| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: resultant and acceleration |
| Difficulty | Standard +0.3 This is a straightforward mechanics problem requiring application of Newton's second law with time-dependent forces, followed by integration with initial conditions. The algebra is simple (exponential and polynomial terms), and all steps are standard M2 techniques with no novel insight required. Slightly easier than average due to the guided structure and routine calculus. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks |
|---|---|
| 3(i) | dv |
| Answer | Marks | Guidance |
|---|---|---|
| dt | M1 | Use Newton’s Second Law |
| Answer | Marks | Guidance |
|---|---|---|
| dt | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| 3(ii) | ∫dv=∫(2t−5e−t)dt | |
| v=t2 +5e−t(+ c) | M1 | Attempt to integrate the |
| Answer | Marks | Guidance |
|---|---|---|
| t = 1 and v = 8 so c = 5.16 | M1 | Attempt to find the constant |
| Answer | Marks |
|---|---|
| v=t2 +5e−t +5.16 or v=t2 +5e−t +7−5e−1 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3(iii) | Evaluates v for t = 0 | M1 |
| V =10.2 ms−1 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 3:
--- 3(i) ---
3(i) | dv
0.4 = 0.8t – 2e−t
dt | M1 | Use Newton’s Second Law
horizontally
dv
= 2t – 5e−t
dt | A1 | AG
2
--- 3(ii) ---
3(ii) | ∫dv=∫(2t−5e−t)dt
v=t2 +5e−t(+ c) | M1 | Attempt to integrate the
equation from part (i)
t = 1 and v = 8 so c = 5.16 | M1 | Attempt to find the constant
of integration, c
v=t2 +5e−t +5.16 or v=t2 +5e−t +7−5e−1 | A1
3
--- 3(iii) ---
3(iii) | Evaluates v for t = 0 | M1
V =10.2 ms−1 | A1
2
Question | Answer | Marks | GuidanceA particle $P$ of mass $0.4 \text{ kg}$ is projected horizontally along a smooth horizontal plane from a point $O$. At time $t \text{ s}$ after projection the velocity of $P$ is $v \text{ ms}^{-1}$. A force of magnitude $0.8t \text{ N}$ directed away from $O$ acts on $P$ and a force of magnitude $2e^{-t} \text{ N}$ opposes the motion of $P$.
\begin{enumerate}[label=(\roman*)]
\item Show that $\frac{dv}{dt} = 2t - 5e^{-t}$. [2]
\item Given that $v = 8$ when $t = 1$, express $v$ in terms of $t$. [3]
\item Find the speed of projection of $P$. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2018 Q3 [7]}}