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共計(jì)3小時(shí)考試時(shí)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(shù)量的小題
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Part A選擇題部分:
Part B簡(jiǎn)答題部分
Solution to Problem 1:
From the red dashed red triangle on the picture one can measure the shooting angle to be θ = (22 ± 3?). Using the scale of the graph and attached to the exam “ruler” one can measure the maximum height above the end of the shooter h =( 9 ± 2) m and the horizontal distance from the end of the shooter to the place of maximum height L = (33 ± 2) m as well as the height H = 35m from the maximum to the drop point and total range? L’ = (83 ± 2) m
From the conservation of energy we have:
mgh + m(vcosθ)2/2 =? mv2/2
v = √2gh/(1-cosθ)2 = ?√2gh/sinθ2 = 35 ± 8 m/s
va = 83/4 = 21 m/s.
The final speed is:
va = (vh+vf)/2? vf = 2va? - vh? = 7 m/s
so the negative acceleration due to air resistance is
vf = vh – at??? a = (vh – vf)/t = 6m/s2
So the force acting on (for example) 1 kg of gravel was about 10 N. It is clear from the photograph that it strongly depend on the size of stones.
There were many ways to solve this problem and all the fully correct ones were awarded full marks and partly correct ones part marks.? ?
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