Companion to the IB-Style Practice SetIB 风格练习题的解析配套
Syllabus A2.1 to A2.7考纲 A2.1 至 A2.7PHYSICS HL
A $8.0\ \mathrm{kg}$ lamp hangs from two cables, each at $25^{\circ}$ to the horizontal ceiling. (a) FBD and equilibrium conditions; (b) tension in each cable; (c) effect of a smaller angle.$8.0\ \mathrm{kg}$ 的灯由两条与水平天花板各成 $25^{\circ}$ 的绳悬挂。(a) 受力图与平衡条件；(b) 每条绳张力；(c) 夹角更小时的影响。
Three forces act at the knot: the weight $W = mg$ vertically down, and the two equal tensions $T$ directed up and outward along each cable at $25^{\circ}$ above the horizontal. (B1)
By symmetry the horizontal components cancel: $\Sigma F_{x} = T\cos 25^{\circ} - T\cos 25^{\circ} = 0$. Vertically: $\Sigma F_{y} = 2T\sin 25^{\circ} - mg = 0$. (A1)
From the vertical condition $2T\sin 25^{\circ} = mg$: (M1)
$$ T = \frac{mg}{2\sin 25^{\circ}} = \frac{(8.0)(9.81)}{2\sin 25^{\circ}} = \frac{78.48}{0.8452} \approx 92.8\ \mathrm{N} \approx 93\ \mathrm{N}. $$(A1)
As the angle to the ceiling decreases, $\sin 25^{\circ}$ is replaced by a smaller value, so the denominator $2\sin\theta$ shrinks and $T$ rises. (A1)
Physically the cables become more nearly horizontal, so each must pull much harder to muster enough vertical component to hold the weight. As $\theta \to 0$ the tension diverges. (R1)
结点受三个力：竖直向下的重力 $W = mg$，以及沿两绳与水平方向成 $25^{\circ}$ 向上、向外的两个相等张力 $T$。(B1)
由对称性水平分量抵消：$\Sigma F_{x} = T\cos 25^{\circ} - T\cos 25^{\circ} = 0$。竖直方向：$\Sigma F_{y} = 2T\sin 25^{\circ} - mg = 0$。(A1)
由竖直条件 $2T\sin 25^{\circ} = mg$：(M1)
$$ T = \frac{mg}{2\sin 25^{\circ}} = \frac{(8.0)(9.81)}{2\sin 25^{\circ}} = \frac{78.48}{0.8452} \approx 92.8\ \mathrm{N} \approx 93\ \mathrm{N}. $$(A1)
当与天花板的夹角减小时，$\sin 25^{\circ}$ 被更小的值取代，分母 $2\sin\theta$ 变小，故 $T$ 增大。(A1)
物理上绳越接近水平，每条绳就必须拉得越猛，才能凑出足够的竖直分量托住重物。当 $\theta \to 0$ 时张力发散。(R1)
A $65\ \mathrm{kg}$ student stands on a scale in a lift. (a) weight and why mass is location-independent; (b) reading when accelerating up at $1.8\ \mathrm{m\,s^{-2}}$; (c) reading when accelerating down at $1.8\ \mathrm{m\,s^{-2}}$ and in free fall.$65\ \mathrm{kg}$ 的学生站在电梯体重计上。(a) 重量及质量为何与地点无关；(b) 以 $1.8\ \mathrm{m\,s^{-2}}$ 向上加速时的读数；(c) 以 $1.8\ \mathrm{m\,s^{-2}}$ 向下加速及自由落体时的读数。
$W = mg = (65)(9.81) = 637.65 \approx 638\ \mathrm{N}$. (A1)
Mass measures the quantity of matter and is the same everywhere; weight is the gravitational force $mg$, which changes with $g$. On the Moon $g$ is smaller, so $W$ falls while $m$ stays $65\ \mathrm{kg}$. (R1)
Take up as positive. Newton's second law on the student: $N - mg = ma$, so $N = m(g + a)$. (M1)
