Unit A.2: Forces and Momentum单元 A.2：力与动量

Identify. Four forces act on the box: weight $W$ down, normal $N$ up, tension $T = 40\ \mathrm{N}$ (say east), friction $f$ (west, opposing impending motion). The box is in equilibrium (at rest).

识别。箱子受四个力：重力 $W$ 向下、法向力 $N$ 向上、张力 $T = 40\ \mathrm{N}$（设向东）、摩擦力 $f$（向西，阻碍即将发生的运动）。箱子处于平衡（静止）。

Set Up. Take east and up as positive. Apply $\Sigma F_{x} = 0$ and $\Sigma F_{y} = 0$.

列式。取东、上为正。用 $\Sigma F_{x} = 0$ 与 $\Sigma F_{y} = 0$。

Execute (vertical). $N - W = 0 \Rightarrow N = mg = (12)(9.81) \approx 118\ \mathrm{N}$.

求解（竖直）。$N - W = 0 \Rightarrow N = mg = (12)(9.81) \approx 118\ \mathrm{N}$。

Execute (horizontal). $T - f = 0 \Rightarrow f = T = 40\ \mathrm{N}$, directed west.

求解（水平）。$T - f = 0 \Rightarrow f = T = 40\ \mathrm{N}$，方向向西。

Evaluate. The friction here is static friction, taking whatever value (up to its maximum) is needed to keep $\Sigma F_{x} = 0$. It is not $\mu N$ unless the box is on the verge of slipping.

评估。这里的摩擦力是静摩擦力，它会取（不超过最大值的）任何使 $\Sigma F_{x} = 0$ 成立的数值。除非箱子即将滑动，否则它并不等于 $\mu N$。

Identify. By symmetry the two tensions are equal, $T$. Three forces meet at the knot: weight $W = mg$ down, and the two tensions up and outward at $30^{\circ}$.

识别。由对称性两张力相等，记为 $T$。三个力交于结点：重力 $W = mg$ 向下，两张力以 $30^{\circ}$ 向上、向外。

Set Up (vertical equilibrium). Each tension contributes a vertical component $T \sin 30^{\circ}$.

列式（竖直平衡）。每条张力贡献竖直分量 $T \sin 30^{\circ}$。

Execute.

求解。

Evaluate. Shallower cables (smaller angle) need larger tension, because the vertical component $T\sin\theta$ shrinks. As $\theta \to 0$, $T \to \infty$: a perfectly horizontal cable cannot support a weight.

评估。绳越接近水平（角度越小）所需张力越大，因为竖直分量 $T\sin\theta$ 变小。当 $\theta \to 0$ 时 $T \to \infty$：完全水平的绳无法承托重物。

Identify. Forces on the person: weight $mg$ down, normal force $N$ (scale reading) up. The acceleration is $a = 2.0\ \mathrm{m\,s^{-2}}$ upward.

识别。人受力：重力 $mg$ 向下、法向力 $N$（即读数）向上。加速度 $a = 2.0\ \mathrm{m\,s^{-2}}$ 向上。

Set Up. Take up as positive. Newton's second law: $N - mg = ma$.

列式。取上为正。牛顿第二定律：$N - mg = ma$。

Execute.

求解。

Evaluate. The apparent weight exceeds the true weight ($686\ \mathrm{N}$) because the floor must both support the person and accelerate them upward. If the lift accelerated downward, $N = m(g - a) < mg$; in free fall ($a = g$), $N = 0$ — weightlessness.

评估。视重大于真实重量（$686\ \mathrm{N}$），因为地板既要支撑人、又要使其向上加速。若电梯向下加速，$N = m(g - a) < mg$；自由落体（$a = g$）时 $N = 0$——失重。

Identify. The two pulls are perpendicular, so combine by Pythagoras.

识别。两拉力互相垂直，用勾股定理合成。

Set Up. Resultant force magnitude $F = \sqrt{12^{2} + 5.0^{2}}$. Then $a = F/m$.

列式。合力大小 $F = \sqrt{12^{2} + 5.0^{2}}$，再用 $a = F/m$。

Execute.

