| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Inverse power force - non-gravitational context |
| Difficulty | Standard +0.8 This M3 variable force question requires applying F=ma with an inverse-square force, integrating to find the relationship between velocity and position, then using given conditions to find k and the rest position. While it involves multiple steps and careful algebraic manipulation, it follows a standard M3 pattern for inverse-square force problems with no particularly novel insights required. |
| Spec | 6.02c Work by variable force: using integration6.02i Conservation of energy: mechanical energy principle6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(F = ma = 0.8\frac{dv}{dx} = -\frac{k}{x^2}\) | M1 | |
| \(\therefore \int 4v\,dv = \int -\frac{5k}{x^2}\,dx\) | M1 | |
| giving \(2v^2 = \frac{5k}{x} + c\) | A1 | |
| \(x = 2, v = 5\): \(50 = \frac{5k}{2} + c\) | M1 A1 | |
| \(x = 4, v = 3\): \(18 = \frac{5k}{4} + c\) | M1 | |
| solve simul. \(32 = k(\frac{5}{2} - \frac{5}{4}) = \frac{5}{4}k\) | M1 | |
| \(\therefore k = \frac{32 \times 4}{5} = \frac{128}{5}\) | A1 | |
| (b) \(c = 50 - \frac{5k}{2} = -14\) \(\therefore v^2 = \frac{64}{x} - 7\) | M1 A1 | |
| rest when \(v = 0\): \(\frac{64}{x} = 7\) so \(x = \frac{64}{7}\) m | M1 A1 | (12) |
(a) $F = ma = 0.8\frac{dv}{dx} = -\frac{k}{x^2}$ | M1 |
$\therefore \int 4v\,dv = \int -\frac{5k}{x^2}\,dx$ | M1 |
giving $2v^2 = \frac{5k}{x} + c$ | A1 |
$x = 2, v = 5$: $50 = \frac{5k}{2} + c$ | M1 A1 |
$x = 4, v = 3$: $18 = \frac{5k}{4} + c$ | M1 |
solve simul. $32 = k(\frac{5}{2} - \frac{5}{4}) = \frac{5}{4}k$ | M1 |
$\therefore k = \frac{32 \times 4}{5} = \frac{128}{5}$ | A1 |
(b) $c = 50 - \frac{5k}{2} = -14$ $\therefore v^2 = \frac{64}{x} - 7$ | M1 A1 |
rest when $v = 0$: $\frac{64}{x} = 7$ so $x = \frac{64}{7}$ m | M1 A1 | (12)5. A particle of mass 0.8 kg is moving along the positive $x$-axis at a speed of $5 \mathrm {~ms} ^ { - 1 }$ away from the origin $O$. When the particle is 2 metres from $O$ it becomes subject to a single force directed towards $O$. The magnitude of the force is $\frac { k } { x ^ { 2 } } \mathrm {~N}$ when the particle is $x$ metres from $O$.
Given that when the particle is 4 m from $O$ its speed has been reduced to $3 \mathrm {~ms} ^ { - 1 }$,
\begin{enumerate}[label=(\alph*)]
\item show that $k = \frac { 128 } { 5 }$,
\item find the distance of the particle from $O$ when it comes to instantaneous rest. (4 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q5 [12]}}