Unit A3 · Space, Time and MotionUnit A3 · 空间、时间与运动
Work, Energy and Power功、能量与功率
IB-Style Practice QuestionsIB 风格练习题
MEDIUMHARDPaper 1Paper 1BPaper 2HL ONLY
Syllabus A3.1 to A3.6考纲 A3.1 至 A3.6PHYSICS 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 a clear "before" and "after" state and a zero of potential energy before applying conservation. 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。应用守恒前，先写明清晰的"之前"与"之后"状态及势能零点（zero of potential energy）。数值答案保留适当的有效数字。
Q1MEDIUMPaper 1work at an angle + area under F-s graph斜向力做的功与 F-s 图面积[6 marks]
A worker drags a crate $8.0\ \mathrm{m}$ across a level floor.一名工人将一个箱子在水平地面上拖动 $8.0\ \mathrm{m}$。
(a)The worker pulls with a constant force of $120\ \mathrm{N}$ directed at $25^{\circ}$ above the horizontal. Calculate the work done by this pulling force, and state one sentence why the full $120\ \mathrm{N}$ is not used in the calculation.工人以 $120\ \mathrm{N}$ 的恒力沿与水平方向成 $25^{\circ}$ 的方向拉动。计算该拉力做的功，并用一句话说明为何计算中不直接用整个 $120\ \mathrm{N}$。[3]
(b)In a second trial the horizontal force is not constant: it increases linearly with displacement from $0\ \mathrm{N}$ at $s = 0$ to $30\ \mathrm{N}$ at $s = 6.0\ \mathrm{m}$. Using the area under the force-displacement graph, calculate the work done over this $6.0\ \mathrm{m}$.在第二次试验中水平力不恒定：它随位移从 $s = 0$ 处的 $0\ \mathrm{N}$ 线性增大到 $s = 6.0\ \mathrm{m}$ 处的 $30\ \mathrm{N}$。利用力-位移图下的面积，计算这 $6.0\ \mathrm{m}$ 内做的功。[3]
A car of mass $1500\ \mathrm{kg}$ travelling at $20\ \mathrm{m\,s^{-1}}$ brakes and comes to rest. The road exerts a constant friction force of $6000\ \mathrm{N}$ on the car.一辆质量 $1500\ \mathrm{kg}$、以 $20\ \mathrm{m\,s^{-1}}$ 行驶的汽车刹车至停。路面对汽车施加恒定摩擦力 $6000\ \mathrm{N}$。
(a)State the work-energy theorem in words.用文字陈述动能定理。[1]
(b)Calculate the kinetic energy of the car before braking.计算汽车刹车前的动能。[2]
(c)Hence determine the braking distance of the car.由此求汽车的刹车距离。[2]
Q3HARDPaper 1Hooke's law + elastic PE to KE胡克定律与弹性势能转动能[6 marks]
A light spring hangs vertically. When a mass of $0.50\ \mathrm{kg}$ is hung from it, the spring extends by $0.10\ \mathrm{m}$.一根轻弹簧竖直悬挂。当挂上 $0.50\ \mathrm{kg}$ 的物体时，弹簧伸长 $0.10\ \mathrm{m}$。
(a)Calculate the spring constant $k$.计算弹簧的劲度系数 $k$。[2]
(b)Calculate the elastic potential energy stored in the spring at this extension.计算弹簧在此伸长量下储存的弹性势能。[2]
(c)The same spring is laid horizontally, compressed by $0.20\ \mathrm{m}$ and used to launch a $0.10\ \mathrm{kg}$ ball. Assuming all the stored elastic energy becomes kinetic energy, calculate the launch speed of the ball.将同一弹簧水平放置，压缩 $0.20\ \mathrm{m}$ 用以发射一个 $0.10\ \mathrm{kg}$ 的小球。假设储存的弹性能全部转为动能，计算小球的发射速率。[2]
Q4HARDPaper 1power and efficiency of a hoist提升机的功率与效率[6 marks]
An electric motor raises a load of mass $250\ \mathrm{kg}$ vertically at a constant speed of $0.50\ \mathrm{m\,s^{-1}}$. The motor draws electrical power at a rate of $1.8\ \mathrm{kW}$.一台电动机以恒定速率 $0.50\ \mathrm{m\,s^{-1}}$ 竖直吊起质量 $250\ \mathrm{kg}$ 的负载。电动机以 $1.8\ \mathrm{kW}$ 的速率消耗电功率。
(a)Calculate the useful output power delivered to the load. State why the lift force equals the weight here.计算输送给负载的有用输出功率。说明此处提升力为何等于重力。[2]
(b)Calculate the efficiency of the motor, expressing your answer as a percentage.计算电动机的效率，并以百分比表示。[2]
(c)Determine the rate at which energy is dissipated by the motor, and state the most likely form this wasted energy takes.求电动机耗散能量的速率，并说明这部分浪费能量最可能的形式。[2]
