| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2005 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Variable force along axis work-energy |
| Difficulty | Standard +0.8 This M3 question requires setting up and evaluating a work-energy integral with a variable force (inverse square law), then solving for a constant and finding a turning point. It demands fluency with integration, the work-energy principle, and algebraic manipulation across multiple steps, placing it moderately above average difficulty but within reach of well-prepared M3 students. |
| Spec | 6.02c Work by variable force: using integration6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{3}\ddot{x} = -\frac{k}{(x+1)^2}\) | M1 | |
| \(\frac{1}{3}v\frac{dv}{dx} = -\frac{k}{(x+1)^2}\) | M1 | |
| \(\int v\, dv = \int -\frac{3k}{(x+1)^2}\, dx\) | Separating variables | |
| \(\frac{1}{2}v^2 = \frac{3k}{x+1}\) \((+C)\) | M1 A1=A1 | Attempting integration of both sides |
| \(v^2 = \frac{6k}{x+1} + A\) | ||
| Using boundary values to obtain two simultaneous equations | M1 | |
| \((1, 4)\): \(16 = 3k + A\) | A1 | |
| \((8, \sqrt{2})\): \(2 = \frac{2k}{3} + A\) | A1 | |
| \(14 = \frac{7}{3}k \Rightarrow k = 6\) | M1 A1 | (10) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(A = -2\) | B1 | |
| \(v^2 = \frac{36}{x+1} - 2 = 0\) | M1 | |
| \(x = 17\) (m) | M1 A1 | (4) |
## Question 7:
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{3}\ddot{x} = -\frac{k}{(x+1)^2}$ | M1 | |
| $\frac{1}{3}v\frac{dv}{dx} = -\frac{k}{(x+1)^2}$ | M1 | |
| $\int v\, dv = \int -\frac{3k}{(x+1)^2}\, dx$ | | Separating variables |
| $\frac{1}{2}v^2 = \frac{3k}{x+1}$ $(+C)$ | M1 A1=A1 | Attempting integration of both sides |
| $v^2 = \frac{6k}{x+1} + A$ | | |
| Using boundary values to obtain two simultaneous equations | M1 | |
| $(1, 4)$: $16 = 3k + A$ | A1 | |
| $(8, \sqrt{2})$: $2 = \frac{2k}{3} + A$ | A1 | |
| $14 = \frac{7}{3}k \Rightarrow k = 6$ | M1 A1 | **(10)** |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A = -2$ | B1 | |
| $v^2 = \frac{36}{x+1} - 2 = 0$ | M1 | |
| $x = 17$ (m) | M1 A1 | **(4)** |
**Total: [14]**7. A particle $P$ of mass $\frac { 1 } { 3 } \mathrm {~kg}$ moves along the positive $x$-axis under the action of a single force. The force is directed towards the origin $O$ and has magnitude $\frac { k } { ( x + 1 ) ^ { 2 } } \mathrm {~N}$, where $O P = x$ metres and $k$ is a constant. Initially $P$ is moving away from $O$. At $x = 1$ the speed of $P$ is $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, and at $x = 8$ the speed of $P$ is $\sqrt { } 2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $k$.
\item Find the distance of $P$ from $O$ when $P$ first comes to instantaneous rest.\\
(Total 14 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2005 Q7 [14]}}