv²=(1)/(2)×2×16−(1)/(2)×2×4−(1)/(2)×5×0.64=10.4J
当板固定时　 fL=(1)/(2)mv02-(1)/(2)mv″²
解得：v″=√(((1)/(2)×2×16−10.4)/((1)/(2)×2))=2.37m/s
答：滑块离开木板时的速度大小2.37m/s (1)/(2)111222mv′²2-(1)/(2)Mv²=(1)/(2)×2×16−(1)/(2)×2×4−(1)/(2)×5×0.64=10.4J
当板固定时　 fL=(1)/(2)mv02-(1)/(2)mv″²
解得：v″=√(((1)/(2)×2×16−10.4)/((1)/(2)×2))=2.37m/s
答：滑块离开木板时的速度大小2.37m/s (1)/(2)111222Mv²2=(1)/(2)×2×16−(1)/(2)×2×4−(1)/(2)×5×0.64=10.4J
当板固定时　 fL=(1)/(2)mv02-(1)/(2)mv″²
解得：v″=√(((1)/(2)×2×16−10.4)/((1)/(2)×2))=2.37m/s
答：滑块离开木板时的速度大小2.37m/s (1)/(2)111222×2×16−(1)/(2)111222×2×4−(1)/(2)111222×5×0.64=10.4J
当板固定时　 fL=(1)/(2)mv02-(1)/(2)mv″²
解得：v″=√(((1)/(2)×2×16−10.4)/((1)/(2)×2))=2.37m/s
答：滑块离开木板时的速度大小2.37m/s (1)/(2)111222mv02-(1)/(2)mv″²
解得：v″=√(((1)/(2)×2×16−10.4)/((1)/(2)×2))=2.37m/s
答：滑块离开木板时的速度大小2.37m/s v02-(1)/(2)mv″²
解得：v″=√(((1)/(2)×2×16−10.4)/((1)/(2)×2))=2.37m/s
答：滑块离开木板时的速度大小2.37m/s 02-(1)/(2)mv″²
解得：v″=√(((1)/(2)×2×16−10.4)/((1)/(2)×2))=2.37m/s
答：滑块离开木板时的速度大小2.37m/s 2-(1)/(2)mv″²
解得：v″=√(((1)/(2)×2×16−10.4)/((1)/(2)×2))=2.37m/s
答：滑块离开木板时的速度大小2.37m/s (1)/(2)111222mv″²2
解得：v″=√(((1)/(2)×2×16−10.4)/((1)/(2)×2))=2.37m/s
答：滑块离开木板时的速度大小2.37m/s √(((1)/(2)×2×16−10.4)/((1)/(2)×2))

((1)/(2)×2×16−10.4)/((1)/(2)×2)

((1)/(2)×2×16−10.4)/((1)/(2)×2)((1)/(2)×2×16−10.4)/((1)/(2)×2)(1)/(2)×2×16−10.4(1)/(2)×2×16−10.4(1)/(2)111222×2×16−10.4(1)/(2)×2(1)/(2)×2(1)/(2)111222×2=2.37m/s
答：滑块离开木板时的速度大小2.37m/s