$$ N = 65(9.81 + 1.8) = 65(11.61) \approx 755\ \mathrm{N}. $$(A1)
Now $a$ is downward, so $N = m(g - a) = 65(9.81 - 1.8) = 65(8.01) \approx 521\ \mathrm{N}$. (M1)
In free fall $a = g$, giving $N = m(g - g) = 0$: the scale reads zero, the sensation of weightlessness. (A1)
$W = mg = (65)(9.81) = 637.65 \approx 638\ \mathrm{N}$。(A1)
质量是物质的量度，处处相同；重量是引力 $mg$，随 $g$ 改变。月球上 $g$ 更小，故 $W$ 减小而 $m$ 仍为 $65\ \mathrm{kg}$。(R1)
取上为正。对学生用牛顿第二定律：$N - mg = ma$，故 $N = m(g + a)$。(M1)
$$ N = 65(9.81 + 1.8) = 65(11.61) \approx 755\ \mathrm{N}. $$(A1)
此时 $a$ 向下，故 $N = m(g - a) = 65(9.81 - 1.8) = 65(8.01) \approx 521\ \mathrm{N}$。(M1)
自由落体时 $a = g$，得 $N = m(g - g) = 0$：体重计读数为零，即失重感。(A1)
$m_{1} = 2.0\ \mathrm{kg}$ and $m_{2} = 3.0\ \mathrm{kg}$ over a frictionless pulley, released from rest. (a) state N3L and the partner of the string's pull on $m_{2}$; (b) acceleration; (c) tension and why it is not a weight.$m_{1} = 2.0\ \mathrm{kg}$ 与 $m_{2} = 3.0\ \mathrm{kg}$ 跨过无摩擦滑轮，从静止释放。(a) 陈述牛顿第三定律及绳对 $m_{2}$ 拉力的配对力；(b) 加速度；(c) 张力及它为何不等于某个重量。
If body A exerts a force on body B, then B exerts an equal-magnitude, opposite-direction force on A, of the same type, acting on a different body. (B1)
The partner of "the string pulls up on $m_{2}$" is "$m_{2}$ pulls down on the string" with equal magnitude. (Note it is not the weight of $m_{2}$, nor the tension on $m_{1}$.) (A1)
Treat the two masses as one system driven by the weight difference, resisted by the total inertia: $a = \dfrac{(m_{2} - m_{1})g}{m_{1} + m_{2}}$. (M1)
$$ a = \frac{(3.0 - 2.0)(9.81)}{2.0 + 3.0} = \frac{9.81}{5.0} \approx 1.96\ \mathrm{m\,s^{-2}}. $$(A1)
Apply Newton's second law to the lighter rising mass, $T - m_{1}g = m_{1}a$, so $T = m_{1}(g + a)$: (M1)
$$ T = 2.0(9.81 + 1.96) = 2.0(11.77) \approx 23.5\ \mathrm{N}. $$The tension lies between the two weights ($m_{1}g = 19.6\ \mathrm{N}$ and $m_{2}g = 29.4\ \mathrm{N}$) because the string must both support and accelerate each block; it equals a weight only when $a = 0$. (A1)
若 A 对 B 施力，则 B 对 A 施加大小相等、方向相反、同类型且作用于另一物体的力。(B1)
"绳向上拉 $m_{2}$"的配对力是"$m_{2}$ 向下拉绳"，大小相等。（注意它不是 $m_{2}$ 的重量，也不是作用于 $m_{1}$ 的张力。）(A1)
把两质量视为一个系统，由重量差驱动、受总惯性阻碍：$a = \dfrac{(m_{2} - m_{1})g}{m_{1} + m_{2}}$。(M1)
$$ a = \frac{(3.0 - 2.0)(9.81)}{2.0 + 3.0} = \frac{9.81}{5.0} \approx 1.96\ \mathrm{m\,s^{-2}}. $$(A1)
对较轻、上升的物块用牛顿第二定律，$T - m_{1}g = m_{1}a$，故 $T = m_{1}(g + a)$：(M1)
$$ T = 2.0(9.81 + 1.96) = 2.0(11.77) \approx 23.5\ \mathrm{N}. $$张力介于两重量之间（$m_{1}g = 19.6\ \mathrm{N}$ 与 $m_{2}g = 29.4\ \mathrm{N}$），因为绳既要支撑又要使每个物块加速；只有 $a = 0$ 时它才等于某个重量。(A1)
