physics position and motion

point a :。

point b :。

point c :。

point a至point b 係14.5cm point b至point c係24.1cm

用strobeoscope紀錄

個ball falling freely from rest

the flashing light of the stroscobe flashes at a fixed frequency of 10Hz

問個acceleration due to gravity 係??

ans:9.6ms^-2

點解唔可以用s=ut+1/2 at^2

14.5/100=0*0.1+1/2 a*0.1^2

a=29ms^2

用s=ut+1/2 at^2,揾唔到答案

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如果個a=9.6ms^-2,用s=ut+1/2 at^2

=0.*0.1+1/2 *9.6*0.1^2

=0.048m-->唔係14.5cm 點解???

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如果個a=9.6ms^-2,用v=u+at

=0+9.6*0.1

=0.96ms^1

但係個ans話係point a個v係1.45ms^1 點解???

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2 個解答

評分
  • 7 年前
    最愛解答

    你不可以這樣計,因為你無法得知個Ball是在何時何地開始Fall freely from rest,也無法得知個Ball在Point A的速度

    但你在Sub數時,已假設個Ball是在Point A fall freely from rest,也就是事先假設個Ball在Point A的速度是零

    不過題目沒有指明這個條件,而且驗算時這個速度也不是零,因此你的方法會出錯

    正確的方法是用兩組s與t(A至B 與 A至C),此時兩組的a與u都相同(因此不能用B至C),可以聯立兩組 s=ut+1/2*at^2 來求得u與a:

    0.145 = u*0.1 + 1/2*a*0.1^2

    0.145+0.241 = u*0.2 + 1/2*a*0.2^2

    即 0.145 = 0.1u + 0.005a -(1)

    0.386 = 0.2u + 0.02a -(2)

    (2)-2*(1), 0.386 - 0.145*2 = (0.2u + 0.02a) - 2(0.1u + 0.005a)

    0.096 = 0.2u + 0.02a - 0.2u - 0.01a = 0.01a

    得出正解a = 9.6 m/s^2

    2013-06-16 11:15:10 補充:

    以上即acceleration due to gravity = 9.6 m/s^2

    4*(1)-(2), 0.145*4-0.386 = 4(0.1u+0.005a) - (0.2u+0.02a)

    0.194 = 0.4u+0.02a-0.2u-0.02a = 0.2u

    u = 0.97 m/s

    Ball在Point A的速度是0.97 m/s(向下)

    2013-06-16 11:20:07 補充:

    若不滿意解答或打算移除問題,請通知我

  • 7 年前

    Stage 1 (From Point A to Point B)

    By s = ut + (1/2)at^2

    0.145 = u(1/10) + (1/2)a(1/10)^2

    14.5 = 10u + (1/2)a

    29 = 20u + a..........(1)

    Stage 2 (Whole Motion, from Point A to Point C)

    Again, by s = ut+(1/2)at^2

    (0.145+0.241) = u[(1/10)(2)] + (1/2)(a)[(1/10)(2)]^2

    0.386 = u/5 + a/50

    19.3 = 10u + a.............(2)

    (1) - (2) :

    29 - 19.3 = 10u

    u = 0.97ms^-1

    Sub u =0.97 to (1)

    a = 9.6ms^-2//

    2013-06-17 22:23:47 補充:

    As above calculation has stated, the initial speed of the ball (at Point A) IS NOT 0ms^-1. So you cannot find the correct answer.

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