| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance kv - vertical motion |
| Difficulty | Standard +0.3 This is a standard M2 differential equation problem with air resistance proportional to velocity. Part (i) requires applying Newton's second law (routine setup) and solving a first-order linear DE using standard separation of variables. Part (ii) involves integrating velocity to find displacement. While it requires multiple steps and familiarity with the technique, it follows a well-established template for this topic with no novel insights needed, making it slightly easier than average overall. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(0.05dv/dt = 0.05g - 0.01v\) | M1 | Uses Newton's Second Law |
| \(dv/dt = 10 - 0.2v\) | AG A1 | |
| \(\int dv/(10 - 0.2v) = \int dt\) | M1 | |
| \(-\ln(10 - 0.2v)/0.2 = t (+ c)\) | A1 | |
| \(t = 0, v = 0\), hence \(c = -5\ln10\) | M1 | \(-4.60517\ldots\) |
| \(\ln(10 - 0.2v)/10 = 0.2t, 1 - 0.02v = e^{-0.2t}\) | A1 | [6] |
| \(v = 50 - 50e^{-0.2t}\) | ||
| (ii) \(dv/dt = 50 - 50e^{-0.2t}\) | ||
| \(x = \int (50 - 50e^{-0.2})dt\) | M1 | |
| \(x = 50t + 50e^{-0.2}/(0.2) (+c)\) | A1 | |
| \(h = [50t + 50e^{-0.2}/0.2]_2^0\) | M1 | Or uses \(h = 0, t = 0\) to evaluate \(c = (-250)\) and then finds \(h(2)\) |
| \(h = 17.6\) | A1 | [4] |
**(i)** $0.05dv/dt = 0.05g - 0.01v$ | M1 | Uses Newton's Second Law
$dv/dt = 10 - 0.2v$ | AG A1
$\int dv/(10 - 0.2v) = \int dt$ | M1
$-\ln(10 - 0.2v)/0.2 = t (+ c)$ | A1
$t = 0, v = 0$, hence $c = -5\ln10$ | M1 | $-4.60517\ldots$
$\ln(10 - 0.2v)/10 = 0.2t, 1 - 0.02v = e^{-0.2t}$ | A1 | [6]
$v = 50 - 50e^{-0.2t}$ | |
**(ii)** $dv/dt = 50 - 50e^{-0.2t}$ | |
$x = \int (50 - 50e^{-0.2})dt$ | M1
$x = 50t + 50e^{-0.2}/(0.2) (+c)$ | A1
$h = [50t + 50e^{-0.2}/0.2]_2^0$ | M1 | Or uses $h = 0, t = 0$ to evaluate $c = (-250)$ and then finds $h(2)$
$h = 17.6$ | A1 | [4]
---5 A ball of mass 0.05 kg is released from rest at a height $h \mathrm {~m}$ above the ground. At time $t \mathrm {~s}$ after its release, the downward velocity of the ball is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Air resistance opposes the motion of the ball with a force of magnitude $0.01 \nu \mathrm {~N}$.\\
(i) Show that $\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 - 0.2 v$. Hence find $v$ in terms of $t$.\\
(ii) Given that the ball reaches the ground when $t = 2$, calculate $h$.
\hfill \mbox{\textit{CAIE M2 2011 Q5 [10]}}