| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2017 |
| Session | June |
| Marks | 12 |
| 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 setting up and evaluating a work-energy integral with a variable force (inverse square law), then applying conservation of energy principles. While the integration itself is straightforward, students must correctly handle the negative work done by an opposing force and solve the resulting equation. This is above-average difficulty due to the conceptual setup and multi-step reasoning, but remains a standard M3 exercise rather than requiring novel insight. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(0.4\ddot{x} = -\frac{k}{x^2}\) | M1 | Form equation of motion, minus sign may be missing |
| \(0.4v\frac{dv}{dx} = -\frac{k}{x^2}\) | M1 | Write acceleration in form \(v\frac{dv}{dx}\). These two M marks may be awarded together. Can be implied by \(\frac{1}{2}mv^2\) after integrating |
| \(0.2v^2 = \int -kx^{-2} \, dx\) | ||
| \(0.2v^2 = \frac{k}{x}(+c)\) | dM1A1ft | dM1: Attempt to integrate both sides wrt \(x\), depends on both M marks; A1ft: Correct integration with correct signs. Constant may be missing. Follow through missing minus sign |
| \(x=2, v=5 \Rightarrow 5 = \frac{k}{2} + c\) | dM1 | Substitute either \(x=2, v=5\) or \(x=5, v=2\). Depends on all M marks above |
| \(x=5, v=2 \Rightarrow 0.8 = \frac{k}{5} + c\) | A1 | Both substitutions made and 2 correct equations in \(k\) and \(c\) found |
| \(4.2 = k\left(\frac{1}{2} - \frac{1}{5}\right)\) | dM1 | Solve simultaneous equations to obtain value of \(k\). Solving 1 linear equation (as \(c\) was omitted) scores M0. Depends on all M marks above |
| \(k = 14\) | A1 cso (8) | Correct value of \(k\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(c = 5 - 7 = -2\) | ||
| \(0.2v^2 = \frac{14}{x} - 2\) | M1A1ft | M1: Obtain value of \(c\) and form expression for \(v^2\); A1ft: Correct expression for \(v^2\), follow through \(k = -14\) giving \(c = -5\) |
| \(v = 0 \Rightarrow x = 7\) | dM1A1 cso (4) | dM1: Substitute \(v=0\) in expression for \(v^2\) and solve for \(x\); A1cso: Correct value of \(x\) |
# Question 5(a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $0.4\ddot{x} = -\frac{k}{x^2}$ | M1 | Form equation of motion, minus sign may be missing |
| $0.4v\frac{dv}{dx} = -\frac{k}{x^2}$ | M1 | Write acceleration in form $v\frac{dv}{dx}$. These two M marks may be awarded together. Can be implied by $\frac{1}{2}mv^2$ after integrating |
| $0.2v^2 = \int -kx^{-2} \, dx$ | | |
| $0.2v^2 = \frac{k}{x}(+c)$ | dM1A1ft | dM1: Attempt to integrate both sides wrt $x$, depends on both M marks; A1ft: Correct integration with correct signs. Constant may be missing. Follow through missing minus sign |
| $x=2, v=5 \Rightarrow 5 = \frac{k}{2} + c$ | dM1 | Substitute either $x=2, v=5$ or $x=5, v=2$. Depends on all M marks above |
| $x=5, v=2 \Rightarrow 0.8 = \frac{k}{5} + c$ | A1 | Both substitutions made and 2 correct equations in $k$ and $c$ found |
| $4.2 = k\left(\frac{1}{2} - \frac{1}{5}\right)$ | dM1 | Solve simultaneous equations to obtain value of $k$. Solving 1 linear equation (as $c$ was omitted) scores M0. Depends on all M marks above |
| $k = 14$ | A1 cso (8) | Correct value of $k$ |
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# Question 5(b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $c = 5 - 7 = -2$ | | |
| $0.2v^2 = \frac{14}{x} - 2$ | M1A1ft | M1: Obtain value of $c$ and form expression for $v^2$; A1ft: Correct expression for $v^2$, follow through $k = -14$ giving $c = -5$ |
| $v = 0 \Rightarrow x = 7$ | dM1A1 cso (4) | dM1: Substitute $v=0$ in expression for $v^2$ and solve for $x$; A1cso: Correct value of $x$ |5. A particle $P$ of mass 0.4 kg moves on the positive $x$-axis under the action of a single force. The force is directed towards the origin $O$ and has magnitude $\frac { k } { x ^ { 2 } }$ newtons, where $O P = x$ metres and $k$ is a constant. Initially $P$ is moving away from $O$. At $x = 2$ the speed of $P$ is $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and at $x = 5$ the speed of $P$ is $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $k$.
The particle first comes to instantaneous rest at the point $A$.
\item Find the value of $x$ at $A$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2017 Q5 [12]}}