| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: resultant and acceleration |
| Difficulty | Standard +0.8 This is a variable force mechanics problem requiring Newton's second law with time-dependent forces, followed by integration to find velocity and displacement. Part (i) is routine force equation setup, but parts (ii) and (iii) require understanding initial conditions and integrating exponential/polynomial expressions—moderately challenging for M2 level but follows standard techniques without requiring novel insight. |
| Spec | 4.10b Model with differential equations: kinematics and other contexts6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| 7(i) | 0.2dv/dt = 0.2 g + 0.6t – ke−t | M1 |
| dv/dt = 10 + 3t – 5ke−t AG | A1 | |
| Total: | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| 7(ii) | dv/dt = 10 – 5ke0 = 0 | M1 |
| M1 | Attempts to solve the equation | |
| 7(ii) | k = 2 | A1 |
| Total: | 3 | |
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 7(iii) | ∫dv = ∫(10 + 3t – 5ke−t)dt | M1 |
| Answer | Marks |
|---|---|
| v = 10t + 3t2/2 + 5e−t – 5 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| x = 5t2 + t3/2 – 5e−t – 5t + c | M1 | Attempts to integrate again. Allow their k or just k not replaced |
| Answer | Marks |
|---|---|
| x = 5 × 22 + 23/2 – 5e−2 – 5 × 2 + 5 | M1 |
| Height = 18.3 m | A1 |
| Total: | 5 |
Question 7:
--- 7(i) ---
7(i) | 0.2dv/dt = 0.2 g + 0.6t – ke−t | M1 | Use Newton's Second Law downwards
dv/dt = 10 + 3t – 5ke−t AG | A1
Total: | 2
--- 7(ii) ---
7(ii) | dv/dt = 10 – 5ke0 = 0 | M1 | Recognise that dv/dt = 0 when t = 0
M1 | Attempts to solve the equation
7(ii) | k = 2 | A1
Total: | 3
Question | Answer | Marks | Guidance
--- 7(iii) ---
7(iii) | ∫dv = ∫(10 + 3t – 5ke−t)dt | M1 | Attempts to integrate the equation from part i with k not replaced
[v = 10t + 3t2 /2 + 5e−t + c, v = 0, t = 0 so c = – 5]
v = 10t + 3t2/2 + 5e−t – 5 | A1
∫dx = ∫(10t + 3t2/2 + 5e−t – 5)dt
x = 5t2 + t3/2 – 5e−t – 5t + c | M1 | Attempts to integrate again. Allow their k or just k not replaced
x = 0, t = 0, so c = 5 and substitutes t = 2
x = 5 × 22 + 23/2 – 5e−2 – 5 × 2 + 5 | M1
Height = 18.3 m | A1
Total: | 5A particle $P$ of mass 0.2 kg is released from rest at a point $O$ above horizontal ground. At time $t$ s after its release the velocity of $P$ is 7.5 m s$^{-1}$ downwards. A vertically downwards force of magnitude 0.6t N acts on $P$. A vertically upwards force of magnitude $ke^{-t}$ N, where $k$ is a constant, also acts on $P$.
\begin{enumerate}[label=(\roman*)]
\item Show that $\frac{dv}{dt} = 10 - 5ke^{-t} + 3t$. [2]
\item Find the greatest value of $k$ for which $P$ does not initially move upwards. [3]
\item Given that $k = 1$, and that $P$ strikes the ground when $t = 2$, find the height of $O$ above the ground. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2018 Q7 [10]}}