$15\ \mathrm{kg}$ crate on a floor, $\mu_{s} = 0.45$, $\mu_{k} = 0.30$. (a) does it move under $60\ \mathrm{N}$, and the friction force; (b) acceleration under $90\ \mathrm{N}$; (c) why it lurches at the start of sliding.地面上 $15\ \mathrm{kg}$ 箱子，$\mu_{s} = 0.45$、$\mu_{k} = 0.30$。(a) $60\ \mathrm{N}$ 下是否移动及摩擦力；(b) $90\ \mathrm{N}$ 下的加速度；(c) 刚滑动时为何猛冲。
On a level floor $N = mg = (15)(9.81) = 147.15\ \mathrm{N}$. Maximum static friction is $f_{s,\max} = \mu_{s}N$: (M1)
$$ f_{s,\max} = (0.45)(147.15) \approx 66.2\ \mathrm{N}. $$Since the applied $60\ \mathrm{N} < 66.2\ \mathrm{N}$, the crate does not move. (A1)
The static friction therefore takes exactly the value needed for equilibrium, $f = 60\ \mathrm{N}$ (not $66\ \mathrm{N}$, not $\mu_{k}N$). (A1)
Now $90\ \mathrm{N} > 66.2\ \mathrm{N}$, so the crate slides and kinetic friction acts: $f_{k} = \mu_{k}N = (0.30)(147.15) = 44.1\ \mathrm{N}$. Newton's second law gives $a = (F - f_{k})/m$: (M1)
$$ a = \frac{90 - 44.1}{15} = \frac{45.9}{15} \approx 3.1\ \mathrm{m\,s^{-2}}. $$(A1)
At the instant of slipping, the friction drops from its static maximum $\mu_{s}N$ to the smaller kinetic value $\mu_{k}N$. The driving force, having just overcome the larger value, now exceeds the smaller one, so there is a sudden surge of net force and acceleration. (R1)
水平地面上 $N = mg = (15)(9.81) = 147.15\ \mathrm{N}$。最大静摩擦力为 $f_{s,\max} = \mu_{s}N$：(M1)
$$ f_{s,\max} = (0.45)(147.15) \approx 66.2\ \mathrm{N}. $$由于所加 $60\ \mathrm{N} < 66.2\ \mathrm{N}$，箱子不移动。(A1)
故静摩擦力恰取维持平衡所需之值，$f = 60\ \mathrm{N}$（不是 $66\ \mathrm{N}$，也不是 $\mu_{k}N$）。(A1)
此时 $90\ \mathrm{N} > 66.2\ \mathrm{N}$，箱子滑动，由动摩擦力作用：$f_{k} = \mu_{k}N = (0.30)(147.15) = 44.1\ \mathrm{N}$。牛顿第二定律给出 $a = (F - f_{k})/m$：(M1)
$$ a = \frac{90 - 44.1}{15} = \frac{45.9}{15} \approx 3.1\ \mathrm{m\,s^{-2}}. $$(A1)
滑动瞬间，摩擦力从静摩擦最大值 $\mu_{s}N$ 骤降至较小的动摩擦值 $\mu_{k}N$。驱动力刚克服较大值，此刻已超过较小值，于是净力与加速度突然激增。(R1)
$0.25\ \mathrm{kg}$ ball on a light string in a vertical circle, $r = 0.90\ \mathrm{m}$. (a) why "centripetal force" is not a new force and the net-force direction; (b) minimum speed at the top; (c) tension at the bottom when $v = 4.0\ \mathrm{m\,s^{-1}}$.$0.25\ \mathrm{kg}$ 小球用轻绳在竖直圆中旋转，$r = 0.90\ \mathrm{m}$。(a) 为何"向心力"非新力及合力方向；(b) 圆顶最小速率；(c) $v = 4.0\ \mathrm{m\,s^{-1}}$ 时最低点张力。
"Centripetal force" is simply the name for the net (resultant) of the real forces, here gravity and tension, that happens to point toward the centre. It is not an extra force to add to the free-body diagram. (A1)
In uniform circular motion the net force, and hence the acceleration, points toward the centre of the circle at every instant. (A1)
