| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2006 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Given acceleration function find velocity |
| Difficulty | Challenging +1.2 This question requires using the chain rule for acceleration (a = v dv/dx) and separating variables to integrate, which is a standard M2 technique but goes beyond routine mechanics. Part (ii) requires integrating dt/dx = 1/v and part (iii) involves substituting t=10 into the derived relationship. The fractional powers add minor algebraic complexity, but the overall approach follows a well-established method for variable force problems. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration6.02d Mechanical energy: KE and PE concepts6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v\dfrac{dv}{dx} = 0.6x^{0.2}\) | B1 | |
| M1 | For separating and integrating | |
| \(0.5v^2 = 0.5x^{1.2}\ (+C_1)\) | A1 | |
| \(v = x^{0.6}\) | A1 (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int x^{-0.6}\,dx = \int dt\) | M1 | For using \(v = dx/dt\), separating and integrating |
| \(x^{0.4}/0.4 = t\ (+C_2)\) | A1 | |
| \(t = 2.5x^{0.4}\) | A1 (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = (10/2.5)^{2.5}\) | M1 | For substituting for \(t\) and solving for \(x\) |
| Distance is \(32\)m | A1 (2) |
## Question 7:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v\dfrac{dv}{dx} = 0.6x^{0.2}$ | B1 | |
| | M1 | For separating and integrating |
| $0.5v^2 = 0.5x^{1.2}\ (+C_1)$ | A1 | |
| $v = x^{0.6}$ | A1 (4) | |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int x^{-0.6}\,dx = \int dt$ | M1 | For using $v = dx/dt$, separating and integrating |
| $x^{0.4}/0.4 = t\ (+C_2)$ | A1 | |
| $t = 2.5x^{0.4}$ | A1 (3) | |
### Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = (10/2.5)^{2.5}$ | M1 | For substituting for $t$ and solving for $x$ |
| Distance is $32$m | A1 (2) | |
---7 A cyclist starts from rest at a point $O$ and travels along a straight path. At time $t \mathrm {~s}$ after starting, the displacement of the cyclist from $O$ is $x \mathrm {~m}$, and the acceleration of the cyclist is $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$, where $a = 0.6 x ^ { 0.2 }$.\\
(i) Find an expression for the velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ of the cyclist in terms of $x$.\\
(ii) Show that $t = 2.5 x ^ { 0.4 }$.\\
(iii) Find the distance travelled by the cyclist in the first 10 s of the journey.
\hfill \mbox{\textit{CAIE M2 2006 Q7 [9]}}