| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2017 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance kv² - falling from rest or projected downward |
| Difficulty | Standard +0.8 This is a multi-part differential equations problem requiring v dv/dx formulation, separation of variables, and integration involving ln(a-bv²). While the setup and showing the given result is straightforward M2 content, part (ii) requires confident manipulation of logarithms and solving for v explicitly, and part (iii) requires careful substitution. This is above-average difficulty for M2 but standard for this topic. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)3.02f Non-uniform acceleration: using differentiation and integration3.03r Friction: concept and vector form6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Notes |
| \(0.4a = 0.4g - 0.2v^2\) | M1 | Uses Newton's Second Law vertically |
| \(v\,dv/dx = 10 - 0.5v^2\) | A1 | AG |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Notes |
| \(\int v\,dv/(10 - 0.5v^2) = \int dx\) | M1 | Separates the variables and attempts to integrate |
| \(-\ln(10 - 0.5v^2) = x\ (+c)\) | A1 | |
| \(x = 0,\ v = 0\) hence \(c = -\ln10\) | M1 | Attempts to find \(c\) using \(x = 0,\ v = 0\) |
| \(v = \sqrt{(20 - 20e^{-x})}\) | A1 | \(10 - 0.5v^2 = e^{-x+\ln10} = 10e^{-x}\) |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Notes |
| \(\text{Increase} = \sqrt{(20 - 20e^{-8})} - \sqrt{(20 - 20e^{-4})}\) | M1 | M1 if \(x\) values are substituted into their value for part (ii) |
| \(\text{Increase} = 0.0404\) m s\(^{-1}\) | A1 | Allow 0.04 |
| Total: 2 |
## Question 4:
**Part (i)**
| Answer | Marks | Notes |
|--------|-------|-------|
| $0.4a = 0.4g - 0.2v^2$ | M1 | Uses Newton's Second Law vertically |
| $v\,dv/dx = 10 - 0.5v^2$ | A1 | AG |
| **Total: 2** | | |
**Part (ii)**
| Answer | Marks | Notes |
|--------|-------|-------|
| $\int v\,dv/(10 - 0.5v^2) = \int dx$ | M1 | Separates the variables and attempts to integrate |
| $-\ln(10 - 0.5v^2) = x\ (+c)$ | A1 | |
| $x = 0,\ v = 0$ hence $c = -\ln10$ | M1 | Attempts to find $c$ using $x = 0,\ v = 0$ |
| $v = \sqrt{(20 - 20e^{-x})}$ | A1 | $10 - 0.5v^2 = e^{-x+\ln10} = 10e^{-x}$ |
| **Total: 4** | | |
**Part (iii)**
| Answer | Marks | Notes |
|--------|-------|-------|
| $\text{Increase} = \sqrt{(20 - 20e^{-8})} - \sqrt{(20 - 20e^{-4})}$ | M1 | M1 if $x$ values are substituted into their value for part (ii) |
| $\text{Increase} = 0.0404$ m s$^{-1}$ | A1 | Allow 0.04 |
| **Total: 2** | | |
---4 A small object of mass 0.4 kg is released from rest at a point 8 m above the ground. The object descends vertically and when its downwards displacement from its initial position is $x \mathrm {~m}$ the object has velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. While the object is moving, a force of magnitude $0.2 v ^ { 2 } \mathrm {~N}$ opposes the motion.\\
(i) Show that $v \frac { \mathrm {~d} v } { \mathrm {~d} x } = 10 - 0.5 v ^ { 2 }$.\\
(ii) Express $v$ in terms of $x$.\\
(iii) Find the increase in the value of $v$ during the final 4 m of the descent of the object.\\
\hfill \mbox{\textit{CAIE M2 2017 Q4 [8]}}