| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Acceleration as function of velocity (separation of variables) |
| Difficulty | Standard +0.3 This is a standard M2 variable acceleration question requiring Newton's second law with resistance, followed by separating variables to solve a first-order ODE, then integrating velocity to find displacement. While it involves multiple steps (forming equation, solving ODE, integration), these are routine techniques for M2 students with no novel insight required. Slightly above average difficulty due to the multi-stage process and careful handling of the differential equation. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)6.06a 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]A ball of mass 0.05 kg is released from rest at a height $h$ m above the ground. At time $t$ s after its release, the downward velocity of the ball is $v$ m s$^{-1}$. Air resistance opposes the motion of the ball with a force of magnitude 0.01$v$ N.
\begin{enumerate}[label=(\roman*)]
\item Show that $\frac{dv}{dt} = 10 - 0.2v$. Hence find $v$ in terms of $t$. [6]
\item Given that the ball reaches the ground when $t = 2$, calculate $h$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2011 Q5 [10]}}