Companion to the IB-Style Practice SetIB 风格练习题的解析配套
Syllabus A3.1 to A3.6考纲 A3.1 至 A3.6PHYSICS HL
A crate is dragged $8.0\ \mathrm{m}$. (a) work done by a $120\ \mathrm{N}$ pull at $25^{\circ}$ above the horizontal; (b) work done by a force rising linearly $0 \to 30\ \mathrm{N}$ over $6.0\ \mathrm{m}$, from the area under the $F$-$s$ graph.箱子被拖动 $8.0\ \mathrm{m}$。(a) 与水平成 $25^{\circ}$ 的 $120\ \mathrm{N}$ 拉力做的功；(b) 力在 $6.0\ \mathrm{m}$ 内从 $0$ 线性增至 $30\ \mathrm{N}$ 时由 $F$-$s$ 图面积求做的功。
Work is the scalar product of force and displacement: $W = Fs\cos\theta$, where $\theta$ is the angle between the force and the displacement. (M1)
$$ W = (120)(8.0)\cos 25^{\circ} = 960 \times 0.9063 \approx 870\ \mathrm{J}. $$(A1)
Only the horizontal component $F\cos\theta$ lies along the displacement; the vertical component is perpendicular to the motion and does no work, so the full $120\ \mathrm{N}$ is not used. (R1)
For a variable force the work is the area under the force-displacement graph. The graph is a straight line from $(0,\,0)$ to $(6.0,\,30)$, so the area is a triangle. (M1)
$$ W = \tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2}(6.0)(30). $$(A1)
$$ W = 90\ \mathrm{J}. $$(A1)
功是力与位移的标量积：$W = Fs\cos\theta$，其中 $\theta$ 为力与位移的夹角。(M1)
$$ W = (120)(8.0)\cos 25^{\circ} = 960 \times 0.9063 \approx 870\ \mathrm{J}. $$(A1)
只有水平分量 $F\cos\theta$ 沿位移方向；竖直分量与运动垂直，不做功，故不直接用整个 $120\ \mathrm{N}$。(R1)
对变力，功是力-位移图下的面积。该图是从 $(0,\,0)$ 到 $(6.0,\,30)$ 的直线，故面积为三角形。(M1)
$$ W = \tfrac{1}{2} \times \text{底} \times \text{高} = \tfrac{1}{2}(6.0)(30). $$(A1)
$$ W = 90\ \mathrm{J}. $$(A1)
A $1500\ \mathrm{kg}$ car at $20\ \mathrm{m\,s^{-1}}$ brakes to rest under a constant friction force of $6000\ \mathrm{N}$. (a) state the work-energy theorem; (b) initial kinetic energy; (c) braking distance.一辆 $1500\ \mathrm{kg}$、以 $20\ \mathrm{m\,s^{-1}}$ 行驶的汽车在 $6000\ \mathrm{N}$ 恒定摩擦力下刹车至停。(a) 陈述动能定理；(b) 初动能；(c) 刹车距离。
The net (total) work done on a body equals the change in its kinetic energy: $W_{\text{net}} = \Delta E_k$. (B1)
$E_k = \tfrac{1}{2}mv^{2}$ with $m = 1500\ \mathrm{kg}$ and $v = 20\ \mathrm{m\,s^{-1}}$: (M1)
$$ E_k = \tfrac{1}{2}(1500)(20)^{2} = \tfrac{1}{2}(1500)(400) = 3.0\times 10^{5}\ \mathrm{J}. $$(A1)
Friction is the only force doing work; it removes all the kinetic energy: $F s = E_k$. (M1)
$$ s = \frac{E_k}{F} = \frac{3.0\times 10^{5}}{6000} = 50\ \mathrm{m}. $$(A1)
物体所受的合（总）功等于其动能的变化：$W_{\text{net}} = \Delta E_k$。(B1)
