STPM 物理问题马拉松！

sylvia_r

A particle P with mass m moves with speed 6u towards another stationary paticle Q of mass 4m. If th collision between P and Q is an elastic collision, what is the speed of Q after the collision?

哈李发

if the collision is elastic, then the energy b4 and after collision must be conserved.
we have 0.5m(6)U^2+0.5(0.4m)u=0.5mV^2+0.5(4m)v^2------equation 1
where U is the initial velocity of P,u is initial velocity of Q  and V is final velocity of P and v is final velocity of Q.

ifthe collision is elastic , then the momentum can be calculated by using the following equation.
6mU+4mu=6mV+amv--------equation 2

using this 2 equation will help you to solve problem already...
by the way i have forgotten how to solve liao, cz long time no read book liao.
hope that my info will help you....
斑竹不要扣分阿。。

hamilan911

Since it is elastic collision,那么就用principle of conservation of momentum 和 principle of conservation of energy
m(6u) + 4m(0) = m(v1) + 4m(v2)
simplify and get  6u = v1 + 4v2     - (1)
0.5m(6u)^2 + 0.5(4m)(0) = 0.5m(v1)^2 + 0.5(4m)(v2)^2   
simplify and get 36u^2 = (v1)^2 + 4(v2)^2     -(2)
from (1) , v1 = 6u - 4v2    -(3)
substitute (3) into (2)
36u^2 = (6u - 4v2)^2 + 4(v2)^2
      = 36u^2 - 48uv2 + 20(v2)^2
so 20(v2)^2 - 48uv2 = 0
4v2(5v2 - 12u) = 0
therefore v2 = 12u/5 (v2 =/= 0)
substitute v2 = 12u/5 into (3)
v1 = 6u - 4(12u/5)
   = -18u/5

懒惰想题目，楼下的代劳

sylvia_r

答对了！
哈李发你放心，我不会扣你分的，因为我也不知道要怎样扣
不过你的concept都是对的
再一题：
A particle X moving wth constant velocity collides elastically with an identical stationary particle  Y. If the direction of X after collision makes an angle of 30 with the original directionof motion X, what is the angle between the direction of motion of Y with the original direction ofmotion of X?
A.0
B.30
C.60
D.90

再见是为了明天

magnetic field question :

show that

q       2V
---  = --------       help: Eq=Bqv ,1/2mv^2 =qV
m      B^2 r^2

再见是为了明天

原帖由 sylvia_r 于 6-6-2007 08:55 AM 发表
答对了！
哈李发你放心，我不会扣你分的，因为我也不知道要怎样扣
不过你的concept都是对的
再一题：
A particle X moving wth constant velocity collides elastically with an identical ...

i have combined the energy conserved and momentum conserved equation and finally i got :

Ux + Vx cos30 = Vy cos @

        Ux + Vx cos30
cos@ = ---------------      ....i have only 2 equation and so many unknowns
           Vy                   how to do?

sylvia_r

你画vector的三角形就看到了。
since it is elastic, it conserve kinetic energy:、

1/2 mu^2= 1/2 mv1^2+ 1/2 mv2^2

u^2 = v1^2 + v2^2
这是直角三角形的equation对吗。。。

所以另外一个angle是 90-30= 60

再见是为了明天

原帖由 sylvia_r 于 7-6-2007 06:00 PM 发表
你画vector的三角形就看到了。
since it is elastic, it conserve kinetic energy:、

1/2 mu^2= 1/2 mv1^2+ 1/2 mv2^2

u^2 = v1^2 + v2^2
这是直角三角形的equation对吗。。。

所以另外一个angle ...

这样都可以 。。。太神奇了
哈哈。。。

gsan

原帖由 再见是为了明天 于 6-6-2007 09:23 PM 发表
magnetic field question :

show that

q       2V
---  = --------       help: Eq=Bqv ,1/2mv^2 =qV
m      B^2 r^2

Bqv等于mv^2/r，也把qV＝1/2mv^2和mv^2/r的关系和起来，酱就可以prove到了。

UnitedDream

好冷清oh.......................
给点basic你们想

Kirchhoff's laws 应该读了吧 ？？

每条wire in the cube 都有one ohm 的resistance , when connected by the two end , what is the totol resistance .

