| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Inverse power force - gravitational/escape velocity context |
| Difficulty | Standard +0.3 This is a straightforward M3 inverse-square law gravity question. Part (a) is simple substitution, part (b) uses standard energy methods (work-energy theorem or integration of F=ma), and part (c) applies constant acceleration equations. All techniques are routine for M3 students with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks |
|---|---|
| (a) \(30 = \frac{k}{(6.37\times 10^6)^2}\) | M1 A1 A1 |
| \(k = 4.06\times 10^{14}\) | |
| Units N m² kg⁻¹ or m³ s⁻² | |
| (b) \(mv\frac{dv}{dx} = -\frac{km}{x^2}\) | M1 A1 A1 |
| \(\frac{v^2}{2} = \frac{k}{x} + c\) | |
| \(v = 0, x = 12.74 \times 10^6\) | |
| \(c = -3.19 \times 10^7\) | M1 A1 |
| \(\frac{v^2}{2} = \frac{4.06\times 10^{14}}{x} - 3.19 \times 10^7\) | M1 A1 |
| When \(x = 6.37 \times 10^6\), \(v = 7.98 \times 10^3\) m s⁻¹ | |
| (c) \(v^2 = 0 + 2 \times 10 \times d\) | M1 A1 A1 |
| \(v^2 = 20d\) | |
| \(d = 3.18\) | 13 marks total |
**(a)** $30 = \frac{k}{(6.37\times 10^6)^2}$ | M1 A1 A1
$k = 4.06\times 10^{14}$ |
Units N m² kg⁻¹ or m³ s⁻² |
**(b)** $mv\frac{dv}{dx} = -\frac{km}{x^2}$ | M1 A1 A1
$\frac{v^2}{2} = \frac{k}{x} + c$ |
$v = 0, x = 12.74 \times 10^6$ |
$c = -3.19 \times 10^7$ | M1 A1
$\frac{v^2}{2} = \frac{4.06\times 10^{14}}{x} - 3.19 \times 10^7$ | M1 A1
When $x = 6.37 \times 10^6$, $v = 7.98 \times 10^3$ m s⁻¹ |
**(c)** $v^2 = 0 + 2 \times 10 \times d$ | M1 A1 A1
$v^2 = 20d$ |
$d = 3.18$ | 13 marks totalA particle of mass $m$ kg, at a distance $x$ m from the centre of the Earth, experiences a force of magnitude $\frac{km}{x^2}$ N towards the centre of the Earth, where $k$ is a constant. Given that the radius of the Earth is $6.37 \times 10^6$ m, and that a 3 kg mass experiences a force of 30 N at the surface of the Earth,
\begin{enumerate}[label=(\alph*)]
\item calculate the value of $k$, stating the units of your answer. [3 marks]
\end{enumerate}
The 3 kg mass falls from rest at a distance $x = 12.74 \times 10^6$ m from the centre of the Earth. Ignoring air resistance,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item show that it reaches the surface of the Earth with speed $7.98 \times 10^3$ ms$^{-1}$. [7 marks]
\end{enumerate}
In a simplified model, the particle is assumed to fall with a constant acceleration 10 ms$^{-2}$. According to this model it attains the same speed as in (b), $7.98 \times 10^3$ ms$^{-1}$, at a distance $(12.74 - d) \times 10^6$ m from the centre of the Earth.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the value of $d$. [3 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q5 [13]}}