| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance kv - vertical motion |
| Difficulty | Standard +0.3 This is a standard M2 variable force question with air resistance proportional to velocity. Part (i) requires straightforward application of F=ma with two forces (weight and resistance), while part (ii) involves separating variables and integrating a simple differential equation to find terminal velocity behavior. The mathematics is routine for M2 students who have practiced this topic, though slightly above average difficulty due to the differential equations component. |
| Spec | 3.03d Newton's second law: 2D vectors6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.4g - 0.08v = 0.4a\) | M1 | For using Newton's second law |
| Acceleration is \(10 - 0.2v\) | A1 | [2 marks] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int \frac{dv}{50-v} = \int 0.2\, dt\) | M1 | For using \(a = \frac{dv}{dt}\), separating variables and attempting to integrate |
| \(-\ln(50-v) = 0.2t\ (+C)\) | A1 | |
| \(-\ln(50-v) = 0.2t - \ln 50\) | M1 | For using \(v(0) = 0\) to find \(C\) |
| \(50 - v = 50e^{-3}\) | M1 | For substituting \(t = 15\) and solving for \(v\) |
| Speed is \(47.5 \text{ ms}^{-1}\) | A1 | [5 marks] [Total: 7] |
## Question 4:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.4g - 0.08v = 0.4a$ | M1 | For using Newton's second law |
| Acceleration is $10 - 0.2v$ | A1 | **[2 marks]** |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{dv}{50-v} = \int 0.2\, dt$ | M1 | For using $a = \frac{dv}{dt}$, separating variables and attempting to integrate |
| $-\ln(50-v) = 0.2t\ (+C)$ | A1 | |
| $-\ln(50-v) = 0.2t - \ln 50$ | M1 | For using $v(0) = 0$ to find $C$ |
| $50 - v = 50e^{-3}$ | M1 | For substituting $t = 15$ and solving for $v$ |
| Speed is $47.5 \text{ ms}^{-1}$ | A1 | **[5 marks]** **[Total: 7]** |
---4 A particle of mass 0.4 kg is released from rest and falls vertically. A resisting force of magnitude $0.08 v \mathrm {~N}$ acts upwards on the particle during its descent, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the velocity of the particle at time $t \mathrm {~s}$ after its release.\\
(i) Show that the acceleration of the particle is $( 10 - 0.2 v ) \mathrm { m } \mathrm { s } ^ { - 2 }$.\\
(ii) Find the velocity of the particle when $t = 15$.
\hfill \mbox{\textit{CAIE M2 2007 Q4 [7]}}