| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Particle on slope then horizontal |
| Difficulty | Standard +0.3 This is a straightforward energy conservation problem with standard mechanics techniques. Part (a) requires basic PE to KE conversion (energy is conserved on smooth surfaces), and part (b) applies work-energy theorem with friction over a horizontal surface. The geometry is simple (one-sixth circle gives clear height drop), calculations are routine, and it's a typical M2 textbook exercise requiring no novel insight. |
| Spec | 3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Height of \(P\) above \(Q\): \(r - r\cos 60° = 4 - 4\times\frac{1}{2} = 2\) m | M1 | Find height dropped from \(P\) to \(Q\) |
| \(P\) is 2 m above \(Q\), so \(P\) is 2 m above \(QR\) (horizontal) | ||
| KE at \(R\) = loss in PE from \(P\) to \(R\) \(= mgh = 32\times 9.8\times 2\) | M1 A1 | Use energy conservation \(P\) to \(R\) |
| \(= 627.2\) J | A1 | Accept \(640\) J if \(g=10\) used |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2}mv^2 = 627.2\) | M1 | |
| \(v^2 = \frac{2\times 627.2}{32} = 39.2\) | ||
| \(v = 6.26\) m s\(^{-1}\) | A1 | Accept \(6.32\) if \(g=10\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Work-energy: KE at \(Q\) = KE at \(R\) + Work done against friction | M1 | |
| KE at \(Q = 32\times 9.8\times 2 = 627.2\) J | Height from \(P\) to \(Q\) = 2 m | |
| Friction force \(F = \mu \times 32g\) | M1 | Normal reaction \(= mg\) on horizontal |
| \(627.2 = \mu\times 32\times 9.8\times 5\) | M1 | Work done by friction \(= F\times 5\) |
| \(\mu = \frac{627.2}{1568} = 0.4\) | A1 |
# Question 3:
## Part (a)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Height of $P$ above $Q$: $r - r\cos 60° = 4 - 4\times\frac{1}{2} = 2$ m | M1 | Find height dropped from $P$ to $Q$ |
| $P$ is 2 m above $Q$, so $P$ is 2 m above $QR$ (horizontal) | | |
| KE at $R$ = loss in PE from $P$ to $R$ $= mgh = 32\times 9.8\times 2$ | M1 A1 | Use energy conservation $P$ to $R$ |
| $= 627.2$ J | A1 | Accept $640$ J if $g=10$ used |
## Part (a)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}mv^2 = 627.2$ | M1 | |
| $v^2 = \frac{2\times 627.2}{32} = 39.2$ | | |
| $v = 6.26$ m s$^{-1}$ | A1 | Accept $6.32$ if $g=10$ |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Work-energy: KE at $Q$ = KE at $R$ + Work done against friction | M1 | |
| KE at $Q = 32\times 9.8\times 2 = 627.2$ J | | Height from $P$ to $Q$ = 2 m |
| Friction force $F = \mu \times 32g$ | M1 | Normal reaction $= mg$ on horizontal |
| $627.2 = \mu\times 32\times 9.8\times 5$ | M1 | Work done by friction $= F\times 5$ |
| $\mu = \frac{627.2}{1568} = 0.4$ | A1 | |3 A diagram shows a children's slide, $P Q R$.\\
\includegraphics[max width=\textwidth, alt={}, center]{691c50b4-50b2-4e3a-a7e0-60f8ec35ee3c-06_352_640_338_699}
Simon, a child of mass 32 kg , uses the slide, starting from rest at $P$. The curved section of the slide, $P Q$, is one sixth of a circle of radius 4 metres so that the child is travelling horizontally at point $Q$. The centre of this circle is at point $O$, which is vertically above point $Q$. The section $Q R$ is horizontal and of length 5 metres.
Assume that air resistance may be ignored.
\begin{enumerate}[label=(\alph*)]
\item Assume that the two sections of the slide, $P Q$ and $Q R$, are both smooth.
\begin{enumerate}[label=(\roman*)]
\item Find the kinetic energy of Simon when he reaches the point $R$.
\item Hence find the speed of Simon when he reaches the point $R$.
\end{enumerate}\item In fact, the section $Q R$ is rough.
Assume that the section $P Q$ is smooth.\\
Find the coefficient of friction between Simon and the section $Q R$ if Simon comes to rest at the point $R$.\\[0pt]
[4 marks]
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{691c50b4-50b2-4e3a-a7e0-60f8ec35ee3c-06_923_1707_1784_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2015 Q3 [9]}}