题型分析 — Linear Motion Under Variable Force

Type 1：建立与求解微分方程

Example 1: A particle of mass 0.5 kg moves in a straight line under the action of a force $F = 2 - 0.5v$ N, where $v$ is the speed. Initially the particle is at rest. Find an expression for $v$ in terms of $t$.

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$F = ma = m\frac{dv}{dt}$ B1

$0.5\frac{dv}{dt} = 2 - 0.5v$ M1

$\frac{dv}{dt} = 4 - v$

$\int \frac{1}{4 - v} dv = \int 1 dt$ M1

$-\ln|4 - v| = t + C$ A1

When $t = 0$, $v = 0$: $-\ln 4 = C$ M1

$-\ln(4 - v) = t - \ln 4$ $\ln(4 - v) = \ln 4 - t$ $4 - v = 4e^{-t}$ M1 $v = 4(1 - e^{-t})$ A1

[Total: 8]

Example 2: A body of mass 2 kg moves under a force $F = 6 - 3x$ N, where $x$ is displacement from a fixed point O. When $x = 0$, velocity $v = 0$. Find $v$ as a function of $x$.

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$F = m\frac{dv}{dt} = mv\frac{dv}{dx}$ M1

$2v\frac{dv}{dx} = 6 - 3x$ M1

$\int 2v dv = \int (6 - 3x) dx$ M1

$v^2 = 6x - \frac{3}{2}x^2 + C$ A1

When $x = 0$, $v = 0 \Rightarrow C = 0$ M1

$v^2 = 6x - \frac{3}{2}x^2$ $v = \sqrt{6x - 1.5x^2}$ A1

[Total: 6]

Example 3: A particle of mass $m$ moves along the $x$-axis under a force directed towards O of magnitude $\frac{mk}{x^3}$, where $k$ is a constant. Initially the particle is at $x = a$ with speed $v_0$ towards O. Find the speed when $x = a/2$.

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Force towards O: $F = -\frac{mk}{x^3}$ (negative since towards O) M1

$mv\frac{dv}{dx} = -\frac{mk}{x^3}$ M1

$\int v dv = -\int \frac{k}{x^3} dx$ M1

$\frac{1}{2}v^2 = \frac{k}{2x^2} + C$ A1

When $x = a$, $v = v_0$: $\frac{1}{2}v_0^2 = \frac{k}{2a^2} + C \Rightarrow C = \frac{1}{2}v_0^2 - \frac{k}{2a^2}$ M1

When $x = a/2$: $\frac{1}{2}v^2 = \frac{k}{2(a/2)^2} + \frac{1}{2}v_0^2 - \frac{k}{2a^2}$ $\frac{1}{2}v^2 = \frac{2k}{a^2} + \frac{1}{2}v_0^2 - \frac{k}{2a^2}$ M1 $\frac{1}{2}v^2 = \frac{1}{2}v_0^2 + \frac{3k}{2a^2}$ $v^2 = v_0^2 + \frac{3k}{a^2}$ $v = \sqrt{v_0^2 + \frac{3k}{a^2}}$ A1

[Total: 8]

Type 2：空气阻力（$kv$ 形式）

Example 1: A particle of mass 0.1 kg falls from rest under gravity, with air resistance proportional to velocity: $R = 0.2v$ N. Find the terminal velocity and the velocity after 2 seconds.

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$m\frac{dv}{dt} = mg - kv$, where $k = 0.2$ M1

Terminal velocity $v_T$: when $mg = kv$ M1 $0.1 \times 9.8 = 0.2v_T$ $v_T = \frac{0.98}{0.2} = 4.9\ \text{m}\,\text{s}^{-1}$ A1

$0.1\frac{dv}{dt} = 0.98 - 0.2v$ $\frac{dv}{dt} = 9.8 - 2v$

$\int \frac{dv}{9.8 - 2v} = \int dt$ M1

$-\frac{1}{2}\ln|9.8 - 2v| = t + C$ A1

At $t = 0$, $v = 0$: $-\frac{1}{2}\ln 9.8 = C$

$-\frac{1}{2}\ln(9.8 - 2v) = t - \frac{1}{2}\ln 9.8$ $\ln(9.8 - 2v) = -2t + \ln 9.8$ $9.8 - 2v = 9.8e^{-2t}$ $v = \frac{9.8}{2}(1 - e^{-2t})$ M1

At $t = 2$: $v = 4.9(1 - e^{-4}) = 4.9(1 - 0.0183) = 4.81\ \text{m}\,\text{s}^{-1}$ A1

[Total: 9]

Example 2: A boat of mass 500 kg is moving at $8\ \text{m}\,\text{s}^{-1}$ when its engine is cut. The resistance is $50v$ N. Find the time taken for the speed to halve.

