| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Inverse-square gravitational force |
| Difficulty | Standard +0.8 This M3 question requires applying work-energy principles to variable gravitational force (inverse square law), integrating F with respect to distance, and finding escape velocity. While the setup is given and the integration is standard for M3, it requires careful manipulation of the energy equation and understanding of the escape condition (v→0 as x→∞). More demanding than typical M3 questions but follows a recognizable pattern for this topic. |
| 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) \(F = ma = M\frac{\text{d}v}{\text{d}x} = -\frac{4.90 \times 10^{15}}{x^2}\) | M1 A1 | |
| \(\int v \text{ d}v = -\int(4.90 \times 10^{15}) \int x^{-2} \text{ d}x\) | M1 | |
| \(\frac{1}{2}v^2 = (4.90 \times 10^{15})x^{-1} + c\) | A1 | |
| \(x = (1.74 \times 10^6)\), \(v = u\) \(\therefore c = \frac{1}{2}u^2 - \frac{4.90 \times 10^{15}}{1.74 \times 10^6}\) | M1 A1 | |
| so \(v^2 = u^2 + (9.80 \times 10^{12})x^{-1} - (5.63 \times 10^9)\) | A1 | |
| (b) we require v > 0 as x → ∞; \(x^{-1} \to 0\) \(\therefore u^2 - (5.63 \times 10^9) > 0\) giving \(u_{\text{min}} = 2400 \text{ m s}^{-1}\) (2sf) | M1; M1 A1; A1 | (11) |
(a) $F = ma = M\frac{\text{d}v}{\text{d}x} = -\frac{4.90 \times 10^{15}}{x^2}$ | M1 A1 |
$\int v \text{ d}v = -\int(4.90 \times 10^{15}) \int x^{-2} \text{ d}x$ | M1 |
$\frac{1}{2}v^2 = (4.90 \times 10^{15})x^{-1} + c$ | A1 |
$x = (1.74 \times 10^6)$, $v = u$ $\therefore c = \frac{1}{2}u^2 - \frac{4.90 \times 10^{15}}{1.74 \times 10^6}$ | M1 A1 |
so $v^2 = u^2 + (9.80 \times 10^{12})x^{-1} - (5.63 \times 10^9)$ | A1 |
(b) we require v > 0 as x → ∞; $x^{-1} \to 0$ $\therefore u^2 - (5.63 \times 10^9) > 0$ giving $u_{\text{min}} = 2400 \text{ m s}^{-1}$ (2sf) | M1; M1 A1; A1 | (11)
---5. When a particle of mass $M$ is at a distance of $x$ metres from the centre of the moon, the gravitational force, $F$ N, acting on it and directed towards the centre of the moon is given by
$$F = \frac { \left( 4.90 \times 10 ^ { 12 } \right) M } { x ^ { 2 } }$$
A rocket is projected vertically into space from a point on the surface of the moon with initial speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Given that the radius of the moon is $\left( 1.74 \times 10 ^ { 6 } \right) \mathrm { m }$,
\begin{enumerate}[label=(\alph*)]
\item show that the speed of the rocket, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, when it is $x$ metres from the centre of the moon is given by
$$v ^ { 2 } = u ^ { 2 } + \frac { a } { x } - b$$
where $a$ and $b$ are constants which should be found correct to 3 significant figures.
\item Find, correct to 2 significant figures, the minimum value of $u$ needed for the rocket to escape the moon's gravitational attraction.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q5 [11]}}