| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Inverse-square gravitational force |
| Difficulty | Standard +0.3 Part (a) is straightforward substitution using F=mg at Earth's surface. Part (b) requires energy conservation with inverse-square gravity, which is a standard M3 technique but involves algebraic manipulation of the energy equation. This is a typical textbook-style M3 question with clear structure and standard methods, making it slightly easier than average overall. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| (At surface) \(\frac{k}{R^2} = mg \Rightarrow k = mgR^2\) | M1A1 | (2) |
| Answer | Marks |
|---|---|
| M1 | for \(\frac{k}{R^2} = mg\). If not made clear that this applies at the surface of the Earth award M0 or \(\frac{k}{x^2} = mg\) and \(x = R\) |
| A1 cso | for \(k = mgR^2\) * |
| Answer | Marks | Guidance |
|---|---|---|
| \(m\ddot{x} = -\frac{mgR^2}{x^2}\) | ||
| \(v\frac{dv}{dx} = -\frac{gR^2}{x^2}\) | M1 | |
| \(\int v \frac{dv}{dx} dx = -gR^2 \int \frac{1}{x^2} dx\) or \(\int d\left(\frac{1}{2}v^2\right) dx\) | DM1A1 | |
| \(\frac{1}{2}v^2 = \frac{gR^2}{x} (+c)\) | DM1A1 | |
| \(x = \frac{5R}{4}, v = \sqrt{\frac{gR}{2}} \Rightarrow c = -\frac{11gR}{20}\) | DM1A1 | |
| \(v = 0 \Rightarrow \frac{gR^2}{x} - \frac{11gR}{20}\) | DM1 | |
| \(x = \frac{20R}{11}\) | A1 | (7) |
| Answer | Marks |
|---|---|
| \(GM_E = gR^2 \Rightarrow F = \frac{mgR^2}{x^2} \Rightarrow F = \frac{k}{x^2}\) with \(k = mgR^2\) * | M1 Complete method A1 Correct answer |
| Answer | Marks |
|---|---|
| Work done against gravity \(= \int_R^x \frac{mgR^2}{x^2} dx = \int_x mgR x^{-2} dx\) | M1 |
| \(= \frac{4mgR}{5} - \frac{mgR^2}{z}\) | DM1(integration)A1(correct) |
| Work-energy equation: \(\frac{mgR}{4} = \frac{4mgR}{5} - \frac{mgR^2}{z}\) | DM1A1 |
| \(z = \frac{20R}{11}\) | DM1A1 |
| Answer | Marks |
|---|---|
| M1 | for using accel \(= v\frac{dv}{dx}\) oe in NL2 with or w/o m. Minus sign not required. |
| M1 dep | for attempting to integrate both sides - minus not needed |
| A1 | for fully correct integration, with or w/o the constant. Must have included the minus sign from the start. |
| M1 dep | for using \(x = \frac{5R}{4}\), \(v = \sqrt{\frac{gR}{2}}\) to obtain a value for the constant. Use of \(x = \frac{R}{4}\) scores M0. Depends on both previous M marks |
| A1 | for \(c = -\frac{11gR}{20}\) |
| M1 dep | for setting \(v = 0\) and solving for \(x\). Depends on 1st and 2nd M marks, but not 3rd |
| A1 cso | for \(x = \frac{20R}{11}\) |
| Answer | Marks |
|---|---|
| DM1 | Using limits \(x = \frac{5R}{4}\), \(v = \sqrt{\frac{gR}{2}}\) |
| DM1 | Using limit \(v = 0\) |
| A1 | Correct substitution |
| A1 cso | for \(x = \frac{20R}{11}\) |
**(a)**
(At surface) $\frac{k}{R^2} = mg \Rightarrow k = mgR^2$ | M1A1 | (2)
**Notes for Question 2(a):**
M1 | for $\frac{k}{R^2} = mg$. If not made clear that this applies at the surface of the Earth award M0 or $\frac{k}{x^2} = mg$ and $x = R$ |