At the top both weight and tension point down (toward the centre): $T + mg = \dfrac{mv^{2}}{r}$. The minimum speed is when the string is just about to go slack, $T = 0$, so $mg = \dfrac{mv_{\min}^{2}}{r}$, giving $v_{\min} = \sqrt{gr}$. (M1)
$$ v_{\min} = \sqrt{(9.81)(0.90)} = \sqrt{8.829} \approx 2.97\ \mathrm{m\,s^{-1}}. $$(A1)
At the lowest point tension acts up (toward the centre) and weight acts down: $T - mg = \dfrac{mv^{2}}{r}$, so $T = m\left(\dfrac{v^{2}}{r} + g\right)$. (M1)
$$ T = 0.25\left(\frac{4.0^{2}}{0.90} + 9.81\right) = 0.25(17.78 + 9.81) = 0.25(27.59) \approx 6.9\ \mathrm{N}. $$(A1)
"向心力"只是真实力（此处为重力与张力）合力的名称，它恰好指向圆心。它不是要额外加进受力图的力。(A1)
匀速圆周运动中，合力以及加速度在每一时刻都指向圆心。(A1)
圆顶处重力与张力都向下（指向圆心）：$T + mg = \dfrac{mv^{2}}{r}$。最小速率对应绳恰要松弛，$T = 0$，故 $mg = \dfrac{mv_{\min}^{2}}{r}$，得 $v_{\min} = \sqrt{gr}$。(M1)
$$ v_{\min} = \sqrt{(9.81)(0.90)} = \sqrt{8.829} \approx 2.97\ \mathrm{m\,s^{-1}}. $$(A1)
最低点处张力向上（指向圆心）、重力向下：$T - mg = \dfrac{mv^{2}}{r}$，故 $T = m\left(\dfrac{v^{2}}{r} + g\right)$。(M1)
$$ T = 0.25\left(\frac{4.0^{2}}{0.90} + 9.81\right) = 0.25(17.78 + 9.81) = 0.25(27.59) \approx 6.9\ \mathrm{N}. $$(A1)
$0.60\ \mathrm{kg}$ trolley from rest, force rises $0\to 18\ \mathrm{N}$ over $0$-$0.20\ \mathrm{s}$, constant $18\ \mathrm{N}$ to $0.50\ \mathrm{s}$, falls to $0$ by $0.60\ \mathrm{s}$. (a) meaning of area + impulse-momentum theorem; (b) total impulse; (c) final speed; (d) average force and why it gives the same speed.$0.60\ \mathrm{kg}$ 小车从静止，力在 $0$-$0.20\ \mathrm{s}$ 由 $0$ 升到 $18\ \mathrm{N}$，到 $0.50\ \mathrm{s}$ 保持 $18\ \mathrm{N}$，到 $0.60\ \mathrm{s}$ 降为 $0$。(a) 面积含义与动量定理；(b) 总冲量；(c) 末速度；(d) 平均力及其为何给出相同速度。
The area under a force-time graph equals the impulse $J = \int F\,\mathrm{d}t$. (B1)
The impulse-momentum theorem states that the net impulse on a body equals its change of momentum: $J = \Delta p = mv - mu$. (A1)
Split the graph into a rise triangle, a flat rectangle, and a fall triangle. (M1)
$$ J = \underbrace{\tfrac{1}{2}(0.20)(18)}_{1.8} + \underbrace{(0.30)(18)}_{5.4} + \underbrace{\tfrac{1}{2}(0.10)(18)}_{0.9}. $$(M1 for all three areas)
$$ J = 1.8 + 5.4 + 0.9 = 8.1\ \mathrm{N\,s}. $$(A1)
Since the trolley starts from rest, $J = \Delta p = mv - 0$, so $v = J/m$: (M1)
$$ v = \frac{8.1}{0.60} = 13.5\ \mathrm{m\,s^{-1}}. $$(A1)
The average force over the whole interval is $F_{\text{avg}} = J/\Delta t$: (M1)
$$ F_{\text{avg}} = \frac{8.1}{0.60} = 13.5\ \mathrm{N}. $$(A1)