求解。

Evaluate. Acceleration is parallel to the resultant force, so it too points $22.6^{\circ}$ north of east. The direction of $\vec{a}$ always matches the direction of $\Sigma\vec{F}$, never the direction of any single force.

评估。加速度与合力同向，故也指向东偏北 $22.6^{\circ}$。$\vec{a}$ 的方向总与 $\Sigma\vec{F}$ 一致，而非任何单个力的方向。

Identify. The heavier $m_{2}$ descends, $m_{1}$ rises, with common acceleration magnitude $a$. The string tension $T$ is the same on both sides.

识别。较重的 $m_{2}$ 下降，$m_{1}$ 上升，加速度大小相同，记为 $a$。两侧绳张力 $T$ 相同。

Set Up. Take the direction of motion as positive for each mass. Newton's second law on each:

列式。对每个质量取其运动方向为正。各写牛顿第二定律：

Execute. Add the two equations to eliminate $T$:

求解。两式相加消去 $T$：

Evaluate. The tension ($36.8\ \mathrm{N}$) lies between the two weights ($29.4\ \mathrm{N}$ and $49.1\ \mathrm{N}$), as it must: it exceeds $m_{1}g$ (to accelerate $m_{1}$ up) and is less than $m_{2}g$ (so $m_{2}$ accelerates down).

评估。张力（$36.8\ \mathrm{N}$）介于两重量（$29.4\ \mathrm{N}$ 与 $49.1\ \mathrm{N}$）之间，理应如此：它大于 $m_{1}g$（使 $m_{1}$ 向上加速）、小于 $m_{2}g$（使 $m_{2}$ 向下加速）。

Identify. Both blocks accelerate together at $a$. The push reaches $m_{A}$ only through the contact force $C$ between them.

识别。两块以相同 $a$ 一起加速。推力只通过两块间的接触力 $C$ 传给 $m_{A}$。

Set Up & Execute (whole system). $a = F / (m_{A} + m_{B}) = 20 / 5.0 = 4.0\ \mathrm{m\,s^{-2}}$.

列式与求解（整体）。$a = F / (m_{A} + m_{B}) = 20 / 5.0 = 4.0\ \mathrm{m\,s^{-2}}$。

Contact force (isolate $m_{A}$). The only horizontal force on $m_{A}$ is $C$: $C = m_{A} a = (2.0)(4.0) = 8.0\ \mathrm{N}$.

接触力（孤立 $m_{A}$）。$m_{A}$ 上唯一的水平力是 $C$：$C = m_{A} a = (2.0)(4.0) = 8.0\ \mathrm{N}$。

Evaluate. By Newton's third law, $m_{A}$ pushes back on $m_{B}$ with $8.0\ \mathrm{N}$. Check $m_{B}$: $F - C = 20 - 8.0 = 12\ \mathrm{N} = m_{B} a = (3.0)(4.0)$. Consistent.

评估。由牛顿第三定律，$m_{A}$ 以 $8.0\ \mathrm{N}$ 反推 $m_{B}$。校验 $m_{B}$：$F - C = 20 - 8.0 = 12\ \mathrm{N} = m_{B} a = (3.0)(4.0)$，一致。

Identify. On a horizontal floor $N = mg = (20)(9.81) = 196\ \mathrm{N}$.

识别。水平地面上 $N = mg = (20)(9.81) = 196\ \mathrm{N}$。

Set Up. Maximum static friction: $f_{s,\max} = \mu_{s} N = (0.50)(196) = 98\ \mathrm{N}$. Compare with the applied $80\ \mathrm{N}$.

列式。最大静摩擦力：$f_{s,\max} = \mu_{s} N = (0.50)(196) = 98\ \mathrm{N}$。与所加 $80\ \mathrm{N}$ 比较。

Execute. Since $80\ \mathrm{N} < 98\ \mathrm{N}$, static friction can match the push: the crate does not move. Friction here equals $80\ \mathrm{N}$ (not $98\ \mathrm{N}$, not $\mu_{k} N$), and $a = 0$.

