a-t graph

2007-04-07 7:21 pm
A metal sphere is released from a height of 0.45m onto a marble floor .It rebounds to height of 0.2m

Sketch a graph showing the variation of the acceleration of the sphere with time
(take accleration in the downward direction to be postive)

ans:http://img211.imageshack.us/img211/8066/55506829vi3.jpg

可唔可以解釋一下幅圖
下面(-ve)果個acceleration magnitude 係咪要大過10ms-2?
點解係-ve?唔係一直+ve?

回答 (3)

2007-04-07 9:58 pm
✔ 最佳答案
回答
因為地心吸力的關係﹐除了在物體與地面碰撞外﹐基本上物體的加速度就是重力加速度g﹐因為向下為正﹐所以數值是10 ms^-2
現在求物體在碰撞地面前的速度
由s=ut+1/2gt^2
0.45=5t^2
t=0.3 s
v=u+at=3 m/s
物體在碰撞地面前的速度是3 m/s
接著求物體在碰撞地面後的速度
由2as=v^2-u^2
2(-10)(0.2)=0-u^2
u=-2 m/s (因為初速向上)
物體在碰撞地面後的速度是2 m/s
所以物體在與地面碰撞時的加速度
=[-2-3]/t
=(-5/t) ms^-2
所以加速度是負值﹐是否大過10ms-2要看碰撞時間而定。

不過碰撞時間大部分都小於0.5 s
所以應該大於10ms-2 吧
2007-04-09 9:09 pm
在第 1 part 中,Take g = 10,for consistency,第 2 part 的 g take 10 較 −10 好,這不是吹毛求庛,而是做物理計算時應注意的地方。BTW,第 2 part 中的 s 是 displacement 反而應是負數,是因為取向下的方向為正。
2007-04-09 8:12 am
The force acting on the sphere is mg (where m is its mass and g the acceleration due to gravity), thus the accleration is g and its direction downward (+ve).
When the sphere strikes the floor, the direction of force F acting on it must be upward (i.e. -ve), and hence the acceleration is also -ve (since F = ma)
The magnitude of force F must be greater than mg, otherwise the sphere wouldn't move up (rebound). Hence (-F)+mg must be negative. (F upward, thus -ve; g downward, thus +ve).



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