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$m\frac{dv}{dt} = -kv$, where $k = 50$ M1

$500\frac{dv}{dt} = -50v$ $\frac{dv}{dt} = -0.1v$ A1

$\int \frac{dv}{v} = -\int 0.1 dt$ M1

$\ln v = -0.1t + C$ A1

At $t = 0$, $v = 8$: $\ln 8 = C$

$\ln v = -0.1t + \ln 8$ $\ln\left(\frac{v}{8}\right) = -0.1t$ M1

When $v = 4$: $\ln(0.5) = -0.1t$ $t = -10\ln(0.5) = 10\ln 2 = 6.93\ \text{s}$ (3 s.f.) A1

[Total: 7]

Example 3: A particle of mass $m$ is projected vertically upwards with speed $u$ in a medium where the resistance is $mkv$. Find the maximum height reached.

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Equation of motion (upward positive): $-mg - mkv = m\frac{dv}{dt}$ M1

Using $a = v\frac{dv}{dx}$: $v\frac{dv}{dx} = -g - kv$ M1

$v\frac{dv}{dx} = -(g + kv)$

$\frac{v}{g + kv} dv = -dx$ M1

$\int \frac{v}{g + kv} dv = -\int dx$

Rewrite: $\frac{v}{g + kv} = \frac{1}{k} - \frac{g}{k(g + kv)}$ M1

$\int \left(\frac{1}{k} - \frac{g}{k(g + kv)}\right) dv = -\int dx$

$\frac{v}{k} - \frac{g}{k^2}\ln(g + kv) = -x + C$ A1

When $x = 0$, $v = u$: $C = \frac{u}{k} - \frac{g}{k^2}\ln(g + ku)$ M1

At max height, $v = 0$: $0 - \frac{g}{k^2}\ln g = -H + \frac{u}{k} - \frac{g}{k^2}\ln(g + ku)$

$H = \frac{u}{k} - \frac{g}{k^2}\ln\left(\frac{g + ku}{g}\right)$ A1

[Total: 9]

Type 3：空气阻力（$kv^2$ 形式）

Example 1: A particle falls from rest under gravity with air resistance $0.5v^2$ N per unit mass. Find the terminal velocity and the velocity after falling a distance $x$.

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Per unit mass: $\frac{dv}{dt} = g - 0.5v^2$ M1

Terminal velocity: $g = 0.5v_T^2 \Rightarrow v_T = \sqrt{2g} = \sqrt{19.6} = 4.43\ \text{m}\,\text{s}^{-1}$ M1 A1

Using $v\frac{dv}{dx} = g - 0.5v^2$ M1

$\int \frac{v}{g - 0.5v^2} dv = \int dx$ M1

$-\ln(g - 0.5v^2) = x + C$ (up to constant factor) A1

At $x = 0$, $v = 0$: $-\ln g = C$

$-\ln(g - 0.5v^2) = x - \ln g$ $\ln\left(\frac{g}{g - 0.5v^2}\right) = x$ $g - 0.5v^2 = ge^{-x}$ $v^2 = 2g(1 - e^{-x})$ $v = \sqrt{2g(1 - e^{-x})}$ A1

[Total: 7]

Example 2: A projectile of mass $m$ is fired vertically upwards with speed $u$ in a medium where resistance is $mkv^2$. Show that the time to reach the highest point is $\frac{1}{\sqrt{kg}}\tan^{-1}\left(u\sqrt{\frac{k}{g}}\right)$.

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Upward motion: $m\frac{dv}{dt} = -mg - mkv^2$ M1 $\frac{dv}{dt} = -g - kv^2$

$\int \frac{dv}{g + kv^2} = -\int dt$ M1

$\frac{1}{\sqrt{kg}}\tan^{-1}\left(v\sqrt{\frac{k}{g}}\right) = -t + C$ A1

At $t = 0$, $v = u$: $C = \frac{1}{\sqrt{kg}}\tan^{-1}\left(u\sqrt{\frac{k}{g}}\right)$ M1

At highest point, $v = 0$: $0 = -t + \frac{1}{\sqrt{kg}}\tan^{-1}\left(u\sqrt{\frac{k}{g}}\right)$ M1

$t = \frac{1}{\sqrt{kg}}\tan^{-1}\left(u\sqrt{\frac{k}{g}}\right)$ A1

[Total: 6]

Example 3: A particle moves in a straight line under a force $F = -kv^2$ (retarding force). Initially $v = V$. Find the distance travelled before the speed reduces to $V/2$.

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$m\frac{dv}{dt} = -kv^2$ M1

Using $a = v\frac{dv}{dx}$: $mv\frac{dv}{dx} = -kv^2$ M1

$\int \frac{m}{v} dv = -\int k dx$ M1

$m\ln v = -kx + C$ A1

At $x = 0$, $v = V$: $C = m\ln V$

$m\ln v = -kx + m\ln V$ $\ln\left(\frac{v}{V}\right) = -\frac{kx}{m}$ M1

When $v = V/2$: $\ln\left(\frac{1}{2}\right) = -\frac{kx}{m}$ $x = \frac{m\ln 2}{k}$ A1

[Total: 6]