| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Variable force (position x) - find velocity |
| Difficulty | Standard +0.3 This is a standard M2 variable force question using v dv/dx = F/m. Part (i) requires straightforward application of Newton's second law, part (ii) involves routine integration and substitution, and part (iii) requires comparing two force expressions to identify the additional force. While it involves multiple steps and calculus, the techniques are standard M2 fare with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation6.02a Work done: concept and definition6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.25v\,dv/dx = 2 + 0.3x^2\) | M1 | |
| \(v\,dv/dx = 1.2x^2 + 8\) | A1 AG | Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int v\,dv = \int(1.2x^2 + 8)\,dx\) | M1 | |
| \(v^2/2 = 0.4x^3 + 8x\,(+\,c)\) | A1 | Allow \(c = 0\) without working |
| \(v = 5.17\) | A1 | Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.25v\,dv/dx = 0.3x^2 + 1.5 - 0.75x\) | M1 | |
| Force is \(0.5 + 0.75x\) N towards O | A1 | Total: 2 marks |
## Question 3:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.25v\,dv/dx = 2 + 0.3x^2$ | M1 | |
| $v\,dv/dx = 1.2x^2 + 8$ | A1 AG | Total: 2 marks |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int v\,dv = \int(1.2x^2 + 8)\,dx$ | M1 | |
| $v^2/2 = 0.4x^3 + 8x\,(+\,c)$ | A1 | Allow $c = 0$ without working |
| $v = 5.17$ | A1 | Total: 3 marks |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.25v\,dv/dx = 0.3x^2 + 1.5 - 0.75x$ | M1 | |
| Force is $0.5 + 0.75x$ N towards O | A1 | Total: 2 marks |
---3 A small block $B$ of mass 0.25 kg is released from rest at a point $O$ on a smooth horizontal surface. After its release the velocity of $B$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when its displacement is $x \mathrm {~m}$ from $O$. The force acting on $B$ has magnitude $\left( 2 + 0.3 x ^ { 2 } \right) \mathrm { N }$ and is directed horizontally away from $O$.\\
(i) Show that $v \frac { \mathrm {~d} v } { \mathrm {~d} x } = 1.2 x ^ { 2 } + 8$.\\
(ii) Find the velocity of $B$ when $x = 1.5$.
An extra force acts on $B$ after $x = 1.5$. It is given that, when $x > 1.5$,
$$v \frac { \mathrm {~d} v } { \mathrm {~d} x } = 1.2 x ^ { 2 } + 6 - 3 x$$
(iii) Find the magnitude of this extra force and state the direction in which it acts.
\hfill \mbox{\textit{CAIE M2 2016 Q3 [7]}}