| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2015 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Variable coefficient of friction |
| Difficulty | Challenging +1.2 This is a multi-part mechanics problem requiring resolution of forces, limiting friction, Newton's second law with variable force, and integration. While it involves several steps and the time-dependent force adds complexity, the techniques are standard M2 content with clear signposting. The 'show that' in part (ii) provides the equation to verify, reducing problem-solving demand. Slightly above average due to the multi-stage nature and need to track when the particle slips vs loses contact, but well within reach of competent M2 students. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(R = 0.2g - 0.4 \times 2\sin30\) | M1 | Resolving vertically, 3 terms |
| \(F_R = 0.4 \times 2\cos30\) | M1 | Use \(F = \mu R\) |
| \(\mu = 0.433\) | A1 | |
| \(0.2g = 0.4\ t\sin30\) | M1 | Solves for \(t\) when \(R = 0\) |
| \(t = 10\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(0.2\,dv/dt = 0.4t\cos30 - 0.433(0.2g - 0.4\ t\sin30)\) | M1, A1 | Newton's Second Law with both forces \(f(t)\) |
| \(dv/dt = 2.165t - 4.33(0)\) AG | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\int dv = \int(2.165t - 4.33)\,dt\) | M1 | Attempts to integrate |
| \(v = 2.165t^2/2 - 4.33t\ (+c)\) | A1 | |
| \(v = 0,\ t = 2\ [c = 4.33]\) | M1 | Must use \(t = 2\) |
| \(v = 2.165 \times 10^2/2 - 4.33 \times 10 + 4.33 \Rightarrow v = 69.3\) | A1 | Puts \(t\) (i) in integrand |
## Question 7:
### Part (i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $R = 0.2g - 0.4 \times 2\sin30$ | M1 | Resolving vertically, 3 terms |
| $F_R = 0.4 \times 2\cos30$ | M1 | Use $F = \mu R$ |
| $\mu = 0.433$ | A1 | |
| $0.2g = 0.4\ t\sin30$ | M1 | Solves for $t$ when $R = 0$ |
| $t = 10$ | A1 | |
**Total: 5 marks**
### Part (ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $0.2\,dv/dt = 0.4t\cos30 - 0.433(0.2g - 0.4\ t\sin30)$ | M1, A1 | Newton's Second Law with both forces $f(t)$ |
| $dv/dt = 2.165t - 4.33(0)$ AG | A1 | |
**Total: 3 marks**
### Part (iii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\int dv = \int(2.165t - 4.33)\,dt$ | M1 | Attempts to integrate |
| $v = 2.165t^2/2 - 4.33t\ (+c)$ | A1 | |
| $v = 0,\ t = 2\ [c = 4.33]$ | M1 | Must use $t = 2$ |
| $v = 2.165 \times 10^2/2 - 4.33 \times 10 + 4.33 \Rightarrow v = 69.3$ | A1 | Puts $t$ (i) in integrand |
**Total: 4 marks**7 A force of magnitude $0.4 t \mathrm {~N}$, applied at an angle of $30 ^ { \circ }$ above the horizontal, acts on a particle $P$, where $t \mathrm {~s}$ is the time since the force starts to act. $P$ is at rest on rough horizontal ground when $t = 0$. The mass of $P$ is 0.2 kg and the coefficient of friction between $P$ and the ground is $\mu$.\\
(i) Given that $P$ is about to slip when $t = 2$, find $\mu$ and the value of $t$ for the instant when $P$ loses contact with the ground.\\
(ii) While $P$ is moving on the ground, it has velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at time $t \mathrm {~s}$. Show that
$$\frac { \mathrm { d } v } { \mathrm {~d} t } = 2.165 t - 4.330$$
where the coefficients are correct to 4 significant figures.\\
(iii) Calculate the speed of $P$ when it loses contact with the ground.
{www.cie.org.uk} after the live examination series.
}
\hfill \mbox{\textit{CAIE M2 2015 Q7 [12]}}