$E_k = \tfrac{1}{2}mv^{2}$，$m = 1500\ \mathrm{kg}$，$v = 20\ \mathrm{m\,s^{-1}}$：(M1)
$$ E_k = \tfrac{1}{2}(1500)(20)^{2} = \tfrac{1}{2}(1500)(400) = 3.0\times 10^{5}\ \mathrm{J}. $$(A1)
摩擦力是唯一做功的力，它移走全部动能：$F s = E_k$。(M1)
$$ s = \frac{E_k}{F} = \frac{3.0\times 10^{5}}{6000} = 50\ \mathrm{m}. $$(A1)
A spring extends $0.10\ \mathrm{m}$ under a $0.50\ \mathrm{kg}$ mass. (a) spring constant; (b) elastic PE at that extension; (c) launch speed of a $0.10\ \mathrm{kg}$ ball from the same spring compressed $0.20\ \mathrm{m}$.弹簧在 $0.50\ \mathrm{kg}$ 物体下伸长 $0.10\ \mathrm{m}$。(a) 劲度系数；(b) 该伸长量下的弹性势能；(c) 同一弹簧压缩 $0.20\ \mathrm{m}$ 时发射 $0.10\ \mathrm{kg}$ 小球的速率。
At equilibrium the spring force balances the weight: $kx = mg$. (M1)
$$ k = \frac{mg}{x} = \frac{(0.50)(9.81)}{0.10} = \frac{4.905}{0.10} \approx 49\ \mathrm{N\,m^{-1}}. $$(A1)
$E_{\text{el}} = \tfrac{1}{2}kx^{2}$ with $x = 0.10\ \mathrm{m}$: (M1)
$$ E_{\text{el}} = \tfrac{1}{2}(49.05)(0.10)^{2} = \tfrac{1}{2}(49.05)(0.010) \approx 0.25\ \mathrm{J}. $$(A1)
Stored elastic energy at $x = 0.20\ \mathrm{m}$ becomes kinetic energy: $\tfrac{1}{2}kx^{2} = \tfrac{1}{2}mv^{2}$. (M1)
$$ E_{\text{el}} = \tfrac{1}{2}(49.05)(0.20)^{2} = 0.981\ \mathrm{J}, \qquad v = \sqrt{\frac{2 E_{\text{el}}}{m}} = \sqrt{\frac{2(0.981)}{0.10}} = \sqrt{19.62} \approx 4.4\ \mathrm{m\,s^{-1}}. $$(A1)
平衡时弹簧力与重力相等：$kx = mg$。(M1)
$$ k = \frac{mg}{x} = \frac{(0.50)(9.81)}{0.10} = \frac{4.905}{0.10} \approx 49\ \mathrm{N\,m^{-1}}. $$(A1)
$E_{\text{el}} = \tfrac{1}{2}kx^{2}$，$x = 0.10\ \mathrm{m}$：(M1)
$$ E_{\text{el}} = \tfrac{1}{2}(49.05)(0.10)^{2} = \tfrac{1}{2}(49.05)(0.010) \approx 0.25\ \mathrm{J}. $$(A1)
$x = 0.20\ \mathrm{m}$ 处储存的弹性能转为动能：$\tfrac{1}{2}kx^{2} = \tfrac{1}{2}mv^{2}$。(M1)
$$ E_{\text{el}} = \tfrac{1}{2}(49.05)(0.20)^{2} = 0.981\ \mathrm{J}, \qquad v = \sqrt{\frac{2 E_{\text{el}}}{m}} = \sqrt{\frac{2(0.981)}{0.10}} = \sqrt{19.62} \approx 4.4\ \mathrm{m\,s^{-1}}. $$(A1)
A motor raises a $250\ \mathrm{kg}$ load at constant $0.50\ \mathrm{m\,s^{-1}}$, drawing $1.8\ \mathrm{kW}$. (a) useful output power; (b) efficiency; (c) rate of energy dissipation and its form.电动机以恒定 $0.50\ \mathrm{m\,s^{-1}}$ 吊起 $250\ \mathrm{kg}$ 负载，消耗 $1.8\ \mathrm{kW}$。(a) 有用输出功率；(b) 效率；(c) 能量耗散速率及其形式。
At constant speed the lift force balances the weight, so $F = mg$; the useful power is $P = Fv = mgv$. (M1)
$$ P_{\text{out}} = (250)(9.81)(0.50) = 1226\ \mathrm{W} \approx 1.2\ \mathrm{kW}. $$(A1)
$\eta = \dfrac{P_{\text{useful}}}{P_{\text{input}}}$ with $P_{\text{input}} = 1800\ \mathrm{W}$: (M1)
