| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2010 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Force depends on time t |
| Difficulty | Standard +0.3 This is a standard M3 variable acceleration problem requiring integration with given initial conditions. Part (a) is a 'show that' requiring one integration with careful attention to the direction (negative acceleration), and part (b) requires finding when velocity equals zero, then integrating again to find displacement. While it involves time-dependent acceleration and two integrations, the techniques are routine for M3 students and the algebraic manipulation is straightforward. |
| Spec | 1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(\frac{d^2x}{dt^2} = -\frac{3}{(t+1)^2}\) | M1 | |
| \(\frac{dx}{dt} = \int -3(t+1)^{-2}\,dt = 3(t+1)^{-1}(+c)\) | M1 A1 | |
| \(t=0,\ v=2\): \(2 = 3 + c\), \(c = -1\) | M1 | |
| \(\frac{dx}{dt} = \frac{3}{t+1} - 1\) ✱ | A1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(x = \int\left(\frac{3}{t+1} - 1\right)dt\) | M1 | |
| \(= 3\ln(t+1) - t\ (+c')\) | A1 | |
| \(t=0,\ x=0 \Rightarrow c'=0\) | B1 | |
| \(x = 3\ln(t+1) - t\) | ||
| \(v=0 \Rightarrow \frac{3}{t+1} = 1\) | M1 | |
| \(t = 2\) | A1 | |
| \(x = 3\ln 3 - 2\) | M1 | |
| \(= 1.30\ \text{m}\) (Allow 1.3) | A1 | (7) [12] |
## Question 6:
### Part (a):
| Answer/Working | Marks | Notes |
|---|---|---|
| $\frac{d^2x}{dt^2} = -\frac{3}{(t+1)^2}$ | M1 | |
| $\frac{dx}{dt} = \int -3(t+1)^{-2}\,dt = 3(t+1)^{-1}(+c)$ | M1 A1 | |
| $t=0,\ v=2$: $2 = 3 + c$, $c = -1$ | M1 | |
| $\frac{dx}{dt} = \frac{3}{t+1} - 1$ ✱ | A1 | (5) |
### Part (b):
| Answer/Working | Marks | Notes |
|---|---|---|
| $x = \int\left(\frac{3}{t+1} - 1\right)dt$ | M1 | |
| $= 3\ln(t+1) - t\ (+c')$ | A1 | |
| $t=0,\ x=0 \Rightarrow c'=0$ | B1 | |
| $x = 3\ln(t+1) - t$ | | |
| $v=0 \Rightarrow \frac{3}{t+1} = 1$ | M1 | |
| $t = 2$ | A1 | |
| $x = 3\ln 3 - 2$ | M1 | |
| $= 1.30\ \text{m}$ (Allow 1.3) | A1 | (7) **[12]** |
---\begin{enumerate}
\item At time $t = 0$, a particle $P$ is at the origin $O$ moving with speed $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ along the $x$-axis in the positive $x$-direction. At time $t$ seconds $( t > 0 )$, the acceleration of $P$ has magnitude $\frac { 3 } { ( t + 1 ) ^ { 2 } } \mathrm {~m} \mathrm {~s} ^ { - 2 }$ and is directed towards $O$.\\
(a) Show that at time $t$ seconds the velocity of $P$ is $\left( \frac { 3 } { t + 1 } - 1 \right) \mathrm { m } \mathrm { s } ^ { - 1 }$.\\
(b) Find, to 3 significant figures, the distance of $P$ from $O$ when $P$ is instantaneously at rest.\\
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2010 Q6 [12]}}