| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Force depends on time t |
| Difficulty | Standard +0.3 This is a standard M2 variable force question requiring integration of F=ma with time-dependent force. Part (i) involves comparing forces to find when motion starts and deriving acceleration from Newton's second law. Parts (ii) and (iii) require straightforward integration with initial conditions. While it involves multiple steps, the techniques are routine for M2 students with no novel problem-solving required. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(t = 2\) | B1 | AG — From \(0.6t = 0.3 \times 4\) or \(1.5t - 3 = 0\) |
| \(0.4a = 0.6t - 0.3 \times 0.4g\) | M1 | |
| \(a = 1.5t - 3\) | A1 (3) | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v = \int(1.5t - 3)\,dt\) | M1 | |
| \(v = 0.75t^2 - 3t\ (+c)\) | A1 | |
| \(0 = 0.75 \times 2^2 - 3 \times 2 + c\) (so \(c = 3\)) | M1 | Or uses limits with 2 and \(t\) |
| \(v = 0.75t^2 - 3t + 3\) | A1 (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = \int(0.75t^2 - 3t + 3)\,dt\) | M1 | |
| \(x = 0.25t^3 - 1.5t^2 + 3t\ (+k)\) | A1\(\checkmark\) | ft \(c\) from (ii) |
| \(x = 0.25t^3 - 1.5t^2 + 3t - 2\) | A1 (3) |
## Question 5:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $t = 2$ | B1 | **AG** — From $0.6t = 0.3 \times 4$ or $1.5t - 3 = 0$ |
| $0.4a = 0.6t - 0.3 \times 0.4g$ | M1 | |
| $a = 1.5t - 3$ | A1 (3) | **AG** |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v = \int(1.5t - 3)\,dt$ | M1 | |
| $v = 0.75t^2 - 3t\ (+c)$ | A1 | |
| $0 = 0.75 \times 2^2 - 3 \times 2 + c$ (so $c = 3$) | M1 | Or uses limits with 2 and $t$ |
| $v = 0.75t^2 - 3t + 3$ | A1 (4) | |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = \int(0.75t^2 - 3t + 3)\,dt$ | M1 | |
| $x = 0.25t^3 - 1.5t^2 + 3t\ (+k)$ | A1$\checkmark$ | ft $c$ from (ii) |
| $x = 0.25t^3 - 1.5t^2 + 3t - 2$ | A1 (3) | |
---5 A particle $P$ of mass 0.4 kg is placed at rest at a point $A$ on a rough horizontal surface. A horizontal force, directed away from $A$ and with magnitude $0.6 t \mathrm {~N}$, acts on $P$, where $t \mathrm {~s}$ is the time after $P$ is placed at $A$. The coefficient of friction between $P$ and the surface is 0.3 , and $P$ has displacement from $A$ of $x \mathrm {~m}$ at time $t \mathrm {~s}$.\\
(i) Show that $P$ starts to move when $t = 2$. Show also that when $P$ is in motion it has acceleration $( 1.5 t - 3 ) \mathrm { m } \mathrm { s } ^ { - 2 }$.\\
(ii) Express the velocity of $P$ in terms of $t$, for $t \geqslant 2$.\\
(iii) Express $x$ in terms of $t$, for $t \geqslant 2$.
\hfill \mbox{\textit{CAIE M2 2016 Q5 [10]}}