2004CAP Prize Exam加拿大高中物理大奖赛真题答案下载

历年CAP High School Prize Exam加拿大高中物理大奖赛

真题与答案下载

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2004 CAP Prize Exam完整版真题免费下载

共计3小时考试时间

此套试卷由25道选择题以及3道大题组成

每道大题含有不同数量的小题

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2004 CAP Prize Exam完整版真题

2004CAP Prize Exam加拿大物理奥林匹克完整版答案免费下载

部分真题答案预览：

Part A选择题部分：

1. C
2. B
3. C
4. B
5. A

Part B简答题部分

Solution to Problem 1：

Let x be the horizontal distance from the point where the ball was released, to the point where a ray originat­ing from the instantaneous position of the ball at depth y crossesthesurface andreaches the observer.

From Snell’s law we have sin才 = nw sin万, where nw =1.33 is the index of refraction of water.

From geometry, wecanal so write:

x = d − h tanα

tanβ = x/y

Then

 

(a)

The initial acceleration a0 can be calculated from the firs velocity point in the data table. Velocity changed from 0 to0.091m/sin 0.01s,which gives a0 =9.10 m/s² .

The terminal speed can be read off the table as approxi­mately vf =0.916 m/s² .

The total instantaneous downward force F acting on the ballwhen ithas speed v is

F = Mg − ρwVg − bv (1)

where Mg isthegravitationalforce, ρwVg istheupward buoyancy force if the ball has volume V, and bv is the drag force. By Newton’s second law, F = Ma, where a is theacceleration oftheball..

Initially, v is small enough that the drag force can be ig­nored, andwe can write

F = ρballVg − ρwVg = ρballVa0

with ρball the densityof theball. Thus,

ρw

ρ=

ball

1 − a0 /g

1000 kg/m³

= 1 − 9.10/9.80

=1.4 × 10⁴ kg/m³

Ontheotherhand,whenthespeed v has reacheditsfina value vf,theacceleration vanishes andeq.(1) becomes

ρw

0= Mg − ρwVg − bvf = (1 − )Mg − bvfρball

Solvingfor b yields

Mg ρ

w

b =1 − vfρball

(1.00 kg)(9.80 m/s² )1.00 × 10³ kg/m³

=1 −

0.916 m/s1.4 × 10⁴ kg/m³ =9.9 kg/s

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