求解。由于 $80\ \mathrm{N} < 98\ \mathrm{N}$，静摩擦力可与推力相抵：箱子不动。此时摩擦力等于 $80\ \mathrm{N}$（不是 $98\ \mathrm{N}$，也不是 $\mu_{k} N$），$a = 0$。

Evaluate. Had the push been, say, $120\ \mathrm{N} > 98\ \mathrm{N}$, the crate would slide and kinetic friction $f_{k} = \mu_{k} N = (0.40)(196) = 78.4\ \mathrm{N}$ would take over, giving $a = (120 - 78.4)/20 \approx 2.1\ \mathrm{m\,s^{-2}}$.

评估。若推力为 $120\ \mathrm{N} > 98\ \mathrm{N}$，箱子会滑动，由动摩擦力 $f_{k} = \mu_{k} N = (0.40)(196) = 78.4\ \mathrm{N}$ 接管，得 $a = (120 - 78.4)/20 \approx 2.1\ \mathrm{m\,s^{-2}}$。

Identify. At terminal velocity, drag up equals weight down: $b v_{T}^{2} = m g$.

识别。收尾速度时，向上阻力等于向下重力：$b v_{T}^{2} = m g$。

Set Up & Execute.

列式与求解。

Evaluate. Initially (at $v = 0$) drag is zero, so $a = g$; as $v$ grows, drag grows, $a$ falls, and $v$ approaches $v_{T}$ asymptotically. The suvat equations do not apply here because $a$ is not constant.

评估。初始（$v = 0$）时阻力为零，故 $a = g$；随 $v$ 增大，阻力增大，$a$ 减小，$v$ 渐近趋向 $v_{T}$。此处 $a$ 不恒定，suvat 方程不适用。

Identify. On a flat curve, the only horizontal force available to point inward is static friction. It must supply the entire centripetal force.

识别。平直弯道上，唯一能指向圆心的水平力是静摩擦力。它必须提供全部向心力。

Set Up. Centripetal force $F_{c} = m v^{2}/r$. Friction available: up to $\mu_{s} N = \mu_{s} m g$.

列式。向心力 $F_{c} = m v^{2}/r$。可用摩擦力：至多 $\mu_{s} N = \mu_{s} m g$。

Execute.

求解。

Evaluate. Notice mass cancels in the friction condition: whether a car can corner depends on $v$, $r$ and $\mu_{s}$, not on how heavy it is. Double the speed and the required friction quadruples (the $v^{2}$ dependence).

评估。注意摩擦条件中质量约去：汽车能否过弯取决于 $v$、$r$ 与 $\mu_{s}$，与轻重无关。速度加倍，所需摩擦力变为四倍（$v^{2}$ 依赖）。

Identify. At the top, both weight $mg$ and tension $T$ point downward — toward the centre. Together they provide the centripetal force.

识别。在顶端，重力 $mg$ 与张力 $T$ 都向下——指向圆心。二者共同提供向心力。

Set Up. $T + mg = m v^{2}/r$. The minimum speed is when the string is just about to go slack, $T = 0$.

列式。$T + mg = m v^{2}/r$。最小速率对应绳恰要松弛，$T = 0$。

Execute. With $T = 0$: $mg = m v_{\min}^{2}/r \Rightarrow v_{\min} = \sqrt{g r}$.

求解。令 $T = 0$：$mg = m v_{\min}^{2}/r \Rightarrow v_{\min} = \sqrt{g r}$。

Evaluate. Below $\sqrt{gr}$, gravity alone exceeds the required centripetal force at the top, so the ball falls inward and the string slackens. Notice mass again cancels: $v_{\min}$ depends only on $g$ and $r$.

评估。低于 $\sqrt{gr}$ 时，仅重力就超过顶端所需向心力，球向内坠落、绳松弛。注意质量再次约去：$v_{\min}$ 只取决于 $g$ 与 $r$。

Identify. Take the rebound (outgoing) direction as positive. Then the incoming velocity is $-8.0\ \mathrm{m\,s^{-1}}$ and the outgoing is $+6.0\ \mathrm{m\,s^{-1}}$.