$$ \eta = \frac{1226}{1800} = 0.681 \approx 68\%. $$(A1)
The dissipated power is the input power minus the useful output: $1800 - 1226 = 574\ \mathrm{W}$. (M1)
This is about $0.57\ \mathrm{kW}$, dissipated mostly as thermal energy (heat) in the motor windings and bearings, with a little as sound. (A1)
匀速时提升力与重力平衡，故 $F = mg$；有用功率为 $P = Fv = mgv$。(M1)
$$ P_{\text{out}} = (250)(9.81)(0.50) = 1226\ \mathrm{W} \approx 1.2\ \mathrm{kW}. $$(A1)
$\eta = \dfrac{P_{\text{useful}}}{P_{\text{input}}}$，$P_{\text{input}} = 1800\ \mathrm{W}$：(M1)
$$ \eta = \frac{1226}{1800} = 0.681 \approx 68\%. $$(A1)
耗散功率为输入功率减有用输出：$1800 - 1226 = 574\ \mathrm{W}$。(M1)
约为 $0.57\ \mathrm{kW}$，主要以热能（在电机绕组与轴承中）耗散，少量为声。(A1)
Trolley A ($2.0\ \mathrm{kg}$, $4.0\ \mathrm{m\,s^{-1}}$) couples to stationary trolley B ($6.0\ \mathrm{kg}$). (a) conservation law and common velocity; (b) kinetic energy before, after, and lost; (c) elastic or inelastic, and where the energy went.小车 A（$2.0\ \mathrm{kg}$，$4.0\ \mathrm{m\,s^{-1}}$）与静止的小车 B（$6.0\ \mathrm{kg}$）连接。(a) 守恒定律与共同速度；(b) 碰前、碰后及损失的动能；(c) 弹性还是非弹性，能量去向。
With no external force, total momentum is conserved (this holds for every collision). (M1)
$$ m_1 u_1 = (m_1 + m_2)v \;\Rightarrow\; (2.0)(4.0) = (8.0)v. \quad (\text{M1}) $$ $$ v = \frac{8.0}{8.0} = 1.0\ \mathrm{m\,s^{-1}}. $$(A1)
Before: only A moves. After: the $8.0\ \mathrm{kg}$ pair moves at $1.0\ \mathrm{m\,s^{-1}}$. (M1)
$$ E_k^{\text{before}} = \tfrac{1}{2}(2.0)(4.0)^{2} = 16\ \mathrm{J}, \qquad E_k^{\text{after}} = \tfrac{1}{2}(8.0)(1.0)^{2} = 4.0\ \mathrm{J}. $$(A1)
$$ \Delta E_k = 16 - 4.0 = 12\ \mathrm{J}\ \text{lost}. $$(A1)
Kinetic energy is not conserved ($16 \to 4.0\ \mathrm{J}$), so the collision is inelastic; the missing $12\ \mathrm{J}$ becomes heat, sound, and permanent deformation of the coupling. (B1)
无外力时总动量守恒（对每种碰撞都成立）。(M1)
$$ m_1 u_1 = (m_1 + m_2)v \;\Rightarrow\; (2.0)(4.0) = (8.0)v. \quad (\text{M1}) $$ $$ v = \frac{8.0}{8.0} = 1.0\ \mathrm{m\,s^{-1}}. $$(A1)
碰前：仅 A 运动。碰后：$8.0\ \mathrm{kg}$ 的整体以 $1.0\ \mathrm{m\,s^{-1}}$ 运动。(M1)
$$ E_k^{\text{before}} = \tfrac{1}{2}(2.0)(4.0)^{2} = 16\ \mathrm{J}, \qquad E_k^{\text{after}} = \tfrac{1}{2}(8.0)(1.0)^{2} = 4.0\ \mathrm{J}. $$(A1)
$$ \Delta E_k = 16 - 4.0 = 12\ \mathrm{J}\ \text{损失}. $$(A1)
动能不守恒（$16 \to 4.0\ \mathrm{J}$），故为非弹性碰撞；缺失的 $12\ \mathrm{J}$ 转为热、声以及连接处的永久形变。(B1)