Q5HARDPaper 1inelastic collision: energy bookkeeping非弹性碰撞的能量账目[7 marks]
A trolley A of mass $2.0\ \mathrm{kg}$ moving at $4.0\ \mathrm{m\,s^{-1}}$ collides head-on with a stationary trolley B of mass $6.0\ \mathrm{kg}$. The two trolleys couple together on impact.一辆质量 $2.0\ \mathrm{kg}$、以 $4.0\ \mathrm{m\,s^{-1}}$ 运动的小车 A 正面撞上静止的质量 $6.0\ \mathrm{kg}$ 的小车 B。两车碰撞时连接在一起。
(a)State which conservation law you must apply first, and calculate the common velocity of the coupled trolleys after the collision.写出你必须首先应用的守恒定律，并计算碰撞后连接小车的共同速度。[3]
(b)Calculate the total kinetic energy before and after the collision, and hence the kinetic energy lost.计算碰撞前后的总动能，由此求损失的动能。[3]
(c)State whether this collision is elastic or inelastic, and name where the lost kinetic energy has gone.说明该碰撞是弹性还是非弹性，并指出损失的动能去向。[1]
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 student stretches a spring and records the applied force $F$ for several extensions $x$:一名学生拉伸弹簧，记录若干伸长量 $x$ 对应的所施力 $F$：
$x\ /\ \mathrm{m}$
$0.10$
$0.20$
$0.30$
$0.40$
$F\ /\ \mathrm{N}$
$2.5$
$5.0$
$7.5$
$10.0$
(a)State Hooke's law and explain why a graph of $F$ against $x$ should be a straight line through the origin, identifying what the gradient represents.陈述胡克定律，并解释为何 $F$ 对 $x$ 的图应为过原点的直线，指出斜率代表什么。[3]
(b)Calculate the gradient of the line and hence state the spring constant.计算该直线的斜率，由此写出弹簧的劲度系数。[3]
(c)Using the area under the line, determine the elastic potential energy stored in the spring at an extension of $0.40\ \mathrm{m}$.利用图线下的面积，求弹簧在伸长量 $0.40\ \mathrm{m}$ 时储存的弹性势能。[2]
(d)Each force reading has an absolute uncertainty of $\pm 0.2\ \mathrm{N}$. For the point at $x = 0.40\ \mathrm{m}$, calculate the percentage uncertainty in $F$.每个力读数的绝对不确定度为 $\pm 0.2\ \mathrm{N}$。对 $x = 0.40\ \mathrm{m}$ 处的数据点，计算 $F$ 的百分比不确定度。[2]
Q7HARDPaper 1BHL ONLYvariable force: area = integral of work变力：面积即功的积分[12 marks]
A cart of mass $5.0\ \mathrm{kg}$ on a frictionless track is pushed by a horizontal force whose magnitude varies with displacement as follows: it rises linearly from $0\ \mathrm{N}$ at $s = 0$ to $40\ \mathrm{N}$ at $s = 4.0\ \mathrm{m}$, stays constant at $40\ \mathrm{N}$ from $s = 4.0\ \mathrm{m}$ to $s = 8.0\ \mathrm{m}$, then falls linearly to $0\ \mathrm{N}$ at $s = 10.0\ \mathrm{m}$. The cart starts from rest.一辆质量 $5.0\ \mathrm{kg}$ 的小车在无摩擦轨道上受一水平力推动，其大小随位移变化如下：从 $s = 0$ 处的 $0\ \mathrm{N}$ 线性增大到 $s = 4.0\ \mathrm{m}$ 处的 $40\ \mathrm{N}$，从 $s = 4.0\ \mathrm{m}$ 到 $s = 8.0\ \mathrm{m}$ 保持 $40\ \mathrm{N}$ 不变，再从 $s = 8.0\ \mathrm{m}$ 线性减小到 $s = 10.0\ \mathrm{m}$ 处的 $0\ \mathrm{N}$。小车从静止开始。
(a)Explain why the work done by a variable force equals the area under the force-displacement graph, writing down the integral expression for the work.解释变力做的功为何等于力-位移图下的面积，并写出功的积分表达式。[2]
(b)Calculate the work done by the force over the first $4.0\ \mathrm{m}$.计算力在前 $4.0\ \mathrm{m}$ 内做的功。[2]
(c)Calculate the total work done by the force over the full $10.0\ \mathrm{m}$.计算力在整个 $10.0\ \mathrm{m}$ 内做的总功。[3]
(d)Using the work-energy theorem, calculate the speed of the cart after it has travelled the full $10.0\ \mathrm{m}$.用动能定理计算小车行驶完整个 $10.0\ \mathrm{m}$ 后的速率。[3]
(e)State the displacement at which the cart's acceleration is greatest, giving a reason in terms of the force.写出小车加速度最大处的位移，并结合力给出理由。[2]
