physics 疑 难 区

jernie

这 些 应 该 怎 么 做 ??

1. In which group below do all three quantities remain constant when a particle moves in simple harmonic motion (shm)
a. accelaration , force , total energy
b. total energy , amplitude,angular frequency
c.force,total energy, amplitude
d.amplitude,angular frequency, acceleration

为 什 么 答 案 会 是 b?要 怎 么 样 判 断 的 ??

2.A body,initialy at rest,explode into two pieces of mass 2M and 3M respectively having a total kinetic energy E. The kinetic energy of the piece of the piece of mass 2M after the explosion is ___.
a. E/3         b. E/5        c.2E/5         d. 3E/5

ans - d


要 怎 么 算 ?


3.A cylinder of mass 50kg and radius 0.20m has moment of inertia of 1.0 kgm^2 about its axis.When it rolls without slipping on a horizontal plane with linear velocity 2.0 ms^-1,its kinetic energy is ____.

ans -150J

4.A projevtile of mass 2 kg is fired at 60 degree above the horizontal with a speed of 20 ms^-1.What is the minimum momentum value of the projectile during its flight?

ans - 20Ns

5.A rocket is accelerated from rest for 5s at 100ms^-2.It then travels at a constant speedfor 5s.The distance travelled in the first 7s is _____, and its maximum speed is _____.

ans- 2250m , 500 ms^-1

6.A particle performs simple harmonic motion according to equation x=3cos wt where x is measured in cm,and t in s.If the angular frequency w is 3.14rads^-1,what is the total distance travelled by the particle at t=2.5s?
ans- 15cm

7.The force constant of a spring is k.When an object of mass m is suspended at its end,its period of oscillation is T.If the spring is cut into half and connected in parallel and its lower end of the system is still suspended with a total mass m,what will its period of oscillation be?

ans - 1/2T

dunwan2tellu

1. In which group below do all three quantities remain constant when a particle moves in simple harmonic motion (shm)
a. accelaration , force , total energy
b. total energy , amplitude,angular frequency
c.force,total energy, amplitude
d.amplitude,angular frequency, acceleration

在 SHM 里， acceleration 和 displacement 成正比，所以 acceleration 不是 constant . 如果acceleration 不 constant , 那么 force 也不会。淘汰法，a,c,d 都错。剩下 b . b 对因为 total energy = potential + kinetic 一定不变（如果没有 energy lost的话）。Amplitude，angular frequency 当然不变。
因为 在 SHM ; displacement , x = A sin wt (A=amplitude,w=angular frequency)

2.A body,initialy at rest,explode into two pieces of mass 2M and 3M respectively having a total kinetic energy E. The kinetic energy of the piece of the piece of mass 2M after the explosion is ___.
a. E/3         b. E/5        c.2E/5         d. 3E/5
ans - d
要 怎 么 算 ?

kinetic energy of 2M mass after explode = 1/2 (2M) v1^2
kinetic energy of 3M mass after explode = 1/2 (3M)v2^2

Principal of conservation momentum : (2M)v1 + (3M)v2 = 0
=> v2 = - 2/3 v1

E = 1/2 (2M) v1^2 + 1/2 (3M) v2^2
  = 1/2 (2M) v1^2 + 1/2 (3M) (4/9)v2^2
  = 5/3 M v2^2
=> v2^2 = 3E/5M
kinetic energy of 2M mass : 1/2 (2M)(3E/5M) = 3/5 E

3.A cylinder of mass 50kg and radius 0.20m has moment of inertia of 1.0 kgm^2 about its axis.When it rolls without slipping on a horizontal plane with linear velocity 2.0 ms^-1,its kinetic energy is ____.
ans -150J

Total kinetic energy = rotational KE + linear KE
= 1/2 Iw^2 + 1/2 mv^2
=1/2 I (v/r)^2 + 1/2 mv^2
= ....