A1 cso | for $k = mgR^2$ * |
**(b)**
$m\ddot{x} = -\frac{mgR^2}{x^2}$ | |
$v\frac{dv}{dx} = -\frac{gR^2}{x^2}$ | M1 |
$\int v \frac{dv}{dx} dx = -gR^2 \int \frac{1}{x^2} dx$ or $\int d\left(\frac{1}{2}v^2\right) dx$ | DM1A1 |
$\frac{1}{2}v^2 = \frac{gR^2}{x} (+c)$ | DM1A1 |
$x = \frac{5R}{4}, v = \sqrt{\frac{gR}{2}} \Rightarrow c = -\frac{11gR}{20}$ | DM1A1 |
$v = 0 \Rightarrow \frac{gR^2}{x} - \frac{11gR}{20}$ | DM1 |
$x = \frac{20R}{11}$ | A1 | (7)
[9]
**Alternative for Question 2:**
Using $F = \frac{GM_1 M_2}{x^2}$ with $x = R$ and one mass as mass of Earth:
$mg = \frac{GmM_E}{R^2}$
$GM_E = gR^2 \Rightarrow F = \frac{mgR^2}{x^2} \Rightarrow F = \frac{k}{x^2}$ with $k = mgR^2$ * | M1 Complete method A1 Correct answer |
By conservation of energy:
Work done against gravity $= \int_R^x \frac{mgR^2}{x^2} dx = \int_x mgR x^{-2} dx$ | M1 |
$= \frac{4mgR}{5} - \frac{mgR^2}{z}$ | DM1(integration)A1(correct) |
Work-energy equation: $\frac{mgR}{4} = \frac{4mgR}{5} - \frac{mgR^2}{z}$ | DM1A1 |
$z = \frac{20R}{11}$ | DM1A1 |
**Notes for Question 2(b):**
M1 | for using accel $= v\frac{dv}{dx}$ oe in NL2 with or w/o m. Minus sign not required. |
M1 dep | for attempting to integrate both sides - minus not needed |
A1 | for fully correct integration, with or w/o the constant. Must have included the minus sign from the start. |
M1 dep | for using $x = \frac{5R}{4}$, $v = \sqrt{\frac{gR}{2}}$ to obtain a value for the constant. Use of $x = \frac{R}{4}$ scores M0. Depends on both previous M marks |
A1 | for $c = -\frac{11gR}{20}$ |
M1 dep | for setting $v = 0$ and solving for $x$. Depends on 1st and 2nd M marks, but not 3rd |
A1 cso | for $x = \frac{20R}{11}$ |
**ALT:** By definite integration
First 3 marks as above, then
DM1 | Using limits $x = \frac{5R}{4}$, $v = \sqrt{\frac{gR}{2}}$ |
DM1 | Using limit $v = 0$ |
A1 | Correct substitution |
A1 cso | for $x = \frac{20R}{11}$ |
NB: The penultimate A mark has changed position, but must be entered on e-pen in its original position.
---2. A particle $P$ of mass $m$ is fired vertically upwards from a point on the surface of the Earth and initially moves in a straight line directly away from the centre of the Earth. When $P$ is at a distance $x$ from the centre of the Earth, the gravitational force exerted by the Earth on $P$ is directed towards the centre of the Earth and has magnitude $\frac { k } { x ^ { 2 } }$, where $k$ is a constant.
At the surface of the Earth the acceleration due to gravity is $g$. The Earth is modelled as a fixed sphere of radius $R$.
\begin{enumerate}[label=(\alph*)]
\item Show that $k = m g R ^ { 2 }$.
When $P$ is at a height $\frac { R } { 4 }$ above the surface of the Earth, the speed of $P$ is $\sqrt { \frac { g R } { 2 } }$ Given that air resistance can be ignored,
\item find, in terms of $R$, the greatest distance from the centre of the Earth reached by $P$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2014 Q2 [9]}}