识别。取弹回（离去）方向为正。则入射速度为 $-8.0\ \mathrm{m\,s^{-1}}$，离去速度为 $+6.0\ \mathrm{m\,s^{-1}}$。

Set Up. Impulse $J = \Delta p = m v - m u$.

列式。冲量 $J = \Delta p = m v - m u$。

Execute.

求解。

Evaluate. The sign error students make is treating the rebound as $\Delta v = 6 - 8 = -2$. Velocity is a vector: the ball reverses direction, so the speeds add in magnitude. Bouncing produces a larger impulse than stopping dead.

评估。学生常犯的符号错误是把弹回当成 $\Delta v = 6 - 8 = -2$。速度是矢量：球反向，故两速率在大小上相加。弹回比直接停下产生更大的冲量。

Identify. The impulse is the area under the $F$-$t$ graph — here a triangle of base $0.40\ \mathrm{s}$ and height $12\ \mathrm{N}$.

识别。冲量为 $F$-$t$ 图下的面积——此处为底 $0.40\ \mathrm{s}$、高 $12\ \mathrm{N}$ 的三角形。

Set Up & Execute.

列式与求解。

Final speed. Since $J = \Delta p = m v - 0$:

末速度。由 $J = \Delta p = m v - 0$：

Evaluate. Using the area handles a varying force without calculus. The same answer would follow from $J = F_{\text{avg}}\Delta t$ with $F_{\text{avg}} = 6.0\ \mathrm{N}$ (the mean of $0$ and $12$).

评估。用面积可在不借助微积分的情况下处理变力。用 $J = F_{\text{avg}}\Delta t$、$F_{\text{avg}} = 6.0\ \mathrm{N}$（$0$ 与 $12$ 的平均）也得同样结果。

Identify. Perfectly inelastic (they stick). Momentum is conserved; kinetic energy is not.

识别。完全非弹性（粘连）。动量守恒；动能不守恒。

Set Up. $m_{1} u_{1} = (m_{1} + m_{2}) v$.

列式。$m_{1} u_{1} = (m_{1} + m_{2}) v$。

Execute.

求解。

Energy. $KE_{i} = \tfrac{1}{2}(1500)(20)^{2} = 300\,000\ \mathrm{J}$; $KE_{f} = \tfrac{1}{2}(2500)(12)^{2} = 180\,000\ \mathrm{J}$. Fraction lost $= (300\,000 - 180\,000)/300\,000 = 0.40$, i.e. $40\%$.

能量。$KE_{i} = \tfrac{1}{2}(1500)(20)^{2} = 300\,000\ \mathrm{J}$；$KE_{f} = \tfrac{1}{2}(2500)(12)^{2} = 180\,000\ \mathrm{J}$。损失比例 $= (300\,000 - 180\,000)/300\,000 = 0.40$，即 $40\%$。

Evaluate. The "lost" KE went to crumpling metal, heat, and sound. Momentum is still exactly conserved — that is why we solve for $v$ from momentum, never from energy, in an inelastic collision.

评估。"损失"的动能变成了金属变形、热与声。动量仍严格守恒——这正是非弹性碰撞中我们由动量而非能量求 $v$ 的原因。

Identify. Before firing, total momentum is zero. After, it must still be zero (no external horizontal force).

识别。开火前总动量为零。开火后仍须为零（无外加水平力）。

Set Up. $0 = m_{b} v_{b} + m_{r} v_{r}$.

列式。$0 = m_{b} v_{b} + m_{r} v_{r}$。

Execute.

求解。

Evaluate. The minus sign shows the rifle recoils opposite to the bullet. Its speed is far smaller because its mass is much larger — equal and opposite momenta, not equal speeds. (The bullet also carries far more KE, $\propto p^{2}/m$.)

评估。负号表明步枪沿子弹相反方向反冲。其速率远小，因为质量远大——动量大小相等方向相反，而非速率相等。（子弹携带的动能也远多，$\propto p^{2}/m$。）