| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2007 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Given acceleration function find velocity |
| Difficulty | Moderate -0.3 This is a standard M3 variable acceleration question requiring recognition that maximum speed occurs when acceleration equals zero, followed by routine integration using v dv/dx = a. The conceptual insight (max speed when a=0) is straightforward, and the integration technique is a core M3 skill practiced extensively. Slightly easier than average due to the guided structure and standard method. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Maximum speed when accel. = 0 | B1 | |
| (b) \(\frac{1}{12}(30 - x) = v \frac{dv}{dx}\) (accehn = ... + attempt to integrate) | M1 ↓ M1 A1 ↓ M1 | |
| Using of \(v \frac{dv}{dx}\): \(\frac{v^2}{2} = \frac{1}{12}\left(30x - \frac{x^2}{2}\right) (+ c)\) | ||
| Substituting \(x = 30, v = 10\) and finding \(c = (12.5)\), or limits | ||
| \(v^2 = 25 + 5x - \frac{1}{12}x^2\) (o.e.) | A1 (5) | |
| (a) Allow "accel > 0 for x < 30, accel < 0 for x > 30" Also "accelerating for x < 30, decelerating for x > 30" But "accel < 0 for x > 30" only is B0 | Guidance note | |
| (b) 1st M1 will be generous for wrong form of accel (e.g. dv/dx)! 3rd M1 If use limits, they must use them in correct way with correct values Final A1. Have to accept any expression, but it must be for \(v^2\) explicitly (not 1/2v²), and if in separate terms, one can expect like terms to be collected. Hence answer in form as above, or e.g. \(\frac{1}{12}(300 + 60x - x^2)\); also \(100 - \frac{1}{12}(30-x)^2\) | Guidance note |
(a) Maximum speed when accel. = 0 | B1 |
(b) $\frac{1}{12}(30 - x) = v \frac{dv}{dx}$ (accehn = ... + attempt to integrate) | M1 ↓ M1 A1 ↓ M1 |
Using of $v \frac{dv}{dx}$: $\frac{v^2}{2} = \frac{1}{12}\left(30x - \frac{x^2}{2}\right) (+ c)$ | |
Substituting $x = 30, v = 10$ and finding $c = (12.5)$, or limits | |
$v^2 = 25 + 5x - \frac{1}{12}x^2$ (o.e.) | A1 (5) |
(a) Allow "accel > 0 for x < 30, accel < 0 for x > 30" Also "accelerating for x < 30, decelerating for x > 30" But "accel < 0 for x > 30" only is B0 | | Guidance note
(b) 1st M1 will be generous for wrong form of accel (e.g. dv/dx)! 3rd M1 If use limits, they must use them in correct way with correct values Final A1. Have to accept any expression, but it must be for $v^2$ explicitly (not 1/2v²), and if in separate terms, one can expect like terms to be collected. Hence answer in form as above, or e.g. $\frac{1}{12}(300 + 60x - x^2)$; also $100 - \frac{1}{12}(30-x)^2$ | | Guidance note
---\begin{enumerate}
\item A particle $P$ moves along the $x$-axis. At time $t = 0 , P$ passes through the origin $O$, moving in the positive $x$-direction. At time $t$ seconds, the velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $O P = x$ metres. The acceleration of $P$ is $\frac { 1 } { 12 } ( 30 - x ) \mathrm { m } \mathrm { s } ^ { - 2 }$, measured in the positive $x$-direction.\\
(a) Give a reason why the maximum speed of $P$ occurs when $x = 30$.
\end{enumerate}
Given that the maximum speed of $P$ is $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$,\\
(b) find an expression for $v ^ { 2 }$ in terms of $x$.\\
\hfill \mbox{\textit{Edexcel M3 2007 Q1 [6]}}