英语翻译After falling from rest at a height of 29 m,a 0.48 kg ba
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英语翻译
After falling from rest at a height of 29 m,a 0.48 kg ball rebounds upward,reaching a height of 19 m.If the contact between ball and ground lasted 2.5 ms,what average force was exerted on the ball?
After falling from rest at a height of 29 m,a 0.48 kg ball rebounds upward,reaching a height of 19 m.If the contact between ball and ground lasted 2.5 ms,what average force was exerted on the ball?
一个质量为0.48kg的球从29米位置自由下落(初速为0),下落后与地面接触时间为2.5ms,然后反弹至高度为19米处.问:此球与地面接触时地面平均给球施加多大的压力.
首先根据能量守恒定律算出球从29米处下落的地面的速度v1,和球在地面反弹时的速度-v2.(设竖直向下为正方向);
即:1/2mv^2=mgh,
v=(2gh)^(1/2)
求得v1=(2*10*29)^(1/2)=24.1m/s;
v2=(2*10*19)^(1/2)=19.5m/s.(g取10m/s*s)
再有动量守恒定律知:
冲量I=ft=mv1-(-mv2)=m(v1+v2)=0.48*(24.1+19.5)=21.0
则平均压力f=[m(v1+v2)]/t=21.0/0.0025=8371.2N
首先根据能量守恒定律算出球从29米处下落的地面的速度v1,和球在地面反弹时的速度-v2.(设竖直向下为正方向);
即:1/2mv^2=mgh,
v=(2gh)^(1/2)
求得v1=(2*10*29)^(1/2)=24.1m/s;
v2=(2*10*19)^(1/2)=19.5m/s.(g取10m/s*s)
再有动量守恒定律知:
冲量I=ft=mv1-(-mv2)=m(v1+v2)=0.48*(24.1+19.5)=21.0
则平均压力f=[m(v1+v2)]/t=21.0/0.0025=8371.2N
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