$F$ vs $x$ data for a stretched spring. (a) state Hooke's law and why $F$ vs $x$ is linear through the origin; (b) gradient and spring constant; (c) elastic PE at $x = 0.40\ \mathrm{m}$ from the area; (d) percentage uncertainty in $F$ at $x = 0.40\ \mathrm{m}$ given $\pm 0.2\ \mathrm{N}$.拉伸弹簧的 $F$ 对 $x$ 数据。(a) 陈述胡克定律及 $F$ 对 $x$ 为过原点直线的原因；(b) 斜率与劲度系数；(c) 由面积求 $x = 0.40\ \mathrm{m}$ 时的弹性势能；(d) 给定 $\pm 0.2\ \mathrm{N}$ 时 $x = 0.40\ \mathrm{m}$ 处 $F$ 的百分比不确定度。
Hooke's law: the force needed to stretch a spring is proportional to the extension, $F = kx$, while the spring is not overstretched. (M1)
This has the form $F = (\text{gradient})\times x$ with no intercept, so a plot of $F$ against $x$ is a straight line through the origin. (A1)
Comparing with $y = mx$, the gradient equals the spring constant $k$. (A1)
Read the gradient from two well-separated points, $(0.10,\,2.5)$ and $(0.40,\,10.0)$: (M1)
$$ \text{gradient} = \frac{10.0 - 2.5}{0.40 - 0.10} = \frac{7.5}{0.30} = 25\ \mathrm{N\,m^{-1}}. $$(A1)
Since the gradient is the spring constant, $k = 25\ \mathrm{N\,m^{-1}}$. (A1)
The elastic PE is the area under the $F$-$x$ line up to $x = 0.40\ \mathrm{m}$, a triangle of height $F = 10.0\ \mathrm{N}$: (M1)
$$ E_{\text{el}} = \tfrac{1}{2}(0.40)(10.0) = 2.0\ \mathrm{J}. $$(A1) (equivalently $\tfrac{1}{2}kx^{2} = \tfrac{1}{2}(25)(0.40)^{2} = 2.0\ \mathrm{J}$.)
At $x = 0.40\ \mathrm{m}$, $F = 10.0\ \mathrm{N}$ with absolute uncertainty $\pm 0.2\ \mathrm{N}$: (M1)
$$ \frac{0.2}{10.0}\times 100\% = 2\%. $$(A1)
胡克定律：在弹簧未被过度拉伸时，拉伸弹簧所需的力与伸长量成正比，$F = kx$。(M1)
此式形如 $F = (\text{斜率})\times x$，无截距，故 $F$ 对 $x$ 作图为过原点的直线。(A1)
与 $y = mx$ 比较，斜率等于劲度系数 $k$。(A1)
用相距较远的两点读斜率，$(0.10,\,2.5)$ 与 $(0.40,\,10.0)$：(M1)
$$ \text{斜率} = \frac{10.0 - 2.5}{0.40 - 0.10} = \frac{7.5}{0.30} = 25\ \mathrm{N\,m^{-1}}. $$(A1)
因斜率即劲度系数，$k = 25\ \mathrm{N\,m^{-1}}$。(A1)
弹性势能是 $F$-$x$ 图线到 $x = 0.40\ \mathrm{m}$ 处下的面积，是高 $F = 10.0\ \mathrm{N}$ 的三角形：(M1)
$$ E_{\text{el}} = \tfrac{1}{2}(0.40)(10.0) = 2.0\ \mathrm{J}. $$(A1)（等价地 $\tfrac{1}{2}kx^{2} = \tfrac{1}{2}(25)(0.40)^{2} = 2.0\ \mathrm{J}$。）
$x = 0.40\ \mathrm{m}$ 处 $F = 10.0\ \mathrm{N}$，绝对不确定度 $\pm 0.2\ \mathrm{N}$：(M1)
$$ \frac{0.2}{10.0}\times 100\% = 2\%. $$(A1)