A constant force equal to the average delivers the same area, hence the same impulse and the same change of momentum, so it produces the identical final speed. (R1)
力-时间图下的面积等于冲量 $J = \int F\,\mathrm{d}t$。(B1)
动量定理指出：物体所受净冲量等于其动量变化：$J = \Delta p = mv - mu$。(A1)
把图分为上升三角形、平顶矩形、下降三角形。(M1)
$$ J = \underbrace{\tfrac{1}{2}(0.20)(18)}_{1.8} + \underbrace{(0.30)(18)}_{5.4} + \underbrace{\tfrac{1}{2}(0.10)(18)}_{0.9}. $$(列出全部三块面积得 M1)
$$ J = 1.8 + 5.4 + 0.9 = 8.1\ \mathrm{N\,s}. $$(A1)
小车从静止出发，$J = \Delta p = mv - 0$，故 $v = J/m$：(M1)
$$ v = \frac{8.1}{0.60} = 13.5\ \mathrm{m\,s^{-1}}. $$(A1)
整段时间内的平均力为 $F_{\text{avg}} = J/\Delta t$：(M1)
$$ F_{\text{avg}} = \frac{8.1}{0.60} = 13.5\ \mathrm{N}. $$(A1)
等于平均值的恒定力给出相同的面积，因而冲量相同、动量变化相同，故产生完全相同的末速度。(R1)
$0.030\ \mathrm{kg}$ sphere in a liquid; drag $F_{D} = bv^{2}$, data of $F_{D}$ vs $v^{2}$ given. (a) why $F_{D}$ vs $v^{2}$ is linear through the origin and the gradient; (b) gradient and $b$; (c) terminal velocity; (d) percentage uncertainty and a reason for scatter.液体中 $0.030\ \mathrm{kg}$ 小球；阻力 $F_{D} = bv^{2}$，给出 $F_{D}$ 对 $v^{2}$ 数据。(a) 为何 $F_{D}$ 对 $v^{2}$ 过原点直线及斜率；(b) 斜率与 $b$；(c) 收尾速度；(d) 百分比不确定度与散布原因。
The model is $F_{D} = b\,v^{2}$. Writing $X = v^{2}$ gives $F_{D} = b\,X$. (M1)
This has the form $y = mx$ with no constant term, so a plot of $F_{D}$ against $v^{2}$ is a straight line passing through the origin. (A1)
Comparing with $y = mx$, the gradient equals the drag constant $b$. (A1)
Read the gradient from two widely separated points, $(4.0,\,0.030)$ and $(64.0,\,0.480)$: (M1)
$$ \text{gradient} = \frac{0.480 - 0.030}{64.0 - 4.0} = \frac{0.450}{60.0} = 7.5\times 10^{-3}. $$(A1)
Since the gradient is $b$: $b \approx 7.5\times 10^{-3}\ \mathrm{N\,s^{2}\,m^{-2}}$. (A1)
At terminal velocity the drag balances the weight: $mg = b\,v_{T}^{2}$. (M1)
$$ v_{T} = \sqrt{\frac{mg}{b}} = \sqrt{\frac{(0.030)(9.81)}{7.5\times 10^{-3}}}. $$(M1 for substitution)
$$ v_{T} = \sqrt{\frac{0.2943}{0.0075}} = \sqrt{39.24} \approx 6.3\ \mathrm{m\,s^{-1}}. $$(A1)
At $v = 8.0\ \mathrm{m\,s^{-1}}$, $F_{D} = 0.480\ \mathrm{N}$ with $\pm 0.02\ \mathrm{N}$: (M1)
$$ \frac{0.02}{0.480}\times 100\% \approx 4.2\% \approx 4\%. $$(A1)
One experimental reason for scatter: small temperature changes alter the liquid's viscosity, or the sphere may not have reached a truly steady (terminal) speed when each reading was taken. (B1)
模型为 $F_{D} = b\,v^{2}$。令 $X = v^{2}$ 得 $F_{D} = b\,X$。(M1)
此式形如 $y = mx$ 且无常数项，故 $F_{D}$ 对 $v^{2}$ 作图为过原点的直线。(A1)
与 $y = mx$ 比较，斜率等于阻力常数 $b$。(A1)
用相距较远的两点 $(4.0,\,0.030)$ 与 $(64.0,\,0.480)$ 读斜率：(M1)
$$ \text{斜率} = \frac{0.480 - 0.030}{64.0 - 4.0} = \frac{0.450}{60.0} = 7.5\times 10^{-3}. $$(A1)
