| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance kv² - horizontal motion or engine power |
| Difficulty | Standard +0.8 This M3 question requires setting up and solving a differential equation with variable resistance (F = 0.2v²), then integrating to find distance. Part (i) involves separating variables and partial fractions, while part (ii) requires finding distance via ∫v dt. The non-standard resistance law and two-part integration make this moderately challenging, though the techniques are standard M3 material. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.25(\mathrm{d}v/\mathrm{d}t) = -0.2v^2\) | M1, dep | For using Newton's second law with \(a = \mathrm{d}v/\mathrm{d}t\). Allow sign error and/or omitting mass. For separating variables and attempting to integrate (ie get \(v^{-1}\) and \(t\)). |
| \(0.25 \int v^{-2}\mathrm{d}v = -0.2t(+C)\) | M1 | |
| \(-v^{-1}/4 = -t/5 + C\) \([1/4v = t/5 + 1/20]\) | A1, M1 | For using \(v(0) = 5\) to obtain \(C\) |
| \(v = \frac{5}{4t + 1}\) oe | A1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = (5/4)\ln(4t + 1) (+B)\) \(\text{Subst } v = 0.2 \text{ in (i) to find } t\) \(\text{Obtain } x(6) (= 1.25 \ln25 \text{ oe } (4.02359...))\) \(\text{Average speed is } 0.671 \text{ ms}^{-1}\) | M1, A1, M1, M1, A1 [5] | For using \(v = \mathrm{d}x/\mathrm{d}t\) and integrating. Implied by \(t = 6\). May be written as \(\frac{5}{12}\ln 5\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\ln v = -0.8x + B\) \(\text{Subst } v = 0.2 \text{ in (i) to find } t\) \(\text{Obtain } x(0.2) (= 1.25 \ln(5/0.2) \text{ oe } (4.0239...))\) \(\text{Average speed is } 0.671 \text{ ms}^{-1}\) | M1, A1, M1, M1, A1 [5] | For using \(mv(\mathrm{d}v/\mathrm{d}x) = -0.2v^2\), separating variables and integrating. Allow sign error and/or omitting mass. Implied by \(t = 6\). May be written as \(\frac{5}{12}\ln 5\) |
**Part (i):**
$0.25(\mathrm{d}v/\mathrm{d}t) = -0.2v^2$ | M1, dep | For using Newton's second law with $a = \mathrm{d}v/\mathrm{d}t$. Allow sign error and/or omitting mass. For separating variables and attempting to integrate (ie get $v^{-1}$ and $t$). |
$0.25 \int v^{-2}\mathrm{d}v = -0.2t(+C)$ | M1 |
$-v^{-1}/4 = -t/5 + C$ $[1/4v = t/5 + 1/20]$ | A1, M1 | For using $v(0) = 5$ to obtain $C$ |
$v = \frac{5}{4t + 1}$ oe | A1 [5] |
**Part (ii):**
$x = (5/4)\ln(4t + 1) (+B)$ $\text{Subst } v = 0.2 \text{ in (i) to find } t$ $\text{Obtain } x(6) (= 1.25 \ln25 \text{ oe } (4.02359...))$ $\text{Average speed is } 0.671 \text{ ms}^{-1}$ | M1, A1, M1, M1, A1 [5] | For using $v = \mathrm{d}x/\mathrm{d}t$ and integrating. Implied by $t = 6$. May be written as $\frac{5}{12}\ln 5$ |
**Alternative approach:**
$\ln v = -0.8x + B$ $\text{Subst } v = 0.2 \text{ in (i) to find } t$ $\text{Obtain } x(0.2) (= 1.25 \ln(5/0.2) \text{ oe } (4.0239...))$ $\text{Average speed is } 0.671 \text{ ms}^{-1}$ | M1, A1, M1, M1, A1 [5] | For using $mv(\mathrm{d}v/\mathrm{d}x) = -0.2v^2$, separating variables and integrating. Allow sign error and/or omitting mass. Implied by $t = 6$. May be written as $\frac{5}{12}\ln 5$ |A particle $P$ of mass $0.25$ kg is projected horizontally with speed $5$ m s$^{-1}$ from a fixed point $O$ on a smooth horizontal surface and moves in a straight line on the surface. The only horizontal force acting on $P$ has magnitude $0.2v^2$ N, where $v$ m s$^{-1}$ is the velocity of $P$ at time $t$ s after it is projected from $O$. This force is directed towards $O$.
\begin{enumerate}[label=(\roman*)]
\item Find an expression for $v$ in terms of $t$. [5]
\end{enumerate}
The particle $P$ passes through a point $X$ with speed $0.2$ m s$^{-1}$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the average speed of $P$ for its motion between $O$ and $X$. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR M3 2011 Q3 [10]}}