会的话STPM combine resistance and how to determine the flow of current 不用都了

UnitedDream

sorry，没说清楚
用kirchhoff's laws 的 loop 算的话会算死人的
只是用其中的一些而已

哈李发

原帖由 UnitedDream 于 11-6-2007 11:31 PM 发表
好冷清oh.......................
给点basic你们想

Kirchhoff's laws 应该读了吧 ？？

每条wire in the cube 都有one ohm 的resistance , when connected by the two end , what is the totol resistanc ...

哇，救命！！！
我要哭了，我不会！
亏我还要拿物理教育系！！！
呜呜....

UnitedDream

酱ah ， 来点简单点的 ，同样的问题

kentawan

Sand falls vertically at a rate of 0.5 kg/second onto a conveyor belt moving horizontally at a constant velocity of 1.5 m/s . What is the extra force required to move the conveyor belt ?

huhuxboy

F=d(mv)/dt
v=1.5m/s is constant
dm/dt=0.5kg/s
F=v x dm/dt
  = 1.5 x 0.5
  = 0.75N

对吗 ？？？

[ 本帖最后由 huhuxboy 于 6-7-2007 10:00 PM 编辑 ]

剑侠

对吧。

怎么没什么人讨论物理？？

zxteh

原帖由 huhuxboy 于 6-7-2007 09:58 PM 发表
F=d(mv)/dt
v=1.5m/s is constant
dm/dt=0.5kg/s
F=v x dm/dt
  = 1.5 x 0.5
  = 0.75N

对吗 ？？？

对...because
p=mv
dp/dt=m.dv/dt+v.dm/dt
But ,in this case no acceleration
therefore,a=0
F=v.dm/dt
F=[1.5] .[0.5]
F=0.75N

zxteh

原帖由 sylvia_r 于 4-6-2007 11:36 PM 发表
A particle P with mass m moves with speed 6u towards another stationary paticle Q of mass 4m. If th collision between P and Q is an elastic collision, what is the speed of Q after the collision?

i now have see the question of physics...
never mind i solve it now...

if the collision is elastic collision,
By using conservation of momentum,
m1u1+m2u2=m1v1+m2v2
therefore substitute inside the equation,
m(6u)+4m(0)=m(v1)+4m(v2)
v1+4v2=6u------(1)

By using Newton's law of restituition,
v2-v1=u1-u2
therefore substitute inside the equation,
4v2-v1=6u-0
4v2-v1=6u------(2)

(1)+(2),
we get
5v2=12u
v2=12u/5

from(1),we substitute v2=12u/5,
v1+4(12u/5)=6u
v1=-18u/5

therefore v1=18u/5 and moving to the opposite direction that is left hand side

Therefore,v1=18u/5 , v2=12u/5

[ 本帖最后由 zxteh 于 13-7-2007 04:11 PM 编辑 ]

zxteh

原帖由 再见是为了明天 于 6-6-2007 09:23 PM 发表
magnetic field question :

show that

q       2V
---  = --------       help: Eq=Bqv ,1/2mv^2 =qV
m      B^2 r^2

show that

q       2V
---  = --------       help: Eq=Bqv ,1/2mv^2 =qV
m      B^2 r^2 [/quote]

when an particle like proton moving in the circular magnetic path,
they have one force ,we call magnetic force exert on it.
since it have magnetic force and
they must have the centripetal force exert on it,
therfore ,
F(centripetal)=F(magnetic)
mv^2/r=qvB
v^2= {[q^2.B^2.r^2]/m^2} ---(1)

and then by using the conservation of energy,
mv^2/2=qV---(2)

substitute (1) into (2),
[m{[q^2.B^2.r^2]/m^2} ] /2=qV
q/m=2v/B^2.r^2

zxteh

When a conductor of length ,l moving with the velocity,v and cuts the magnectic flux,B with distance, x and the rate changing is t.
Prove that the electromagnetic induction of their induced emf,E=Blv...

[ 本帖最后由 zxteh 于 13-7-2007 05:08 PM 编辑 ]