A $5.0\ \mathrm{kg}$ cart on a frictionless track is pushed by a force: $0 \to 40\ \mathrm{N}$ over $0$ to $4.0\ \mathrm{m}$, constant $40\ \mathrm{N}$ from $4.0$ to $8.0\ \mathrm{m}$, then $40 \to 0\ \mathrm{N}$ from $8.0$ to $10.0\ \mathrm{m}$, from rest. (a) why area $=$ work, with the integral; (b) work over first $4.0\ \mathrm{m}$; (c) total work over $10.0\ \mathrm{m}$; (d) final speed; (e) where acceleration is greatest.无摩擦轨道上 $5.0\ \mathrm{kg}$ 的小车受力推动：$0$ 至 $4.0\ \mathrm{m}$ 从 $0$ 增至 $40\ \mathrm{N}$，$4.0$ 至 $8.0\ \mathrm{m}$ 恒为 $40\ \mathrm{N}$，$8.0$ 至 $10.0\ \mathrm{m}$ 从 $40$ 减至 $0\ \mathrm{N}$，从静止开始。(a) 面积为何等于功及积分；(b) 前 $4.0\ \mathrm{m}$ 的功；(c) $10.0\ \mathrm{m}$ 的总功；(d) 末速率；(e) 加速度最大处。
Over an infinitesimal displacement $ds$ the force is effectively constant and does work $dW = F\,ds$, which is the area of a thin strip under the graph. Summing the strips: (M1)
$$ W = \int_{0}^{10} F(s)\,ds = \text{area under the } F\text{-}s \text{ graph}. $$(A1)
This segment is a triangle from $0$ to $40\ \mathrm{N}$ over $4.0\ \mathrm{m}$: (M1)
$$ W_1 = \tfrac{1}{2}(4.0)(40) = 80\ \mathrm{J}. $$(A1)
Add the rectangle ($4.0$ to $8.0\ \mathrm{m}$) and the final triangle ($8.0$ to $10.0\ \mathrm{m}$) to the first triangle: (M1)
$$ W = \underbrace{\tfrac{1}{2}(4.0)(40)}_{80} + \underbrace{(4.0)(40)}_{160} + \underbrace{\tfrac{1}{2}(2.0)(40)}_{40}. $$(M1 for setting up all three areas)
$$ W = 80 + 160 + 40 = 280\ \mathrm{J}. $$(A1)
The track is frictionless, so all the work becomes kinetic energy. By the work-energy theorem $W = \tfrac{1}{2}mv^{2}$ from rest: (M1)
$$ v = \sqrt{\frac{2W}{m}} = \sqrt{\frac{2(280)}{5.0}}. \quad (\text{M1}) $$ $$ v = \sqrt{112} \approx 11\ \mathrm{m\,s^{-1}}. $$(A1)
The acceleration is greatest where the force is greatest, which is the flat region $4.0\ \mathrm{m} \le s \le 8.0\ \mathrm{m}$ where $F = 40\ \mathrm{N}$. (B1)
By Newton's second law $a = F/m$, so on a frictionless track the acceleration tracks the force exactly. (R1)
在无穷小位移 $ds$ 内力近似恒定，做功 $dW = F\,ds$，即图下一条窄条的面积。把这些窄条相加：(M1)
$$ W = \int_{0}^{10} F(s)\,ds = F\text{-}s \text{ 图下的面积}. $$(A1)
该段是从 $0$ 到 $40\ \mathrm{N}$、跨 $4.0\ \mathrm{m}$ 的三角形：(M1)
$$ W_1 = \tfrac{1}{2}(4.0)(40) = 80\ \mathrm{J}. $$(A1)
把矩形（$4.0$ 至 $8.0\ \mathrm{m}$）与末段三角形（$8.0$ 至 $10.0\ \mathrm{m}$）加到第一个三角形上：(M1)
$$ W = \underbrace{\tfrac{1}{2}(4.0)(40)}_{80} + \underbrace{(4.0)(40)}_{160} + \underbrace{\tfrac{1}{2}(2.0)(40)}_{40}. $$(列出全部三块面积得 M1)
$$ W = 80 + 160 + 40 = 280\ \mathrm{J}. $$(A1)
轨道无摩擦，故全部功转为动能。由动能定理（从静止）$W = \tfrac{1}{2}mv^{2}$：(M1)
$$ v = \sqrt{\frac{2W}{m}} = \sqrt{\frac{2(280)}{5.0}}. \quad (\text{M1}) $$ $$ v = \sqrt{112} \approx 11\ \mathrm{m\,s^{-1}}. $$(A1)
加速度在力最大处最大，即 $F = 40\ \mathrm{N}$ 的平坦区 $4.0\ \mathrm{m} \le s \le 8.0\ \mathrm{m}$。(B1)
由牛顿第二定律 $a = F/m$，故无摩擦轨道上加速度与力完全同步。(R1)
