| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Constant acceleration vector problems |
| Difficulty | Standard +0.3 This is a multi-part mechanics question requiring Newton's second law and integration across three time intervals with changing forces. While it involves non-constant acceleration (0.32t and 0.06t²), the calculus is straightforward and the problem is methodically structured with clear phases. It's slightly above average due to the piecewise nature and need to track velocity across intervals, but remains a standard M2 exercise without requiring novel insight. |
| Spec | 3.03d Newton's second law: 2D vectors6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(0.2a = 0.42\) | M1 | Newton's Second Law and \(v = u + at\) |
| \(v = (2.1 \times 1) = 2.1\) | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(0.2\,dv/dt = 0.42 - 0.32t\) | M1 | Newton's Second Law with \(a = dv/dt\) |
| \([v]_{2.1}^{v} = [2.1t - 0.8t^2]_1^2\) | M1 | |
| \(v = 1.8\) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(v = \int(0.42 - 0.32t + 0.06t^2)\,dt / 0.2\) | M1 | |
| \(v = [0.42t - 0.16t^2 + 0.02t^3]_0^3 / 0.2\) or | M1 | For attempt to integrate and correct limits seen |
| \([v]_{1.8}^{v} = [0.42t - 1.16t^2 + 0.02t^3]_2^3 / 0.2\) | ||
| \(v = 1.8\), so no change | A1 | [3] |
## Question 3:
### Part (i):
| Working | Mark | Guidance |
|---------|------|----------|
| $0.2a = 0.42$ | M1 | Newton's Second Law and $v = u + at$ |
| $v = (2.1 \times 1) = 2.1$ | A1 | [2] |
### Part (ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $0.2\,dv/dt = 0.42 - 0.32t$ | M1 | Newton's Second Law with $a = dv/dt$ |
| $[v]_{2.1}^{v} = [2.1t - 0.8t^2]_1^2$ | M1 | |
| $v = 1.8$ | A1 | [3] |
### Part (iii):
| Working | Mark | Guidance |
|---------|------|----------|
| $v = \int(0.42 - 0.32t + 0.06t^2)\,dt / 0.2$ | M1 | |
| $v = [0.42t - 0.16t^2 + 0.02t^3]_0^3 / 0.2$ or | M1 | For attempt to integrate and correct limits seen |
| $[v]_{1.8}^{v} = [0.42t - 1.16t^2 + 0.02t^3]_2^3 / 0.2$ | | |
| $v = 1.8$, so no change | A1 | [3] |
---3 A small block $B$ of mass 0.2 kg is placed at a fixed point $O$ on a smooth horizontal surface. A horizontal force of magnitude 0.42 N is applied to $B$. At time $t \mathrm {~s}$ after the force is first applied, the velocity of $B$ away from $O$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Find the value of $v$ when $t = 1$.
For $t > 1$ an additional force, of magnitude $0.32 t \mathrm {~N}$ and directed towards $O$, is applied to $B$. The force of magnitude 0.42 N continues to act as before.\\
(ii) Find the value of $v$ when $t = 2$.
For $t > 2$ a third force, of magnitude $0.06 t ^ { 2 } \mathrm {~N}$ and directed away from $O$, is applied to $B$. The other two forces continue to act as before.\\
(iii) Show that the velocity of $B$ is the same when $t = 2$ and when $t = 3$.
\hfill \mbox{\textit{CAIE M2 2014 Q3 [8]}}