因斜率即 $b$：$b \approx 7.5\times 10^{-3}\ \mathrm{N\,s^{2}\,m^{-2}}$。(A1)
收尾时阻力平衡重力：$mg = b\,v_{T}^{2}$。(M1)
$$ v_{T} = \sqrt{\frac{mg}{b}} = \sqrt{\frac{(0.030)(9.81)}{7.5\times 10^{-3}}}. $$(代入得 M1)
$$ v_{T} = \sqrt{\frac{0.2943}{0.0075}} = \sqrt{39.24} \approx 6.3\ \mathrm{m\,s^{-1}}. $$(A1)
$v = 8.0\ \mathrm{m\,s^{-1}}$ 处 $F_{D} = 0.480\ \mathrm{N}$，$\pm 0.02\ \mathrm{N}$：(M1)
$$ \frac{0.02}{0.480}\times 100\% \approx 4.2\% \approx 4\%. $$(A1)
散布的一个实验原因：微小温度变化改变液体黏度，或读数时小球尚未达到真正稳定（收尾）速率。(B1)
$1200\ \mathrm{kg}$ car at $18\ \mathrm{m\,s^{-1}}$ rear-ends a stationary $800\ \mathrm{kg}$ car; they lock together. (a) state momentum conservation and its condition; (b) common velocity; (c) KE before and after, fraction lost, classify; (d) a $5.0\ \mathrm{kg}$ object explodes into $2.0\ \mathrm{kg}$ at $6.0\ \mathrm{m\,s^{-1}}$ and $3.0\ \mathrm{kg}$, find its speed and direction.$1200\ \mathrm{kg}$ 车以 $18\ \mathrm{m\,s^{-1}}$ 追尾静止的 $800\ \mathrm{kg}$ 车，锁在一起。(a) 陈述动量守恒及条件；(b) 共同速度；(c) 碰前碰后动能、损失比例、判型；(d) $5.0\ \mathrm{kg}$ 物体爆炸为 $2.0\ \mathrm{kg}$（$6.0\ \mathrm{m\,s^{-1}}$）与 $3.0\ \mathrm{kg}$，求其速率与方向。
If no net external force acts on a system, its total linear momentum is constant: $\Sigma p_{\text{before}} = \Sigma p_{\text{after}}$. (B1)
Here the impact is brief and external friction is negligible during it, so the two-car system has no net external horizontal force and momentum is conserved. (A1)
Perfectly inelastic: the bodies move off together. $m_{1}u_{1} = (m_{1} + m_{2})v$. (M1)
$$ v = \frac{m_{1}u_{1}}{m_{1} + m_{2}} = \frac{(1200)(18)}{1200 + 800}. $$(M1 for substitution)
$$ v = \frac{21600}{2000} = 10.8\ \mathrm{m\,s^{-1}}. $$(A1)
Before: $KE_{i} = \tfrac{1}{2}(1200)(18)^{2} = 194{,}400\ \mathrm{J}$. (M1)
After: $KE_{f} = \tfrac{1}{2}(2000)(10.8)^{2} = 116{,}640\ \mathrm{J}$. (A1)
$$ \text{fraction lost} = \frac{194{,}400 - 116{,}640}{194{,}400} = \frac{77{,}760}{194{,}400} = 0.40. $$So $40\%$ of the kinetic energy is lost. (A1)
Kinetic energy is not conserved, so the collision is inelastic (here perfectly inelastic, since they stick). (R1)
Total momentum is zero before and after: $0 = m_{a}v_{a} + m_{b}v_{b}$, taking the $2.0\ \mathrm{kg}$ piece's direction as positive. (M1)
$$ v_{b} = -\frac{m_{a}v_{a}}{m_{b}} = -\frac{(2.0)(6.0)}{3.0} = -4.0\ \mathrm{m\,s^{-1}}. $$The $3.0\ \mathrm{kg}$ piece moves at $4.0\ \mathrm{m\,s^{-1}}$ in the opposite direction to the $2.0\ \mathrm{kg}$ piece. (A1)
若系统不受净外力，其总线动量守恒：$\Sigma p_{\text{前}} = \Sigma p_{\text{后}}$。(B1)