A $4.0\ \mathrm{kg}$ block released from rest down a $5.0\ \mathrm{m}$ ramp at $30^{\circ}$ reaches the bottom at $4.0\ \mathrm{m\,s^{-1}}$. (a) vertical height; (b) PE lost; (c) KE gained; (d) thermal energy from friction; (e) average friction force; (f) frictionless speed and why it is greater.$4.0\ \mathrm{kg}$ 木块从 $30^{\circ}$、长 $5.0\ \mathrm{m}$ 斜面由静止释放，到底端速率 $4.0\ \mathrm{m\,s^{-1}}$。(a) 竖直高度；(b) 损失的势能；(c) 获得的动能；(d) 摩擦生热；(e) 平均摩擦力；(f) 无摩擦时速率及为何更大。
The drop is the ramp length times $\sin 30^{\circ}$: $h = 5.0\sin 30^{\circ} = 5.0(0.5) = 2.5\ \mathrm{m}$. (A1)
$\Delta E_p = mgh$: (M1)
$$ \Delta E_p = (4.0)(9.81)(2.5) = 98.1\ \mathrm{J} \approx 98\ \mathrm{J}. $$(A1)
From rest, $E_k = \tfrac{1}{2}mv^{2}$: (M1)
$$ E_k = \tfrac{1}{2}(4.0)(4.0)^{2} = \tfrac{1}{2}(4.0)(16) = 32\ \mathrm{J}. $$(A1)
Energy is conserved, so the PE lost equals the KE gained plus the heat: $Q = \Delta E_p - E_k$. (M1)
$$ Q = 98.1 - 32 = 66.1\ \mathrm{J} \approx 66\ \mathrm{J}. $$(A1)
The heat equals the friction force times the distance moved along the ramp: $Q = F_f \, d$ with $d = 5.0\ \mathrm{m}$. (M1)
$$ F_f = \frac{Q}{d} = \frac{66.1}{5.0} = 13.2\ \mathrm{N} \approx 13\ \mathrm{N}. $$(A1)
Without friction all the PE becomes KE: $mgh = \tfrac{1}{2}mv^{2}$, so $v = \sqrt{2gh} = \sqrt{2(9.81)(2.5)} = \sqrt{49.05} \approx 7.0\ \mathrm{m\,s^{-1}}$. (M1)
It exceeds $4.0\ \mathrm{m\,s^{-1}}$ because, with friction present, part of the gravitational PE is diverted into heat rather than kinetic energy. (A1)
落差为斜面长乘 $\sin 30^{\circ}$：$h = 5.0\sin 30^{\circ} = 5.0(0.5) = 2.5\ \mathrm{m}$。(A1)
$\Delta E_p = mgh$：(M1)
$$ \Delta E_p = (4.0)(9.81)(2.5) = 98.1\ \mathrm{J} \approx 98\ \mathrm{J}. $$(A1)
从静止，$E_k = \tfrac{1}{2}mv^{2}$：(M1)
$$ E_k = \tfrac{1}{2}(4.0)(4.0)^{2} = \tfrac{1}{2}(4.0)(16) = 32\ \mathrm{J}. $$(A1)
能量守恒，故损失的势能等于获得的动能加上热：$Q = \Delta E_p - E_k$。(M1)
$$ Q = 98.1 - 32 = 66.1\ \mathrm{J} \approx 66\ \mathrm{J}. $$(A1)
热等于摩擦力乘沿斜面移动的距离：$Q = F_f \, d$，$d = 5.0\ \mathrm{m}$。(M1)
$$ F_f = \frac{Q}{d} = \frac{66.1}{5.0} = 13.2\ \mathrm{N} \approx 13\ \mathrm{N}. $$(A1)
无摩擦时全部势能转为动能：$mgh = \tfrac{1}{2}mv^{2}$，故 $v = \sqrt{2gh} = \sqrt{2(9.81)(2.5)} = \sqrt{49.05} \approx 7.0\ \mathrm{m\,s^{-1}}$。(M1)
它大于 $4.0\ \mathrm{m\,s^{-1}}$，因为存在摩擦时部分重力势能被转入热而非动能。(A1)
