| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance kv² - projected vertically upward |
| Difficulty | Challenging +1.2 This is a standard M2/Further Mechanics question on variable force with air resistance proportional to v². Part (i) is routine application of F=ma with v(dv/dx), part (ii) requires integration of a separable differential equation (standard technique), and part (iii) requires setting up and solving a similar equation for descent. While it involves multiple steps and calculus, the techniques are well-practiced in M2 syllabi with no novel insight required. |
| Spec | 1.08h Integration by substitution3.02a Kinematics language: position, displacement, velocity, acceleration6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.5v\frac{dv}{dx} = -0.5g - 0.02v^2\) | M1 | |
| \(v\frac{dv}{dx} = -10 - 0.04v^2\) (AG) | A1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_8^0 \frac{v}{-10 - 0.04v^2}\, dv = \int_0^x dx\) | M1 | Separates variables and attempts to integrate |
| \(x = \left[-\ln(10 + 0.04v^2)/0.08\right]_8^0\) | M1 | Uses limits or finds constant |
| \(x = 2.85\) | A1 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v\frac{dv}{dx} = 10 - 0.04v^2\) | B1 | |
| \(\int_0^v \frac{v}{10 - 0.04v^2}\, dv = \int_0^{2.85} dx\) | M1 | Integrates new acceleration |
| \(\ln\left[(10 - 0.04v^2)/10\right]/(-0.08) = 2.85\) | M1 | Uses earlier answer as distance |
| \(v = 7.14 \text{ ms}^{-1}\) | A1 | |
| Total: 4 |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.5v\frac{dv}{dx} = -0.5g - 0.02v^2$ | M1 | |
| $v\frac{dv}{dx} = -10 - 0.04v^2$ (AG) | A1 | |
| **Total: 2** | | |
---
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_8^0 \frac{v}{-10 - 0.04v^2}\, dv = \int_0^x dx$ | M1 | Separates variables and attempts to integrate |
| $x = \left[-\ln(10 + 0.04v^2)/0.08\right]_8^0$ | M1 | Uses limits or finds constant |
| $x = 2.85$ | A1 | |
| **Total: 3** | | |
---
## Question 5(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v\frac{dv}{dx} = 10 - 0.04v^2$ | B1 | |
| $\int_0^v \frac{v}{10 - 0.04v^2}\, dv = \int_0^{2.85} dx$ | M1 | Integrates new acceleration |
| $\ln\left[(10 - 0.04v^2)/10\right]/(-0.08) = 2.85$ | M1 | Uses earlier answer as distance |
| $v = 7.14 \text{ ms}^{-1}$ | A1 | |
| **Total: 4** | | |5 A particle $P$ of mass 0.5 kg is projected vertically upwards from a point on a horizontal surface. A resisting force of magnitude $0.02 v ^ { 2 } \mathrm {~N}$ acts on $P$, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the upward velocity of $P$ when it is a height of $x \mathrm {~m}$ above the surface. The initial speed of $P$ is $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Show that, while $P$ is moving upwards, $v \frac { \mathrm {~d} v } { \mathrm {~d} x } = - 10 - 0.04 v ^ { 2 }$.\\
(ii) Find the greatest height of $P$ above the surface.\\
(iii) Find the speed of $P$ immediately before it strikes the surface after descending.\\
\hfill \mbox{\textit{CAIE M2 Q5 [9]}}