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 a labelled "before" and "after" energy account. 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"之前"与"之后"的能量账目。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 2conservation of energy with friction含摩擦的能量守恒[11 marks]
A block of mass $4.0\ \mathrm{kg}$ is released from rest at the top of a straight ramp of length $5.0\ \mathrm{m}$ inclined at $30^{\circ}$ to the horizontal. It reaches the bottom of the ramp at a speed of $4.0\ \mathrm{m\,s^{-1}}$.一个质量 $4.0\ \mathrm{kg}$ 的木块从一条长 $5.0\ \mathrm{m}$、与水平成 $30^{\circ}$ 的直斜面顶端由静止释放。它到达斜面底端时速率为 $4.0\ \mathrm{m\,s^{-1}}$。
(a)Calculate the vertical height through which the block descends.计算木块下降的竖直高度。[1]
(b)Calculate the gravitational potential energy lost by the block.计算木块损失的重力势能。[2]
(c)Calculate the kinetic energy gained by the block at the bottom.计算木块到达底端时获得的动能。[2]
(d)Hence determine the energy transferred to thermal energy by friction.由此求摩擦转化为热能的能量。[2]
(e)Calculate the average friction force acting on the block along the ramp.计算木块沿斜面所受的平均摩擦力。[2]
(f)State the speed the block would have reached at the bottom if the ramp were frictionless, and explain in one sentence why it is greater than $4.0\ \mathrm{m\,s^{-1}}$.写出若斜面无摩擦时木块到达底端的速率，并用一句话解释为何它大于 $4.0\ \mathrm{m\,s^{-1}}$。[2]
Q9HARDPaper 2HL ONLYelastic collision in two dimensions二维弹性碰撞[9 marks]
A puck A of mass $0.20\ \mathrm{kg}$ moving east at $3.0\ \mathrm{m\,s^{-1}}$ collides elastically with an identical stationary puck B on a frictionless horizontal surface. After the collision puck A moves off at $30^{\circ}$ north of east.在无摩擦水平面上，一个质量 $0.20\ \mathrm{kg}$、以 $3.0\ \mathrm{m\,s^{-1}}$ 向东运动的冰球 A 与一个相同的静止冰球 B 发生弹性碰撞。碰撞后冰球 A 沿东偏北 $30^{\circ}$ 方向飞出。
(a)State the result, special to an elastic collision between equal masses where one is initially at rest, for the angle between the two final velocities.写出等质量、其中一个初始静止的弹性碰撞所特有的结论：两个末速度之间的夹角。[2]
(b)Hence state the direction in which puck B moves after the collision.由此写出冰球 B 碰撞后的运动方向。[2]
(c)By conserving momentum in two perpendicular directions, calculate the speed of each puck after the collision.通过在两个互相垂直方向上的动量守恒，计算每个冰球碰撞后的速率。[3]
(d)Verify that your answers to (c) are consistent with kinetic energy being conserved.验证你在 (c) 中的答案与动能守恒相一致。[2]
Q10HARDPaper 2power, P = Fv and efficiency on an incline功率、P = Fv 与斜坡上的效率[10 marks]
A car of mass $1200\ \mathrm{kg}$ climbs a straight road inclined at $5.0^{\circ}$ to the horizontal at a constant speed of $15\ \mathrm{m\,s^{-1}}$. The total resistive (frictional) force opposing the motion is $500\ \mathrm{N}$.一辆质量 $1200\ \mathrm{kg}$ 的汽车以恒定速率 $15\ \mathrm{m\,s^{-1}}$ 沿与水平成 $5.0^{\circ}$ 的直路上坡行驶。阻碍运动的总阻力（摩擦力）为 $500\ \mathrm{N}$。
(a)Calculate the component of the car's weight acting down the slope.计算汽车重力沿斜坡向下的分量。[2]
(b)State why the driving force equals the sum of the weight component and the resistive force, and calculate this driving force.说明驱动力为何等于重力分量与阻力之和，并计算该驱动力。[2]
(c)Calculate the useful output power of the engine.计算发动机的有用输出功率。[2]
(d)The engine is $25\%$ efficient at converting fuel energy into useful output. Calculate the rate at which the engine consumes fuel energy.发动机将燃料能转化为有用输出的效率为 $25\%$。计算发动机消耗燃料能的速率。[2]
(e)Calculate the power dissipated against the resistive force alone, and state in one sentence why this is less than the total useful output power.仅计算克服阻力所耗散的功率，并用一句话说明为何它小于总有用输出功率。[2]