4.A projevtile of mass 2 kg is fired at 60 degree above the horizontal with a speed of 20 ms^-1.What is the minimum momentum value of the projectile during its flight?
ans - 20Ns

horizontal initial velocity : vx = 20 cos 60 = 10m/s
vertical initial velocity : vy = 20 sin 60 = 17.32m/s

total veloticity at time t , v 受 vx 和 vy 影响。因为 vx 不变，vy minimum = 0 ,所以  minimum velocity = 10m/s .因此 minimum momentum = mv = 2(10)=20Ns

5.A rocket is accelerated from rest for 5s at 100ms^-2.It then travels at a constant speedfor 5s.The distance travelled in the first 7s is _____, and its maximum speed is _____.

ans- 2250m , 500 ms^-1

speed after 5s : v = u + at = 0 + 100(5) = 500m/s .
distance travel first 5s : s = ut + 1/2 at^2 = 1/2 (100)(25) = 1250
distance travel 2 sec later : s = vt = 500(2) = 1000
total distance travel = 2250m

max speed 明显 = 500m/s

6.A particle performs simple harmonic motion according to equation x=3cos wt where x is measured in cm,and t in s.If the angular frequency w is 3.14rads^-1,what is the total distance travelled by the particle at t=2.5s?
ans- 15cm

angular frequency , w = 3.14
=> period , T = 2pi/w = 2 sec
每 2/4=0.5 sec 可以 travel distance = amplitude = 3cm
所以 2.5 sec travel 15cm

7.The force constant of a spring is k.When an object of mass m is suspended at its end,its period of oscillation is T.If the spring is cut into half and connected in parallel and its lower end of the system is still suspended with a total mass m,what will its period of oscillation be?

ans - 1/2T

开始时，T = 2pi sqrt[m/k]
spring 分成一半后，spring constant divide into half . 过后你又把它 connect in parallel 所以计算时又要除 2 （总共除 4)

所以 period ,T2 = 2pi sqrt[m/4k] = 1/2 T

[ 本帖最后由 dunwan2tellu 于 17-9-2006 08:14 PM 编辑 ]

jernie

关 于 doppler effect 的 计 算
1. A police car P sounding a siren of frequency 100HZ is chasing a car T.The speed of P is 30m/s whereas the speed of T is 40m/s. The sound of siren from P reaches T, is reflected by T and returns to P. Wat is the apparent frequency heard by the driver of T,and wat is the apparent frequency of the reflected wave heard by the driver of P ?
(hint : for the reflected wave,treat the apparent frequency heard by T as the true frequency of the reflected wave.)

关 于 wave的
1.The equation of a transverse wave travelling along a string is y=6.0 sin (0.20x + 5.0t )
where x and y are in cm and t in second. Find (i) amplitude (ii)wavelength (iii)frequency (iv)speed (v)direction of propagation (vi) the max transverse speed of a particle on the string,

2.A stationary wave is produced in the air of a tube closed at one end.(i)State if a node or an antinode is obtained at the closed end and the open end. (ii)draw a graph to show the distribution of amplitudes along the tube for the second overtone.(iii) The velovity of the sound wave in air is 300m/s.Wat is the length of the tube if the frequency of the second overtone is 750HZ??

dunwan2tellu

1. A police car P sounding a siren of frequency 100HZ is chasing a car T.The speed of P is 30m/s whereas the speed of T is 40m/s. The sound of siren from P reaches T, is reflected by T and returns to P. Wat is the apparent frequency heard by the driver of T,and wat is the apparent frequency of the reflected wave heard by the driver of P ?
(hint : for the reflected wave,treat the apparent frequency heard by T as the true frequency of the reflected wave.)

frequency heard by T : (v-vo)/(v-vs) f = (330-40)/(330-30) x 100 = 96.67Hz
这里我用 v = speed of sound in air = 330m/s , vo = velocity of observer(T) = 40 , vs = velocity of source(P)

When reflected , vs = velocity of source(这时变成 T) , vo=velocity of observer(这时变成 P) , v=330 照旧
apparent  frequency heard by P is , (v+vo)/(v-vs) x f = (330+30)/(330-40) x 100 = 124.14 Hz

1.The equation of a transverse wave travelling along a string is y=6.0 sin (0.20x + 5.0t )
where x and y are in cm and t in second. Find (i) amplitude (ii)wavelength (iii)frequency (iv)speed (v)direction of propagation (vi) the max transverse speed of a particle on the string,

y = 6 sin(0.20x +5.0t)
General equation : y = A sin(kx +wt)
Amplitude = A , k = 2pi/L , w = angular frequency (L = wavelength)
所以 A = 6cm , k = 2pi/L = 0.2 => L = 31.4 cm , frequenct ,w = 5
speed = wavelength x frequency = L x w = 157 cm/s
In transverse wave , particles vibrate at right angle to the direction of propagation .

