| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Variable resistance: find constant speed |
| Difficulty | Standard +0.3 This is a standard M2 power-resistance question requiring P=Fv at maximum speed, then applying the same principle on an incline with weight component. The algebra is straightforward and the 'show that' structure guides students through each step, making it slightly easier than average for M2. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Resolving vertically: \(T_{AP}\cos20° = 5g\) | M1 | Vertical equilibrium |
| \(T_{AP} = \frac{5g}{\cos20°}\) | A1 | |
| \(T_{AP} = 52.1\) N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Resolving horizontally: \(T_{AP}\sin20° + T_{BP} = \frac{mv^2}{r}\) | M1 | Newton's second law horizontally |
| \(T_{BP} = \frac{5v^2}{0.6} - T_{AP}\sin20°\) | M1 | |
| \(T_{BP} = \frac{25}{3}v^2 - \frac{5g\sin20°}{\cos20°} = \frac{25}{3}v^2 - 5g\tan20°\) | A1 | Correct simplification shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T_{AP} = T_{BP}\): \(\frac{5g}{\cos20°} = \frac{25}{3}v^2 - 5g\tan20°\) | M1 | Setting tensions equal |
| \(\frac{25}{3}v^2 = 5g\tan20° + \frac{5g}{\cos20°}\) | M1 | Rearranging |
| \(v^2 = \frac{3}{25}\left(5g\tan20° + \frac{5g}{\cos20°}\right)\) | A1 | |
| \(v = 2.72\) ms\(^{-1}\) | A1 |
# Question 4:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Resolving vertically: $T_{AP}\cos20° = 5g$ | M1 | Vertical equilibrium |
| $T_{AP} = \frac{5g}{\cos20°}$ | A1 | |
| $T_{AP} = 52.1$ N | A1 | |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Resolving horizontally: $T_{AP}\sin20° + T_{BP} = \frac{mv^2}{r}$ | M1 | Newton's second law horizontally |
| $T_{BP} = \frac{5v^2}{0.6} - T_{AP}\sin20°$ | M1 | |
| $T_{BP} = \frac{25}{3}v^2 - \frac{5g\sin20°}{\cos20°} = \frac{25}{3}v^2 - 5g\tan20°$ | A1 | Correct simplification shown |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T_{AP} = T_{BP}$: $\frac{5g}{\cos20°} = \frac{25}{3}v^2 - 5g\tan20°$ | M1 | Setting tensions equal |
| $\frac{25}{3}v^2 = 5g\tan20° + \frac{5g}{\cos20°}$ | M1 | Rearranging |
| $v^2 = \frac{3}{25}\left(5g\tan20° + \frac{5g}{\cos20°}\right)$ | A1 | |
| $v = 2.72$ ms$^{-1}$ | A1 | |
---4 A car has a maximum speed of $42 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when it is moving on a horizontal road. When the speed of the car is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, it experiences a resistance force of magnitude $30 v$ newtons.
\begin{enumerate}[label=(\alph*)]
\item Show that the maximum power of the car is 52920 W .
\item The car has mass 1200 kg . It travels, from rest, up a slope inclined at $5 ^ { \circ }$ to the horizontal.
\begin{enumerate}[label=(\roman*)]
\item Show that, when the car is travelling at its maximum speed $\mathrm { V } \mathrm { m } \mathrm { s } ^ { - 1 }$ up the slope,
$$V ^ { 2 } + 392 \sin 5 ^ { \circ } V - 1764 = 0$$
\item Hence find $V$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M2 Q4}}