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[求助]大家来帮偶看看这道物理题

iou1985@2005-01-20 07:13

You have a summer job as an assistant in a research group that is designing a device to sample atmospheric pollution. In this device, it is useful to separate fast moving ions from slow moving ones. To do this, the ions are brought into the device in a narrow beam so that they are all going in the same direction. The ion beam then passes between two parallel metal plates. Each plate is 5.0 cm long, 4.0 cm wide, and the two plates are separated by 3.0 cm. A high voltage is applied to the plates, causing the ions passing between them to have a constant acceleration directly toward one of the plates and away from the other plate. Before the ions enter the gap between the plates, they are moving directly toward the center of the gap, parallel to the surface of the plates. After the ions leave the gap between the plates, they are no longer accelerated during the 70 cm journey to the ion detector. Calculate the magnitude of acceleration between the plates necessary to separate ions with a velocity of 200 m/s from those in the beam going 1100 m/s by 2 cm.

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seiyafan@2005-01-20 07:59

This situation only happens when the fast ions and slow ions have opposite charge, because they all enter at the middle of the gap, so there is 1.5cm space to the surface of the plate, and when they come out, their spacing is 2cm, so one of the ion is drawn to the opposite plate. That is very strange to me. I somehow think the problem is wrong, unless someone corrects me.

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JanusWang@2005-01-20 08:18

我觉得题目没有问题啊。画图就很明白了。具体公式我不记得了，太多年了，但是原理还记得的。首先，水平速度不会改变，根据牛顿定理，没有F，不会有a。其次，粒子的垂直偏转应该是一样的。我记得已知电荷数E，还有U，是可以求出F的，在这个力的作用下，离子向两块板的方向做运动。因为所有离子的质量相同，根据F=ma，它们的加速度也是相同的，因为初速度也相同，根据S=Vot+1/2VoT^2，垂直移动相同的距离所花的时间也相同。这样会出现慢的离子在还没有出两块板的区间就已经打在两块板上了，也就是不会走出这个区间。这道题还不够复杂，不用你算两块板上加的电压，也没有告诉你离子的质量。只需要算一下加速度a就可以。不难得。

反正做这一类的题目就是记住速度的独立性。速度是矢量，可以分解的，某一个方向不受力，大小方向就不会改变，比如平抛就是水平速度不变，只有竖直方向有g这个了重力加速度。

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iou1985@2005-01-20 08:37

慢的离子不走出那两块板？

不是两种粒子都走出那两块板，走出了之后再走70CM，两种离子的距离是2CM吗？

就像这个图那样。。。

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JanusWang@2005-01-20 09:08

如果要求所有的粒子都能出板的话，a小一点就是了。当时因为快离子行动的快，水平S=vt，v大，t小，也就是它在板子里的时间小，竖直方向上根据S=Vot+1/2at^2，他的竖直偏移就比慢的小。图上绿色的慢，蓝色的快

假设200m/s的离子为Ia，速度为1100m/s的离子为Ib，所有的下标都用ab表示。设板长度为L，一半间隔D，半离地d。设加速度为a，偏移量为S。加速以后的速度为Vout;
根据速度的独立性，我可以得到：
Ta = L/Va; Tb = L/Vb;
Sa = 1/2aTa^2; Sb = 1/2aTb^2;
Vouta = aTa; Voutb = aTb;

出板以后，水平竖直方向都作匀速运动
a打到地的时间为：Touta = (D+d-Sa)/Vaouta
b打到地的时间为：Toutb = (D+d-Sb)/Vaoutb

然后我们可以得到等式
VbTaoutb - VaTouta = 0.02

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兰瑟@2005-01-21 21:57

这种题也放到国外的大学？？

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1111111@2005-01-22 11:31

Plate: L = 5.0cm, W = 4.0cm, d = 3.0cm
acceleration = constant between plates
vx1 = 200m/s, vx2=1100m/s
x1=x2=70cm, y2-y1=2cm

For slow to travel 70cm, t1 = 70[cm] / 200[m/s] = 0.0035 s = 3.5ms
For fast to travel 70cm, t2 = 70[cm] / 1100[m/s]= 7/11 ms

Time b/w plates:
t1p = 5[cm] / 200[m/s] = 0.25 ms
t2p = 5[cm]/ 1100[m/s] = 5/110 ms

(a*t2p)*t2 - (a*t1p)*t1 = 20 mm
a(t2p*t2 - t1p*t1) = 20 mm
a( 5/110 *7/11 - 0.25*3.5) [us^2]= 20 mm
a= 20 /( 35/1210 - 0.875) [km/s^2]
a = -8699/368 [km/s^2] = -23.6386 km/s^2

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JanusWang@2005-01-22 15:54

楼上，我觉得你的解是不对的

力是加在竖直方向上的，水平方向是没有加速度的， 20cm是水平方向的位移差，所以(a*t2p)*t2 - (a*t1p)*t1 = 20 mm显然不成立，水平方向根本没有a的存在。

而且图上很明显，并不是所有的离子都可以水平运行70cm的。他们的x方向的位移是不同的，因为竖直方向的速度使得一些离子提前打到了板上。

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iou1985@2005-01-23 06:32

其实图画错了，离子不是掉到水平平面上的，而是射到垂直平面上的，在垂直平面上相差2CM

不好意思。。。。。。。

不过总算算到了，谢谢大家帮忙

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1111111@2005-01-23 07:03

I have (a*t2p)*t2 - (a*t1p)*t1 = 20 mm because i used another picture, like iou1985 said at the last post.

I want to see the solution though to see if I am correct or not.

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iou1985@2005-01-23 07:06

The answer is 2.28 X 10^4 m/s^2

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1111111@2005-01-23 14:08

sigh, i was close in number but not right, guess I did wrong as well

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iou1985@2005-01-24 03:07

偶是这样算的

Y1 (slow ion) = (1/2)*a*(0.05/200)^2 + a*(0.05/200)*(0.7/200)

Y2(fast ion) = (1/2)*a*(0.05/1100)^2 + a*(0.05/1100)*(0.7/1100)

Y1-Y2=0.02

然后把a求出来。。。。。。。。。

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water1@2005-01-24 03:51

i hate physics!!!!!!!!!
what the hell are those formulas!!!!!
those are not for highschool is it

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iou1985@2005-01-24 04:24

I hate physics too..........

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