| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2017 |
| Session | March |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Variable force (position x) - find velocity |
| Difficulty | Challenging +1.2 This is a variable force mechanics problem requiring Newton's second law with v(dv/dx) = a, integration of a rational function, and inequality solving. Part (i) is straightforward force equation setup. Part (ii) requires integrating ln(x) terms and solving inequalities with given bounds. Standard M2/Further Mechanics content with multiple steps but follows predictable methods without requiring novel insight. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| 6(i) | Friction = 0.4 x 0.4g | B1 |
| 0.4vdv/dx = –0.4 x 0.4g – 0.8/x | M1 | Uses Newton's Second Law and |
| Answer | Marks |
|---|---|
| vdv/dx = –4 – 2/x | A1 |
| Total: | 3 |
| Answer | Marks |
|---|---|
| 6(ii) | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| x | M1 | Separates the variables and attempts to |
| Answer | Marks | Guidance |
|---|---|---|
| v2/2 = –4x – 2lnx ( + c) | A1 | |
| v = U when x = 1 hence c = U2 /2 + 4 | M1 | Attempts to find c |
| Answer | Marks | Guidance |
|---|---|---|
| [U2 =11.7(677)] | M1 | Put v = 0 and x = 2 |
| Answer | Marks | Guidance |
|---|---|---|
| 3.28 ˂ U ˂ 3.43 | A1 | |
| Total: | 5 | |
| Question | Answer | Marks |
Question 6:
--- 6(i) ---
6(i) | Friction = 0.4 x 0.4g | B1 | Uses F = µR
0.4vdv/dx = –0.4 x 0.4g – 0.8/x | M1 | Uses Newton's Second Law and
a = vdv/dx
vdv/dx = –4 – 2/x | A1
Total: | 3
--- 6(ii) ---
6(ii) | 2
∫vdv = ∫−4− dx
x | M1 | Separates the variables and attempts to
integrate
v2/2 = –4x – 2lnx ( + c) | A1
v = U when x = 1 hence c = U2 /2 + 4 | M1 | Attempts to find c
0= –4 x 2 – 2ln2 + U2/2 + 4 [U2= 10.7(725)]
0= –4 x 2.1 – 2ln2.1 + U2/2 + 4
[U2 =11.7(677)] | M1 | Put v = 0 and x = 2
Put v = 0 and x = 2.1
3.28 ˂ U ˂ 3.43 | A1
Total: | 5
Question | Answer | Marks | Guidance$O$ and $A$ are fixed points on a rough horizontal surface, with $OA = 1 \text{ m}$. A particle $P$ of mass $0.4 \text{ kg}$ is projected horizontally with speed $U \text{ m s}^{-1}$ from $A$ in the direction $OA$ and moves in a straight line. After projection, when the displacement of $P$ from $O$ is $x \text{ m}$, the velocity of $P$ is $v \text{ m s}^{-1}$. The coefficient of friction between the surface and $P$ is $0.4$. A force of magnitude $\frac{0.8}{x} \text{ N}$ acts on $P$ in the direction $PO$.
\begin{enumerate}[label=(\roman*)]
\item Show that, while the particle is in motion, $v \frac{\text{d}v}{\text{d}x} = -4 - \frac{2}{x}$. [3]
\end{enumerate}
It is given that $P$ comes to instantaneous rest between $x = 2.0$ and $x = 2.1$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the set of possible values of $U$. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2017 Q6 [8]}}