| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2006 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Force depends on time t |
| Difficulty | Standard +0.3 This is a straightforward M2 variable force question requiring reading a graph, applying F=ma, then integrating twice with given initial conditions. All steps are standard procedures with no conceptual challenges—slightly easier than average due to the scaffolded structure and explicit 'show that' in part (a). |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.03d Newton's second law: 2D vectors6.02c Work by variable force: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(F = 800 + \frac{1200}{20}t = 800 + 60t\) and \(1200a = 800 + 60t\), so \(a = \frac{800}{1200} - \frac{60}{1200}t = \frac{2}{3} - \frac{t}{20}\) | M1, A1, B1, dM1, A1 | 5 marks |
| (b) \(v = \int\left(\frac{2}{3} + \frac{t}{20}\right)dt = \frac{2t}{3} + \frac{t^2}{40} + c\) with \(v = 0, t = 0 \Rightarrow c = 0\), so \(v = \frac{2t}{3} + \frac{t^2}{40}\) | M1, A1, A1 | 3 marks |
| (c) \(s = \int_0^{20}\left(\frac{2t}{3} + \frac{t^2}{40}\right)dt = \left[\frac{t^2}{3} + \frac{t^3}{120}\right]_0^{20} = 200 \text{ m}\) | M1, A1, dM1, A1 | 4 marks |
| (d) The \(\frac{2t}{3}\) term would change, because only the constant term in the force would change. When integrated this becomes the \(t\) term in the velocity. | B1, B1 | 2 marks |
**(a)** $F = 800 + \frac{1200}{20}t = 800 + 60t$ and $1200a = 800 + 60t$, so $a = \frac{800}{1200} - \frac{60}{1200}t = \frac{2}{3} - \frac{t}{20}$ | M1, A1, B1, dM1, A1 | 5 marks | Finding the gradient of the line; correct gradient; correct intercept; using Newton's second law on two terms; correct result from correct working. (AG)
**(b)** $v = \int\left(\frac{2}{3} + \frac{t}{20}\right)dt = \frac{2t}{3} + \frac{t^2}{40} + c$ with $v = 0, t = 0 \Rightarrow c = 0$, so $v = \frac{2t}{3} + \frac{t^2}{40}$ | M1, A1, A1 | 3 marks | Integrating; correct integral with or without $c$; showing $c = 0$
**(c)** $s = \int_0^{20}\left(\frac{2t}{3} + \frac{t^2}{40}\right)dt = \left[\frac{t^2}{3} + \frac{t^3}{120}\right]_0^{20} = 200 \text{ m}$ | M1, A1, dM1, A1 | 4 marks | Integrating; correct integral, with or without $c$; use of both limits or finding $c$; correct distance
**(d)** The $\frac{2t}{3}$ term would change, because only the constant term in the force would change. When integrated this becomes the $t$ term in the velocity. | B1, B1 | 2 marks | Correct term; correct explanation
**Total for Question 5: 14 marks**
---5 The graph shows a model for the resultant horizontal force on a car, which varies as it accelerates from rest for 20 seconds. The mass of the car is 1200 kg .\\
\includegraphics[max width=\textwidth, alt={}, center]{c02cf013-365b-44e2-8c16-aa8209cbe250-4_373_1203_445_390}
\begin{enumerate}[label=(\alph*)]
\item The acceleration of the car at time $t$ seconds is $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$. Show that
$$a = \frac { 2 } { 3 } + \frac { t } { 20 } , \text { for } 0 \leqslant t \leqslant 20$$
\item Find an expression for the velocity of the car at time $t$.
\item Find the distance travelled by the car in the 20 seconds.
\item An alternative model assumes that the resultant force increases uniformly from 900 to 2100 newtons during the 20 seconds. Which term in your expression for the velocity would change as a result of this modification? Explain why.
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2006 Q5 [14]}}