| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Given velocity function find force |
| Difficulty | Standard +0.8 This is a multi-part mechanics question requiring differentiation of an implicit equation involving logarithms (non-trivial chain rule application), applying Newton's second law to find a variable resisting force, and integration to find time. The logarithmic differentiation and the need to recognize v(dv/dx) = a requires more sophistication than standard M2 questions, though the individual techniques are within the syllabus. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08k Separable differential equations: dy/dx = f(x)g(y)6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([3 - 90/(v+30)](dv/dx) + 1 = 0\) or \(3 - 90/(v+30) + (dx/dv) = 0\) | B1 | |
| M1 | For using \(a = v(dv/dx)\) | |
| Acceleration is \(-\frac{1}{3}(v + 30)\text{ ms}^{-2}\) | A1 | AG Subtotal: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([0.3g + R = 0.3(v+30)/3]\) | M1 | For using Newton's second law |
| Resisting force is \(0.1v\) N | A1 | Subtotal: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int \frac{dv}{v+30} = -\frac{1}{3}\int dt\) | M1 | For using \(a = dv/dt\), separating variables and integrating |
| \(\ln(v + 30) = -t/3\ (+A)\) | A1 | |
| \(\ln 50 = 0 + A\) | B1 | |
| \([\ln 30 = -t/3 + \ln 50]\) | M1 | For finding \(t\) when \(v = 0\) |
| Time taken is 1.53 s | A1 | Subtotal: 5 |
## Question 7:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[3 - 90/(v+30)](dv/dx) + 1 = 0$ or $3 - 90/(v+30) + (dx/dv) = 0$ | B1 | |
| | M1 | For using $a = v(dv/dx)$ |
| Acceleration is $-\frac{1}{3}(v + 30)\text{ ms}^{-2}$ | A1 | AG **Subtotal: 3** |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[0.3g + R = 0.3(v+30)/3]$ | M1 | For using Newton's second law |
| Resisting force is $0.1v$ N | A1 | **Subtotal: 2** |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{dv}{v+30} = -\frac{1}{3}\int dt$ | M1 | For using $a = dv/dt$, separating variables and integrating |
| $\ln(v + 30) = -t/3\ (+A)$ | A1 | |
| $\ln 50 = 0 + A$ | B1 | |
| $[\ln 30 = -t/3 + \ln 50]$ | M1 | For finding $t$ when $v = 0$ |
| Time taken is 1.53 s | A1 | **Subtotal: 5** | **Total: 10** |7 A particle $P$ of mass 0.3 kg is projected vertically upwards from the ground with an initial speed of $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. When $P$ is at height $x \mathrm {~m}$ above the ground, its upward speed is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. It is given that
$$3 v - 90 \ln ( v + 30 ) + x = A ,$$
where $A$ is a constant.\\
(i) Differentiate this equation with respect to $x$ and hence show that the acceleration of the particle is $- \frac { 1 } { 3 } ( v + 30 ) \mathrm { m } \mathrm { s } ^ { - 2 }$.\\
(ii) Find, in terms of $v$, the resisting force acting on the particle.\\
(iii) Find the time taken for $P$ to reach its maximum height.
\hfill \mbox{\textit{CAIE M2 2009 Q7 [10]}}