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[题目]一股射流以10m/s的速度从喷嘴竖直向上喷出,喷嘴截面积为0.5cm~2.有一质量为0.32kg的球,因水对其下侧的冲击而悬在空中.若水冲击球后速度为零,则小球悬在高喷嘴多高处?(g=10m/s~2).原解:设在△t时间内有质量为△m的水与球发生相互作用,相互作用力大小为F.那么,对球有F=Mg,对△m的水应用动量定理,有F△t=△mv.这里v为水对球下侧冲击前的速度,又△m=ρSv△t,从而v=(Mg/ρS)~(1/2).根据v_0~2-v~2=2gh,可得小球停处离喷嘴的高度h=(v_0~2)/(2g)-M/(2ρS)=1.8(m).(见本刊编辑部编《高三物理教学研究》1997年版第13页例4)
[Topic] A jet is ejected vertically upward from the nozzle at a speed of 10m/s. The cross-sectional area of the nozzle is 0.5cm~ 2. A ball with a mass of 0.32kg is suspended in the air due to the impact of the water on its underside. If the velocity of the water after the ball hits the ball is zero, then the ball is suspended at a high nozzle height (g=10m/s~2). Original solution: The water with mass △m interacts with the ball in △t time. , The interaction force is F. Then, there is F=Mg for the ball, and Momentum Theorem for the water of △m, there is FΔt=Δmv. Here, v is the velocity of the water before the impact on the lower side of the ball, and △m =ρSvΔt, so that v=(Mg/ρS)~(1/2). According to v_0~2-v~2=2gh, the height of the ball stop from the nozzle can be obtained h=(v_0~2)/( 2g)-M/(2ρS) = 1.8(m). (See “Journal of Physics Teaching Research for Seniors”, 1997 edition, page 13, page 4)