| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Given velocity function find force |
| Difficulty | Standard +0.8 This M3 question requires separating variables to integrate v = dx/dt = 4/(x+2), then using the result to find acceleration via a = v(dv/dx). While the calculus is standard, it involves multiple connected steps (integration, applying initial conditions, then differentiation/chain rule for acceleration) and careful algebraic manipulation, making it moderately challenging but within typical M3 scope. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(v = \frac{4}{(x+2)} = \frac{dx}{dt}\) | B1 | |
| \(\frac{dt}{dx} = \frac{x+2}{4}\); \(\int_{t=0}^{t=2}1\,dt = \frac{1}{4}\int_{x=0}^{x=X}(x+2)\,dx\), \([t]_0^2 = \frac{1}{4}\left[\frac{x^2}{2}+2x\right]_0^X\) | M1, A1 | |
| \(2 = \frac{X^2}{8} + \frac{X}{2}\) | ||
| \(0 = X^2 + 4X - 16\), \(X = \frac{-4+\sqrt{80}}{2} = 2.47\ \text{(m)}\) | M1depA1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(a\left(=\frac{dv}{dt}\right) = v\frac{dv}{dx}\) | B1 | |
| \(= \frac{4}{(x+2)}\times\frac{-4}{(x+2)^2}\) | M1A1 | |
| \(= \frac{-16}{(2.47+2)^3} = -0.1788...\) | M1dep | their \(X\) |
| \(0.18\ (\text{m s}^{-2})\) towards \(O\) | A1 | (5) [10] |
## Question 4:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $v = \frac{4}{(x+2)} = \frac{dx}{dt}$ | B1 | |
| $\frac{dt}{dx} = \frac{x+2}{4}$; $\int_{t=0}^{t=2}1\,dt = \frac{1}{4}\int_{x=0}^{x=X}(x+2)\,dx$, $[t]_0^2 = \frac{1}{4}\left[\frac{x^2}{2}+2x\right]_0^X$ | M1, A1 | |
| $2 = \frac{X^2}{8} + \frac{X}{2}$ | | |
| $0 = X^2 + 4X - 16$, $X = \frac{-4+\sqrt{80}}{2} = 2.47\ \text{(m)}$ | M1depA1 | **(5)** |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $a\left(=\frac{dv}{dt}\right) = v\frac{dv}{dx}$ | B1 | |
| $= \frac{4}{(x+2)}\times\frac{-4}{(x+2)^2}$ | M1A1 | |
| $= \frac{-16}{(2.47+2)^3} = -0.1788...$ | M1dep | their $X$ |
| $0.18\ (\text{m s}^{-2})$ towards $O$ | A1 | **(5) [10]** |
---\begin{enumerate}
\item A particle $P$ is moving along the positive $x$-axis. At time $t$ seconds, $t \geqslant 0 , P$ is $x$ metres from the origin $O$ and is moving away from $O$ with velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, where $v = \frac { 4 } { ( x + 2 ) }$. When $t = 0 , P$ is at $O$. Find\\
(a) the distance of $P$ from $O$ when $t = 2$\\
(b) the magnitude and direction of the acceleration of $P$ when $t = 2$\\
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2013 Q4 [10]}}