Unit A1 · Space, Time and MotionUnit A1 · 空间、时间与运动
Kinematics运动学
IB-Style Practice QuestionsIB 风格练习题
MEDIUMHARDPaper 1Paper 1BPaper 2HL ONLY
Syllabus A1.1 to A1.6考纲 A1.1 至 A1.6PHYSICS HL
Name:姓名：Date:日期：
PART I · PAPER 1 STYLE第一部分 · 第一卷风格Short structured · calculator · 30 marks短结构题 · 可用计算器 · 30 分
Short Structured Items短结构题
Show all working in the space below each question. Marks are awarded for correct method as well as final answers. State your sign convention before setting up suvat. Give numerical answers to an appropriate number of significant figures.在每题下方空白处写出全部解题过程。方法分（method marks）与最终答案同等重要。列 suvat 方程前先写明正方向约定（sign convention）。数值答案保留适当的有效数字。
Q1MEDIUMPaper 1distance vs displacement路程与位移[4 marks]
A cyclist rides $100\ \mathrm{m}$ due east, then turns and rides $40\ \mathrm{m}$ due north. The whole trip takes $25\ \mathrm{s}$.一名骑行者向正东行驶 $100\ \mathrm{m}$，随后转向并向正北行驶 $40\ \mathrm{m}$。全程用时 $25\ \mathrm{s}$。
(a)State the total distance travelled and calculate the magnitude of the displacement.写出行驶总路程，并计算位移的大小。[2]
(b)Calculate the average speed and the magnitude of the average velocity for the trip.计算全程的平均速率与平均速度的大小。[2]
Q2MEDIUMPaper 1suvat selectionsuvat 选式[6 marks]
A car travelling at $28\ \mathrm{m\,s^{-1}}$ brakes uniformly and comes to rest after travelling $98\ \mathrm{m}$.一辆以 $28\ \mathrm{m\,s^{-1}}$ 行驶的汽车均匀制动，行驶 $98\ \mathrm{m}$ 后停下。
(a)State which suvat equation links $u$, $v$, $a$ and $s$ without involving $t$, and explain in one sentence why it is the efficient choice here.写出仅联系 $u$、$v$、$a$、$s$ 而不含 $t$ 的 suvat 方程，并用一句话说明为何此处选它最高效。[2]
(b)Calculate the deceleration of the car.计算汽车的减速度。[2]
(c)Hence find the time taken to stop.由此求停车所用时间。[2]
A ball is thrown vertically upward from ground level with an initial speed of $21\ \mathrm{m\,s^{-1}}$. Air resistance is negligible. Take upward as the positive direction.一个小球从地面以初速率 $21\ \mathrm{m\,s^{-1}}$ 竖直上抛。忽略空气阻力。取向上为正方向。
(a)Calculate the maximum height reached above the launch point.计算小球相对抛出点所达的最大高度。[2]
(b)Calculate the time taken to reach that maximum height.计算到达该最大高度所需的时间。[2]
(c)Find the velocity of the ball $3.0\ \mathrm{s}$ after launch, and state what the sign of your answer tells you about the direction of motion at that instant.求小球抛出后 $3.0\ \mathrm{s}$ 时的速度，并说明答案的正负号反映该时刻的运动方向如何。[2]
Q4HARDPaper 1v-t graph: gradient and areav-t 图：斜率与面积[6 marks]
A train starts from rest and its velocity-time graph consists of three straight segments: it accelerates uniformly from $0$ to $20\ \mathrm{m\,s^{-1}}$ over the first $8.0\ \mathrm{s}$, travels at a constant $20\ \mathrm{m\,s^{-1}}$ for the next $6.0\ \mathrm{s}$, then decelerates uniformly to rest over a final $4.0\ \mathrm{s}$.一列火车从静止出发，其速度-时间图由三段直线组成：前 $8.0\ \mathrm{s}$ 内从 $0$ 均匀加速至 $20\ \mathrm{m\,s^{-1}}$，随后 $6.0\ \mathrm{s}$ 内以 $20\ \mathrm{m\,s^{-1}}$ 匀速行驶，最后 $4.0\ \mathrm{s}$ 内均匀减速至静止。
(a)Determine the acceleration during the first segment, identifying which graphical feature gives it.求第一段的加速度，并指出由图像的哪一特征得出。[2]
(b)By calculating the area under the graph, find the total distance travelled by the train.通过计算图线下的面积，求火车行驶的总距离。[3]