y = 6sin(0.20x+5.0t)
max transverse speed : v = dy/dt = 30cos(0.20x+5.0t) =< 30 cm/s

2.A stationary wave is produced in the air of a tube closed at one end.(i)State if a node or an antinode is obtained at the closed end and the open end. (ii)draw a graph to show the distribution of amplitudes along the tube for the second overtone.(iii) The velovity of the sound wave in air is 300m/s.Wat is the length of the tube if the frequency of the second overtone is 750HZ??

(i)close end : Node , open end : Antinode
(iii)second ovetone = 3 x fo   [in general the nth ovetone = (2n-1) x fo ]
fo = natural frequency = v/L = 300/L  , L = wavelength
=> 750 = 3 x 300/L => L = 1.2 m
length of tube = 3/4 L = 0.9 m

[ 本帖最后由 dunwan2tellu 于 17-9-2006 11:47 PM 编辑 ]

jernie

kevng

yup...my answer is B too....

jernie

原帖由 kevng 于 23-9-2006 02:49 PM 发表
yup...my answer is B too....

i got it liao..hehez..

jernie

how about tis ler?




WHY III  is wrong???

dunwan2tellu

momentum = mv

v 一直在变，所以 momentum 不是constant

kensai

原帖由 jernie 于 23-9-2006 03:22 PM 发表
how about tis ler?




WHY III  is wrong???

如果你的mass是constant的话，
因为velocity增加,p=mv, momentum也会增加

jernie

QUOTE:
5.A rocket is accelerated from rest for 5s at 100ms^-2.It then travels at a constant speedfor 5s.The distance travelled in the first 7s is _____, and its maximum speed is _____.

ans- 2250m , 500 ms^-1
speed after 5s : v = u + at = 0 + 100(5) = 500m/s .
distance travel first 5s : s = ut + 1/2 at^2 = 1/2 (100)(25) = 1250
distance travel 2 sec later : s = vt = 500(2) = 1000
total distance travel = 2250m




为 什 么 要 用 s = vt ???
不 能 用 s=1/2at^2+  ut 吗 ???

dunwan2tellu

原帖由 jernie 于 23-9-2006 03:46 PM 发表
QUOTE:
5.A rocket is accelerated from rest for 5s at 100ms^-2.It then travels at a constant speedfor 5s.The distance travelled in the first 7s is _____, and its maximum speed is _____.

ans- 225 ...

因为他用 constant velocity 来 travel , 所以没有 acceleration.

kevng

在那段时间是它是平均速率，速率没改变，s=ut+0.5ut^2只能在变速时用到

jernie

how to know the direction??

kevng

|
           |
vx<--------|---------->
          /|
         / |
        /  |  
     v /   |
          vy

就好像找合力的方式一样 。然后用毕氏定理找出v

jernie

原帖由 kevng 于 23-9-2006 07:43 PM 发表
|
           |
vx<--------|---------->
          /|
         / |
        /  |  
     v /   |
          vy

就好像找合力的方式一样 。然后用毕氏定理找出v

找 到 了 。 。 可 是 为 什 么 它 的 方 向 不 是 south -westerly 而 是  north-westerly 呢

kevng

erm..r u sure dat the answer is c..

jernie

原帖由 kevng 于 23-9-2006 08:33 PM 发表
erm..r u sure dat the answer is c..

yup...my teacher give de..but i think should be a..ru u oso think like tis rite?

kevng

oh..ya...i get it now..giv me some time...i think the answer is rite...didn't noticed that it want "acceleration"....

jernie

7.The force constant of a spring is k.When an object of mass m is suspended at its end,its period of oscillation is T.If the spring is cut into half and connected in parallel and its lower end of the system is still suspended with a total mass m,what will its period of oscillation be?

ans - 1/2T
开始时，T = 2pi sqrt[m/k]
spring 分成一半后，spring constant divide into half . 过后你又把它 connect in parallel 所以计算时又要除 2 （总共除 4)

所以 period ,T2 = 2pi sqrt[m/4k] = 1/2 T


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