Puck A ($0.20\ \mathrm{kg}$, $3.0\ \mathrm{m\,s^{-1}}$ east) collides elastically with identical stationary puck B; A leaves at $30^{\circ}$ north of east. (a) the equal-mass elastic angle result; (b) direction of B; (c) speed of each puck; (d) verify kinetic energy is conserved.冰球 A（$0.20\ \mathrm{kg}$，向东 $3.0\ \mathrm{m\,s^{-1}}$）与相同的静止冰球 B 弹性碰撞；A 以东偏北 $30^{\circ}$ 离开。(a) 等质量弹性碰撞的夹角结论；(b) B 的方向；(c) 每个冰球的速率；(d) 验证动能守恒。
When a moving body strikes an identical stationary body elastically, the two bodies separate at exactly $90^{\circ}$ to each other. (B1)
This follows from combining momentum conservation ($\vec{u} = \vec{v}_A + \vec{v}_B$) with kinetic energy conservation ($u^{2} = v_A^{2} + v_B^{2}$), which together force $\vec{v}_A \cdot \vec{v}_B = 0$. (R1)
A leaves at $30^{\circ}$ north of east, so B must leave $90^{\circ}$ away from A: (M1)
B travels at $90^{\circ} - 30^{\circ} = 60^{\circ}$ on the other side of the original line, i.e. $60^{\circ}$ south of east. (A1)
Conserve momentum along the original (east) direction and perpendicular (north) to it, with $u = 3.0\ \mathrm{m\,s^{-1}}$ and equal masses (which cancel): (M1)
$$ \text{north: } 0 = v_A\sin 30^{\circ} - v_B\sin 60^{\circ} \;\Rightarrow\; v_A(0.5) = v_B(0.866). $$ $$ \text{east: } 3.0 = v_A\cos 30^{\circ} + v_B\cos 60^{\circ} = v_A(0.866) + v_B(0.5). \quad (\text{M1}) $$From the north equation $v_A = 1.732\,v_B$; substituting into the east equation gives $3.0 = 1.5\,v_B + 0.5\,v_B = 2.0\,v_B$, so $v_B = 1.5\ \mathrm{m\,s^{-1}}$ and $v_A = 2.598 \approx 2.6\ \mathrm{m\,s^{-1}}$. (A1)
Before: $E_k = \tfrac{1}{2}(0.20)(3.0)^{2} = 0.90\ \mathrm{J}$. After: (M1)
$$ E_k = \tfrac{1}{2}(0.20)\left(2.598^{2} + 1.5^{2}\right) = 0.10\,(6.75 + 2.25) = 0.10(9.00) = 0.90\ \mathrm{J}. $$The totals match, confirming the collision is elastic. (A1)
当运动物体弹性撞上一个相同的静止物体时，两物体以恰好 $90^{\circ}$ 相互分开。(B1)
这由动量守恒（$\vec{u} = \vec{v}_A + \vec{v}_B$）与动能守恒（$u^{2} = v_A^{2} + v_B^{2}$）联立得出，两者共同迫使 $\vec{v}_A \cdot \vec{v}_B = 0$。(R1)
A 以东偏北 $30^{\circ}$ 离开，故 B 必与 A 成 $90^{\circ}$ 离开：(M1)
B 在原方向线另一侧以 $90^{\circ} - 30^{\circ} = 60^{\circ}$ 飞出，即东偏南 $60^{\circ}$。(A1)
沿原方向（东）及其垂直方向（北）分别用动量守恒，$u = 3.0\ \mathrm{m\,s^{-1}}$，等质量（约去）：(M1)
$$ \text{北: } 0 = v_A\sin 30^{\circ} - v_B\sin 60^{\circ} \;\Rightarrow\; v_A(0.5) = v_B(0.866). $$ $$ \text{东: } 3.0 = v_A\cos 30^{\circ} + v_B\cos 60^{\circ} = v_A(0.866) + v_B(0.5). \quad (\text{M1}) $$由北向方程 $v_A = 1.732\,v_B$；代入东向方程得 $3.0 = 1.5\,v_B + 0.5\,v_B = 2.0\,v_B$，故 $v_B = 1.5\ \mathrm{m\,s^{-1}}$，$v_A = 2.598 \approx 2.6\ \mathrm{m\,s^{-1}}$。(A1)
碰前：$E_k = \tfrac{1}{2}(0.20)(3.0)^{2} = 0.90\ \mathrm{J}$。碰后：(M1)