这里碰撞短暂、其间外部摩擦可忽略，故两车系统无净外部水平力，动量守恒。(A1)
完全非弹性：两体一起离开。$m_{1}u_{1} = (m_{1} + m_{2})v$。(M1)
$$ v = \frac{m_{1}u_{1}}{m_{1} + m_{2}} = \frac{(1200)(18)}{1200 + 800}. $$(代入得 M1)
$$ v = \frac{21600}{2000} = 10.8\ \mathrm{m\,s^{-1}}. $$(A1)
碰前：$KE_{i} = \tfrac{1}{2}(1200)(18)^{2} = 194{,}400\ \mathrm{J}$。(M1)
碰后：$KE_{f} = \tfrac{1}{2}(2000)(10.8)^{2} = 116{,}640\ \mathrm{J}$。(A1)
$$ \text{损失比例} = \frac{194{,}400 - 116{,}640}{194{,}400} = \frac{77{,}760}{194{,}400} = 0.40. $$故损失了 $40\%$ 的动能。(A1)
动能不守恒，故碰撞为非弹性（此处为完全非弹性，因两车粘连）。(R1)
爆炸前后总动量均为零：$0 = m_{a}v_{a} + m_{b}v_{b}$，取 $2.0\ \mathrm{kg}$ 块方向为正。(M1)
$$ v_{b} = -\frac{m_{a}v_{a}}{m_{b}} = -\frac{(2.0)(6.0)}{3.0} = -4.0\ \mathrm{m\,s^{-1}}. $$$3.0\ \mathrm{kg}$ 块以 $4.0\ \mathrm{m\,s^{-1}}$ 沿与 $2.0\ \mathrm{kg}$ 块相反的方向运动。(A1)
$1100\ \mathrm{kg}$ car, curve radius $60\ \mathrm{m}$, speed $20\ \mathrm{m\,s^{-1}}$. (a) centripetal acceleration and force; (b) minimum $\mu_{s}$ on a flat curve and why it is mass-independent; (c) banking angle for no friction.$1100\ \mathrm{kg}$ 车，弯道半径 $60\ \mathrm{m}$，速率 $20\ \mathrm{m\,s^{-1}}$。(a) 向心加速度与向心力；(b) 平直弯道最小 $\mu_{s}$ 及为何与质量无关；(c) 无摩擦时的倾斜角。
The centripetal acceleration is $a_{c} = v^{2}/r$: (M1)
$$ a_{c} = \frac{20^{2}}{60} = \frac{400}{60} \approx 6.7\ \mathrm{m\,s^{-2}}. $$(A1)
The required centripetal force is $F_{c} = ma_{c} = (1100)(6.67) \approx 7.3\times 10^{3}\ \mathrm{N}$. (A1)
On a flat curve, static friction supplies the whole centripetal force, and the maximum available is $\mu_{s}N = \mu_{s}mg$: (M1)
$$ \mu_{s}mg \ge \frac{mv^{2}}{r} \;\Rightarrow\; \mu_{s} \ge \frac{v^{2}}{rg}. $$(M1 / A1 for the inequality)
$$ \mu_{s} \ge \frac{20^{2}}{(60)(9.81)} = \frac{400}{588.6} \approx 0.68. $$(A1)
The mass $m$ cancels from both sides, so whether the car can hold the curve depends only on $v$, $r$ and $\mu_{s}$, not on how heavy it is. (R1)
On a frictionless banked curve the horizontal component of the normal force supplies the centripetal force while the vertical component balances gravity, giving $\tan\theta = v^{2}/(rg)$: (M1)
$$ \theta = \arctan\!\left(\frac{20^{2}}{(60)(9.81)}\right) = \arctan(0.6796) \approx 34^{\circ}. $$(A1)
向心加速度为 $a_{c} = v^{2}/r$：(M1)
$$ a_{c} = \frac{20^{2}}{60} = \frac{400}{60} \approx 6.7\ \mathrm{m\,s^{-2}}. $$(A1)
所需向心力为 $F_{c} = ma_{c} = (1100)(6.67) \approx 7.3\times 10^{3}\ \mathrm{N}$。(A1)
平直弯道上，静摩擦力提供全部向心力，可用最大值为 $\mu_{s}N = \mu_{s}mg$：(M1)
$$ \mu_{s}mg \ge \frac{mv^{2}}{r} \;\Rightarrow\; \mu_{s} \ge \frac{v^{2}}{rg}. $$(不等式得 M1 / A1)
$$ \mu_{s} \ge \frac{20^{2}}{(60)(9.81)} = \frac{400}{588.6} \approx 0.68. $$(A1)
