| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2017 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Motion with friction on horizontal surface |
| Difficulty | Standard +0.8 This M2 question requires understanding of limiting friction, Newton's second law with variable force, and integration. While the calculus is straightforward, students must recognize the threshold condition at t=0.5, correctly apply F=ma with friction, then integrate twice. The multi-stage reasoning and careful handling of the friction transition elevates this above routine mechanics problems. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation3.03r Friction: concept and vector form6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| 7(i) | µ = 0.6 × 0.52 / (0.5 g) ( = 0.03) | B1 |
| 0.5dv / dt = 0.6t2 – 0.03 × 0.5g | M1 | Uses Newton's Second Law horizontally |
| dv / dt = 1.2t2 – 0.3 | A1 | |
| Total: | 3 | |
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 7(ii) | ∫dv = ∫(1.2t2 – 0.3) dt | |
| v = 0.4t3 – 0.3t ( + c) | M1 | Separates the variables and attempts to integrate |
| t = 0.5, v = 0 hence c = 0.1 | M1 | Attempts to find c |
| v = 0.4t3 – 0.3t + 0.1 | A1 | |
| Total: | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| 7(iii) | ∫dx = ∫(0.4t3 – 0.3t + 0.1) dt | |
| x = 0.1t4 – 0.15t2 + 0.1t ( + c) | M1 | Attempts to integrate |
| t = 0.5, x = 0 hence c = –0.01875 | M1 | Finds c or substitutes the limits |
| x(1.2) = 0.0926(1) | A1 | |
| Total: | 3 |
Question 7:
--- 7(i) ---
7(i) | µ = 0.6 × 0.52 / (0.5 g) ( = 0.03) | B1 | Uses F = µR
0.5dv / dt = 0.6t2 – 0.03 × 0.5g | M1 | Uses Newton's Second Law horizontally
dv / dt = 1.2t2 – 0.3 | A1
Total: | 3
Question | Answer | Marks | Guidance
--- 7(ii) ---
7(ii) | ∫dv = ∫(1.2t2 – 0.3) dt
v = 0.4t3 – 0.3t ( + c) | M1 | Separates the variables and attempts to integrate
t = 0.5, v = 0 hence c = 0.1 | M1 | Attempts to find c
v = 0.4t3 – 0.3t + 0.1 | A1
Total: | 3
--- 7(iii) ---
7(iii) | ∫dx = ∫(0.4t3 – 0.3t + 0.1) dt
x = 0.1t4 – 0.15t2 + 0.1t ( + c) | M1 | Attempts to integrate
t = 0.5, x = 0 hence c = –0.01875 | M1 | Finds c or substitutes the limits
x(1.2) = 0.0926(1) | A1
Total: | 3A particle $P$ of mass $0.5$ kg is at rest at a point $O$ on a rough horizontal surface. At time $t = 0$, where $t$ is in seconds, a horizontal force acting in a fixed direction is applied to $P$. At time $t$ s the magnitude of the force is $0.6t^2$ N and the velocity of $P$ away from $O$ is $v\,\text{m}\,\text{s}^{-1}$. It is given that $P$ remains at rest at $O$ until $t = 0.5$.
\begin{enumerate}[label=(\roman*)]
\item Calculate the coefficient of friction between $P$ and the surface, and show that
$$\frac{\text{d}v}{\text{d}t} = 1.2t^2 - 0.3 \quad \text{for } t > 0.5.$$ [3]
\item Express $v$ in terms of $t$ for $t > 0.5$. [3]
\item Find the displacement of $P$ from $O$ when $t = 1.2$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2017 Q7 [9]}}