| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2006 |
| Session | June |
| 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 (gravity and resistance), leading to a simple 'show that' result. Part (ii) involves separating variables and integrating a basic linear differential equation, then solving for t when v=0. While it requires multiple techniques (Newton's second law, differential equations, logarithms), these are routine M2 procedures with no novel insight needed, making it slightly easier than average overall. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(-0.4g - 0.1v = 0.4 \frac{dv}{dt}\) | M1 | For using \(a = \frac{dv}{dt}\) and applying Newton's second law |
| \(\frac{dv}{dt} = -0.25(v + 40)\) | A1, 2 |
| Answer | Marks | Guidance |
|---|---|---|
| For obtaining \(C = \ln 56\) or for correct substitution of correct limits | A1, A1 | |
| M1 | For finding \(t\) when \(v = 0\) | |
| \(t = 1.35\) | M1, A1 | 5 |
**(i)** $-0.4g - 0.1v = 0.4 \frac{dv}{dt}$ | M1 | For using $a = \frac{dv}{dt}$ and applying Newton's second law
$\frac{dv}{dt} = -0.25(v + 40)$ | A1, 2 |
**(ii)** $\ln(v + 40) = -0.25t$ $(+C)$
For obtaining $C = \ln 56$ or for correct substitution of correct limits | A1, A1 |
M1 | For finding $t$ when $v = 0$
$t = 1.35$ | M1, A1 | 5 |
---4 An object of mass 0.4 kg is projected vertically upwards from the ground, with an initial speed of $16 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. A resisting force of magnitude $0.1 v$ newtons acts on the object during its ascent, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the speed of the object at time $t \mathrm {~s}$ after it starts to move.\\
(i) Show that $\frac { \mathrm { d } v } { \mathrm {~d} t } = - 0.25 ( v + 40 )$.\\
(ii) Find the value of $t$ at the instant that the object reaches its maximum height.
\hfill \mbox{\textit{CAIE M2 2006 Q4 [7]}}