| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance kv - vertical motion |
| Difficulty | Standard +0.8 This is a standard M2 differential equations problem involving air resistance proportional to velocity. Part (i) requires straightforward application of F=ma with two forces. Part (ii) requires separating variables and integrating to find v(t), which is routine for M2 but involves more steps than typical A-level Core questions. The integration and algebraic manipulation place it moderately above average difficulty. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(a = 10 - 0.45v\) AG | B1 [1] | \(0.2a = 0.2g - 0.09v\) or similar should be seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\int 1/(10-0.45v)\,dv = \int dt\) | M1 | An attempt at integration needed |
| \(-\ln(10 - 0.45v)/0.45 = t\ (+c)\) | A1 | |
| \(t = 0,\ v = 4,\ c = -4.67(58...)\) | DM1 | Attempts to find \(c\) or uses correct limits |
| \(-\ln(10 - 0.45v)/0.45 = 1.5 - 4.676\) | M1 | Uses \(t = 1.5\) and evaluated \(c\) |
| \(v = 12.9\) or \(13.0\) | A1 [5] |
## Question 4:
### Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $a = 10 - 0.45v$ AG | B1 [1] | $0.2a = 0.2g - 0.09v$ or similar should be seen |
### Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int 1/(10-0.45v)\,dv = \int dt$ | M1 | An attempt at integration needed |
| $-\ln(10 - 0.45v)/0.45 = t\ (+c)$ | A1 | |
| $t = 0,\ v = 4,\ c = -4.67(58...)$ | DM1 | Attempts to find $c$ or uses correct limits |
| $-\ln(10 - 0.45v)/0.45 = 1.5 - 4.676$ | M1 | Uses $t = 1.5$ and evaluated $c$ |
| $v = 12.9$ or $13.0$ | A1 [5] | |
**Total: [6]**
---4 A particle of mass 0.2 kg is projected vertically downwards with initial speed $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. A resisting force of magnitude $0.09 v \mathrm {~N}$ acts vertically upwards on the particle during its descent, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the downwards velocity of the particle at time $t \mathrm {~s}$ after being set in motion.\\
(i) Show that the acceleration of the particle is $( 10 - 0.45 v ) \mathrm { m } \mathrm { s } ^ { - 2 }$.\\
(ii) Find $v$ when $t = 1.5$.
\hfill \mbox{\textit{CAIE M2 2013 Q4 [6]}}