质量 $m$ 在两边约去，故汽车能否过弯只取决于 $v$、$r$ 与 $\mu_{s}$，与轻重无关。(R1)
无摩擦倾斜弯道上，法向力的水平分量提供向心力、竖直分量平衡重力，得 $\tan\theta = v^{2}/(rg)$：(M1)
$$ \theta = \arctan\!\left(\frac{20^{2}}{(60)(9.81)}\right) = \arctan(0.6796) \approx 34^{\circ}. $$(A1)
Sand falls onto a belt at $3.0\ \mathrm{kg\,s^{-1}}$; belt moves at constant $1.5\ \mathrm{m\,s^{-1}}$; sand has no horizontal velocity before landing. (a) from $\Sigma F = \mathrm{d}p/\mathrm{d}t$ show $F = v\,\mathrm{d}m/\mathrm{d}t$; (b) the force; (c) motor power vs rate of KE gain, and comment.沙子以 $3.0\ \mathrm{kg\,s^{-1}}$ 落到传送带上；带以恒定 $1.5\ \mathrm{m\,s^{-1}}$ 运动；沙子落前无水平速度。(a) 由 $\Sigma F = \mathrm{d}p/\mathrm{d}t$ 证明 $F = v\,\mathrm{d}m/\mathrm{d}t$；(b) 该力；(c) 电机功率与动能增长率比较并说明。
Each second the belt takes a mass $\mathrm{d}m$ of sand from rest (zero horizontal velocity) up to the belt speed $v$. The horizontal momentum given to the sand is $\mathrm{d}p = v\,\mathrm{d}m$. (M1)
Newton's second law in momentum form: $F = \dfrac{\mathrm{d}p}{\mathrm{d}t}$. (A1)
Since $v$ is constant, $F = \dfrac{\mathrm{d}(mv)}{\mathrm{d}t} = v\,\dfrac{\mathrm{d}m}{\mathrm{d}t}$. (A1)
Substitute $v = 1.5\ \mathrm{m\,s^{-1}}$ and $\mathrm{d}m/\mathrm{d}t = 3.0\ \mathrm{kg\,s^{-1}}$: (M1)
$$ F = v\,\frac{\mathrm{d}m}{\mathrm{d}t} = (1.5)(3.0) = 4.5\ \mathrm{N}. $$(A1)
The motor force moves the belt at $v$, so the motor power is $P = Fv = (4.5)(1.5) = 6.75\ \mathrm{W}$. (M1)
The sand gains kinetic energy at $\dfrac{\mathrm{d}(KE)}{\mathrm{d}t} = \tfrac{1}{2}\dfrac{\mathrm{d}m}{\mathrm{d}t}v^{2} = \tfrac{1}{2}(3.0)(1.5)^{2} = 3.375\ \mathrm{W}$. (A1)
The motor supplies $6.75\ \mathrm{W}$ but the sand gains only $3.375\ \mathrm{W}$ of kinetic energy; exactly half the input is dissipated as heat through the sliding of sand on the belt before it matches belt speed. (R1)
每秒传送带把质量 $\mathrm{d}m$ 的沙子从静止（水平速度为零）带到带速 $v$。给沙子的水平动量为 $\mathrm{d}p = v\,\mathrm{d}m$。(M1)
动量形式的牛顿第二定律：$F = \dfrac{\mathrm{d}p}{\mathrm{d}t}$。(A1)
由于 $v$ 恒定，$F = \dfrac{\mathrm{d}(mv)}{\mathrm{d}t} = v\,\dfrac{\mathrm{d}m}{\mathrm{d}t}$。(A1)
代入 $v = 1.5\ \mathrm{m\,s^{-1}}$ 与 $\mathrm{d}m/\mathrm{d}t = 3.0\ \mathrm{kg\,s^{-1}}$：(M1)
$$ F = v\,\frac{\mathrm{d}m}{\mathrm{d}t} = (1.5)(3.0) = 4.5\ \mathrm{N}. $$(A1)
电机的力以速率 $v$ 带动传送带，故电机功率 $P = Fv = (4.5)(1.5) = 6.75\ \mathrm{W}$。(M1)
沙子的动能增长率为 $\dfrac{\mathrm{d}(KE)}{\mathrm{d}t} = \tfrac{1}{2}\dfrac{\mathrm{d}m}{\mathrm{d}t}v^{2} = \tfrac{1}{2}(3.0)(1.5)^{2} = 3.375\ \mathrm{W}$。(A1)
电机提供 $6.75\ \mathrm{W}$，而沙子只获得 $3.375\ \mathrm{W}$ 的动能；恰好一半的输入在沙子滑到与带同速之前因滑动以热的形式耗散。(R1)