(c)State what the gradient of a velocity-time graph represents and what the area beneath it represents.说明速度-时间图的斜率与图线下面积分别表示什么。[1]
A hailstone of mass $0.080\ \mathrm{kg}$ falls from rest through still air. At high speed the air exerts a drag force of magnitude $F_{D} = b v^{2}$, where $b = 5.0\times 10^{-3}\ \mathrm{N\,s^{2}\,m^{-2}}$ and $v$ is the speed.一颗质量 $0.080\ \mathrm{kg}$ 的冰雹从静止在静止空气中下落。高速时空气施加大小为 $F_{D} = b v^{2}$ 的阻力，其中 $b = 5.0\times 10^{-3}\ \mathrm{N\,s^{2}\,m^{-2}}$，$v$ 为速率。
(a)Explain why the hailstone reaches a terminal velocity, referring to the forces acting on it.结合作用在冰雹上的力，解释它为何会达到收尾速度。[2]
(b)Calculate the terminal velocity of the hailstone.计算冰雹的收尾速度。[3]
(c)Sketch the shape of the velocity-time graph from release until terminal velocity is approached, and explain why the suvat equations cannot be used to describe this motion.画出从释放到接近收尾速度的速度-时间图形状，并解释为何 suvat 方程不能用于描述这段运动。[3]
PART II · PAPER 1B / DATA ANALYSIS第二部分 · 第一卷 B / 数据分析Graphs · data · uncertainties · 22 marks图像 · 数据 · 不确定度 · 22 分
Graph and Data Questions图像与数据题
These items reward careful reading of gradients and areas, and correct handling of uncertainties. Quote uncertainties to one significant figure and round the value to match.这些题考查对斜率与面积的细致读取以及对不确定度的正确处理。不确定度保留 1 位有效数字，并使数值的末位与之对齐。
A trolley is released from rest down a ramp. A student measures the displacement $s$ from the start at several times $t$, and tabulates $s$ against $t^{2}$:一辆小车从静止沿斜面释放。学生测量从起点出发若干时刻 $t$ 对应的位移 $s$，并将 $s$ 对 $t^{2}$ 列表：
$t^{2}\ /\ \mathrm{s^{2}}$
$1.0$
$4.0$
$9.0$
$16.0$
$s\ /\ \mathrm{m}$
$0.80$
$3.20$
$7.20$
$12.80$
(a)Starting from a suvat equation, show that a graph of $s$ against $t^{2}$ should be a straight line through the origin, and state what the gradient represents.从某个 suvat 方程出发，证明 $s$ 对 $t^{2}$ 的图应为过原点的直线，并说明斜率代表什么。[3]
(b)Calculate the gradient of the line and hence determine the acceleration of the trolley.计算该直线的斜率，由此求小车的加速度。[3]
(c)Each value of $s$ has an absolute uncertainty of $\pm 0.10\ \mathrm{m}$. For the data point at $t^{2} = 16.0\ \mathrm{s^{2}}$, calculate the percentage uncertainty in $s$.每个 $s$ 值的绝对不确定度为 $\pm 0.10\ \mathrm{m}$。对 $t^{2} = 16.0\ \mathrm{s^{2}}$ 处的数据点，计算 $s$ 的百分比不确定度。[2]
(d)State one reason why the experimental line might not pass exactly through the origin, and what systematic effect this would indicate.说明实验直线可能不恰好过原点的一个原因，以及这反映出怎样的系统效应。[2]
Q7HARDPaper 1Bhorizontal projectile平抛运动[12 marks]
A ball is projected horizontally with a speed of $18\ \mathrm{m\,s^{-1}}$ from the top of a vertical cliff of height $45\ \mathrm{m}$. It lands on level ground. Air resistance is negligible.一个小球以 $18\ \mathrm{m\,s^{-1}}$ 的速率从高 $45\ \mathrm{m}$ 的竖直悬崖顶端水平抛出，落到等高地面。忽略空气阻力。
(a)State why the horizontal and vertical components of the motion can be treated independently.说明为何可将运动的水平分量与竖直分量分开处理。[1]
(b)Calculate the time of flight before the ball lands.计算小球落地前的飞行时间。[3]
(c)Calculate the horizontal distance from the base of the cliff to the landing point.计算从崖底到落点的水平距离。[2]
(d)Calculate the speed of the ball at the instant it lands.计算小球落地瞬间的速率。[3]
(e)Determine the angle, below the horizontal, at which the ball strikes the ground.求小球击中地面时速度方向与水平方向的夹角（指向水平面以下）。[3]