$$ E_k = \tfrac{1}{2}(0.20)\left(2.598^{2} + 1.5^{2}\right) = 0.10\,(6.75 + 2.25) = 0.10(9.00) = 0.90\ \mathrm{J}. $$总量相等，确认碰撞为弹性。(A1)
A $1200\ \mathrm{kg}$ car climbs a $5.0^{\circ}$ road at constant $15\ \mathrm{m\,s^{-1}}$ against a $500\ \mathrm{N}$ resistive force. (a) weight component down the slope; (b) driving force; (c) useful output power; (d) fuel-energy rate at $25\%$ efficiency; (e) power dissipated against resistance, and why it is less than the output power.$1200\ \mathrm{kg}$ 的汽车以恒定 $15\ \mathrm{m\,s^{-1}}$ 沿 $5.0^{\circ}$ 的路上坡，克服 $500\ \mathrm{N}$ 阻力。(a) 重力沿坡向下的分量；(b) 驱动力；(c) 有用输出功率；(d) $25\%$ 效率时燃料能速率；(e) 克服阻力耗散的功率及其为何小于输出功率。
The component of weight along an incline is $mg\sin\theta$: (M1)
$$ mg\sin\theta = (1200)(9.81)\sin 5.0^{\circ} = 11772 \times 0.0872 \approx 1.03\times 10^{3}\ \mathrm{N}. $$(A1)
At constant speed the resultant force is zero, so the driving force balances both the weight component and the resistive force. (M1)
$$ F = mg\sin\theta + F_{\text{res}} = 1026 + 500 = 1526\ \mathrm{N} \approx 1.5\times 10^{3}\ \mathrm{N}. $$(A1)
$P = Fv$ with the driving force and the constant speed: (M1)
$$ P_{\text{out}} = (1526)(15) = 22890\ \mathrm{W} \approx 22.9\ \mathrm{kW}. $$(A1)
$\eta = \dfrac{P_{\text{useful}}}{P_{\text{input}}}$, so $P_{\text{input}} = \dfrac{P_{\text{out}}}{\eta}$: (M1)
$$ P_{\text{input}} = \frac{22890}{0.25} = 91560\ \mathrm{W} \approx 92\ \mathrm{kW}. $$(A1)
The power spent against the resistive force is $P = F_{\text{res}}\,v = (500)(15) = 7500\ \mathrm{W} = 7.5\ \mathrm{kW}$. (M1)
It is less than the $22.9\ \mathrm{kW}$ output because the remaining $\approx 15.4\ \mathrm{kW}$ goes into raising the car's gravitational PE as it climbs. (A1)
重力沿斜面的分量为 $mg\sin\theta$：(M1)
$$ mg\sin\theta = (1200)(9.81)\sin 5.0^{\circ} = 11772 \times 0.0872 \approx 1.03\times 10^{3}\ \mathrm{N}. $$(A1)
匀速时合力为零，故驱动力同时平衡重力分量与阻力。(M1)
$$ F = mg\sin\theta + F_{\text{res}} = 1026 + 500 = 1526\ \mathrm{N} \approx 1.5\times 10^{3}\ \mathrm{N}. $$(A1)
用驱动力与恒定速率代入 $P = Fv$：(M1)
$$ P_{\text{out}} = (1526)(15) = 22890\ \mathrm{W} \approx 22.9\ \mathrm{kW}. $$(A1)
$\eta = \dfrac{P_{\text{useful}}}{P_{\text{input}}}$，故 $P_{\text{input}} = \dfrac{P_{\text{out}}}{\eta}$：(M1)
$$ P_{\text{input}} = \frac{22890}{0.25} = 91560\ \mathrm{W} \approx 92\ \mathrm{kW}. $$(A1)
克服阻力所耗功率为 $P = F_{\text{res}}\,v = (500)(15) = 7500\ \mathrm{W} = 7.5\ \mathrm{kW}$。(M1)
它小于 $22.9\ \mathrm{kW}$ 的输出，因为其余约 $15.4\ \mathrm{kW}$ 用于在上坡时增加汽车的重力势能。(A1)