2012CAP 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

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2012 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

2012CAP 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. E
2. B
3. C
4. A
5. E

Part B简答题部分

Solution to Problem 2：

(a) From the first statement, we can calculate the combined force of friction and air resistance, Ff. We have to assume that it is independent of speed, so we basically neglect the air resistance. The bike’s kinetic energy is converted to work of the force of friction on the distance d =50 m.

Ffd = mv² /2 v =20 km/h = 5.56 m/s F = mv² /2d =32.7N

So the corresponding coefficient of friction is

µ= F/mg =0.0315.

To bike at 20 km/h, we need mechanical power

P1 = Fv =182 W,

which at efficiency ǫ =85%gives us electricalpower

P1ǫ = P1 /ǫ=283 W.

Electricalpower isaproductofthevoltage V times the current I, so P1ǫ = VI1 . We know the voltage V =48 V, so finally

I1 = P1ǫ/V = Fv/ǫV =4.45 A.

(b) The battery capacity is 8 Ah, so is will supply 4.45 A for a time equal to:

8Ah/ 4.45A =1.8h.

During this time, at 20 km/h, the cyclist will bike for 36 km.

(c) At this constant speed, the net force on the bike and net torque must be zero, as there is no acceleration orangularaccelerationof thewheel. [Wealsogavepointstothestudentswhocalculated thetorquerelated to the force driving the bike: τ =32.7 N * 0.33 m = 10.8 Nm.]

(d) Now we have to calculate the total force acting on the bike. There are three forces: -The force driving the bike, F -The force of friction, −µmgcos(θ) -The component of the force of gravity, −mgsin(θ) We assume constant acceleration a = v/t1 =(5.56 m/s)/ 20 s. From Newton’s Law

 

ma = F − µmgcos(θ)− mgsin(θ),

so we need the force F:

F =(mv/t1 + µmgcos(θ)+mgsin(θ)). P2 = v(mv/t1 + µmgcos(θ)+mgsin(θ))/2 =424W.

The average electrical power needed is

P2ǫ = P2 /ǫ=499 W,

so the average current is I2 = P2ǫ/V =10.4 A.

(e) We again assume constant acceleration. Maximum power is needed at the end, to reach the maximum speed of20km/h. Fromstatement2wehavetheaveragemechanicalpower likein(d),but nowtheacceleration is(v2 − v1 )/t2 where v2 =20 km/h, v1 =5 km/h and t2 =6 s. Therefore the acceleration is a =0.69 m/s. At the top speed the power needed is

P3 = v2 (m(v2 − v1 )/t2 + µmgcos(θ)− mgsin(θ))/2 =1093W.

(f) Assuming speed v going up the slope, we have

P3 = v(µmgcos(θ)+mgsin(θ))

 

One can substitute cos(θ) with1− sin² (θ) and get a quadratic equation for sin(θ), or solve the problem graphically for a given velocity. v = 20 km/h gives θ =9^◦ . At 5 km/h one gets about 45^◦, which is probably impossible -the wheels will slide.

(g) I3 = P3 /V =26.8A

(h) We assumed constant friction force and constant acceleration.

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