PART III · PAPER 2 STYLE第三部分 · 第二卷风格Extended structured · calculator · 30 marks长结构题 · 可用计算器 · 30 分
Extended Structured Problems长结构问题
Set up each problem with a clear diagram and labelled axes. Method marks dominate the longer items; carry intermediate values to extra figures and round only the final answer.每题先画清晰的示意图并标注坐标轴。长题中方法分占比最大；中间值多保留几位，仅在最终答案处取舍有效数字。
Q8HARDPaper 2projectile range and height抛体射程与高度[12 marks]
A projectile is launched from ground level with an initial speed of $30\ \mathrm{m\,s^{-1}}$ at an angle of $40^{\circ}$ above the horizontal. Air resistance is negligible.一枚弹丸以初速率 $30\ \mathrm{m\,s^{-1}}$、与水平方向夹角 $40^{\circ}$ 自地面发射。忽略空气阻力。
(a)Calculate the horizontal and vertical components of the initial velocity.计算初速度的水平分量与竖直分量。[2]
(b)Calculate the total time of flight.计算总飞行时间。[3]
(c)Calculate the horizontal range and the maximum height reached.计算水平射程与所达最大高度。[3]
(d)Calculate the speed of the projectile $1.5\ \mathrm{s}$ after launch.计算发射后 $1.5\ \mathrm{s}$ 时弹丸的速率。[2]
(e)State, with a reason, the launch angle that would give the maximum range for this launch speed.写出在该发射速率下能给出最大射程的发射角，并说明理由。[2]
A river of width $80\ \mathrm{m}$ flows due east at a constant $2.5\ \mathrm{m\,s^{-1}}$. A boat can travel at $4.0\ \mathrm{m\,s^{-1}}$ relative to the water.一条宽 $80\ \mathrm{m}$ 的河以恒定速率 $2.5\ \mathrm{m\,s^{-1}}$ 向正东流动。一艘船相对水的速率为 $4.0\ \mathrm{m\,s^{-1}}$。
(a)The boat is steered due north (straight across, pointing at right angles to the bank). Calculate the time taken to cross the river.船头指向正北（垂直于河岸笔直横渡）。计算横渡所需时间。[2]
(b)Calculate how far downstream the boat lands relative to the point directly opposite its start, and find the magnitude of its resultant velocity relative to the bank.计算船相对出发点正对岸的位置向下游漂移的距离，并求船相对河岸的合速度大小。[4]
(c)The pilot now wants to reach the point directly opposite the start. Determine the direction, measured from the north (straight-across) line, in which the boat must be steered.现在驾驶者希望到达出发点的正对岸。求船头相对正北（垂直横渡）方向应偏转的角度。[2]
(d)Hence calculate the time taken to cross by this second route, and state in one sentence why it exceeds the time found in (a).由此计算第二种航线横渡所需时间，并用一句话说明为何它比 (a) 中的时间更长。[2]
A skydiver of total mass $75\ \mathrm{kg}$ falls through the air. Before the parachute opens, model the air resistance as a linear drag $F_{D} = k v$ with $k = 15\ \mathrm{N\,s\,m^{-1}}$. The diver starts from rest.一名总质量 $75\ \mathrm{kg}$ 的跳伞者在空中下落。降落伞张开前，将空气阻力建模为线性阻力 $F_{D} = k v$，其中 $k = 15\ \mathrm{N\,s\,m^{-1}}$。跳伞者从静止开始下落。
(a)Write down Newton's second law for the diver while falling, taking downward as positive, and hence state the acceleration at the instant of release.取向下为正，写出跳伞者下落时的牛顿第二定律，并由此写出释放瞬间的加速度。[2]
(b)Calculate the terminal velocity of the diver.计算跳伞者的收尾速度。[2]
(c)Calculate the acceleration of the diver at the moment when the speed is $20\ \mathrm{m\,s^{-1}}$.计算速率为 $20\ \mathrm{m\,s^{-1}}$ 时跳伞者的加速度。[2]
(d)State how the magnitude of the acceleration changes as the speed increases from zero towards the terminal velocity, and explain this in terms of the forces.说明当速率从零增大到收尾速度时加速度大小如何变化，并结合受力加以解释。[2]