2012CAP Prize Exam加拿大高中物理大獎賽真題答案下載

歷年CAP High School Prize Exam加拿大高中物理大獎賽

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2012 CAP Prize Exam完整版真題免費下載

共計3小時考試時間

此套試卷由25道選擇題以及3道大題組成

每道大題含有不同數量的小題

2012CAP Prize Exam加拿大物理奧林匹克完整版答案免費下載

部分真題答案預覽：

Part A選擇題部分：

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

Part B簡答題部分

Solution to Problem 2：

(a) From the ?rst 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 coe?cient of friction is

μ= F/mg =0.0315.

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

P1 = Fv =182 W,

which at e?ciency ? =85%gives us electricalpower

P1? = P1 /?=283 W.

Electricalpower isaproductofthevoltage V times the current I, so P1? = VI1 . We know the voltage V =48 V, so ?nally

I1 = P1?/V = Fv/?V =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

P2? = P2 /?=499 W,

so the average current is I2 = P2?/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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