Edexcel M3 2022 January — Question 3 8 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Year2022
SessionJanuary
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAdvanced work-energy problems
TypeVariable force along axis work-energy
DifficultyStandard +0.8 This M3 question requires setting up and solving a work-energy equation with variable force integration (∫x² dx), combined with resolving forces on an inclined plane. While the integration itself is routine, students must correctly identify that work done by gravity (mgx sin α) minus work against the resistance force equals kinetic energy, then solve the resulting equation. Part (b) requires recognizing that at instantaneous rest, all gravitational PE converts to work against resistance, leading to a cubic equation. This is above-average difficulty due to the combination of mechanics concepts with calculus, though it follows a standard M3 template.
Spec6.06a Variable force: dv/dt or v*dv/dx methods
  1. A particle \(P\) of mass \(m \mathrm {~kg}\) is initially held at rest at the point \(O\) on a smooth inclined plane. The plane is inclined at an angle \(\alpha\) to the horizontal, where \(\sin \alpha = \frac { 2 } { 5 }\)
The particle is released from rest and slides down the plane against a force which acts towards \(O\). The force has magnitude \(\frac { 1 } { 3 } m x ^ { 2 } \mathrm {~N}\), where \(x\) metres is the distance of \(P\) from \(O\).
  1. Find the speed of \(P\) when \(x = 2\) The particle first comes to instantaneous rest at the point \(A\).
  2. Find the distance \(O A\).
Question 3(a):
AnswerMarks Guidance
Working/AnswerMarks Guidance
\(mv\dfrac{dv}{dx} = mg\sin\alpha - \frac{1}{3}mx^2\)M1A1 M1: Attempt equation of motion parallel to plane, acceleration in any form including \(a\). A1: Correct equation with acceleration in \(v\dfrac{dv}{dx}\) form.
\(\frac{1}{2}v^2 = xg\sin\alpha - \frac{1}{9}x^3\ (+c)\)DM1A1 DM1: Attempt integration, powers increase by 1 in 2 terms (constant may be missing). Acceleration must be in \(v\dfrac{dv}{dx}\) form. A1: Correct integration.
\(x=2\): \(\frac{1}{2}v^2 = 2g\sin\alpha - \frac{8}{9}\)DM1 Sub \(x=2\) into expression for \(v^2\). Depends on all previous M marks. \(+c\) must be dealt with. Must be 2 or 3 sf.
\(v = 3.7\) or \(3.73\ \text{ms}^{-1}\)A1cso (6) Correct value.
ALT (energy method):
AnswerMarks Guidance
Working/AnswerMarks Guidance
\(mg\sin\alpha\, x = \int \frac{1}{3}mx^2\,dx + \frac{1}{2}mv^2\)M1A1 M1: Attempt 3-term energy equation (KE, GPE, work done). Integral form not required. A1: Fully correct with integral form for work done.
\(xg\sin\alpha = \frac{1}{9}x^3 + \frac{1}{2}v^2\ (+c)\)DM1A1 Rest as main scheme.
\(x=2\): \(\frac{1}{2}v^2 = 2g\sin\alpha - \frac{8}{9}\)DM1
\(v = 3.7\) or \(3.73\ \text{ms}^{-1}\)A1
Question 3(b):
AnswerMarks Guidance
Working/AnswerMarks Guidance
\(v=0 \Rightarrow x^2 = 9g\sin\alpha = 9g \times \frac{2}{5}\ (x\neq 0)\)M1 Use \(v=0\) in expression for \(v\) to obtain value of \(x\).
\(x = 5.939\ldots \Rightarrow OA = 5.9\) or \(5.94\ \text{m}\)M1A1 (2) A1: Correct value of \(OA\). Allow if \(x\) given instead of \(OA\). Must be 2 or 3 sf (unless already penalised in (a)). 
# Question 3(a):

| Working/Answer | Marks | Guidance |
|---|---|---|
| $mv\dfrac{dv}{dx} = mg\sin\alpha - \frac{1}{3}mx^2$ | M1A1 | M1: Attempt equation of motion parallel to plane, acceleration in any form including $a$. A1: Correct equation with acceleration in $v\dfrac{dv}{dx}$ form. |
| $\frac{1}{2}v^2 = xg\sin\alpha - \frac{1}{9}x^3\ (+c)$ | DM1A1 | DM1: Attempt integration, powers increase by 1 in 2 terms (constant may be missing). Acceleration must be in $v\dfrac{dv}{dx}$ form. A1: Correct integration. |
| $x=2$: $\frac{1}{2}v^2 = 2g\sin\alpha - \frac{8}{9}$ | DM1 | Sub $x=2$ into expression for $v^2$. Depends on all previous M marks. $+c$ must be dealt with. Must be 2 or 3 sf. |
| $v = 3.7$ or $3.73\ \text{ms}^{-1}$ | A1cso (6) | Correct value. |

**ALT (energy method):**

| Working/Answer | Marks | Guidance |
|---|---|---|
| $mg\sin\alpha\, x = \int \frac{1}{3}mx^2\,dx + \frac{1}{2}mv^2$ | M1A1 | M1: Attempt 3-term energy equation (KE, GPE, work done). Integral form not required. A1: Fully correct with integral form for work done. |
| $xg\sin\alpha = \frac{1}{9}x^3 + \frac{1}{2}v^2\ (+c)$ | DM1A1 | Rest as main scheme. |
| $x=2$: $\frac{1}{2}v^2 = 2g\sin\alpha - \frac{8}{9}$ | DM1 | |
| $v = 3.7$ or $3.73\ \text{ms}^{-1}$ | A1 | |

# Question 3(b):

| Working/Answer | Marks | Guidance |
|---|---|---|
| $v=0 \Rightarrow x^2 = 9g\sin\alpha = 9g \times \frac{2}{5}\ (x\neq 0)$ | M1 | Use $v=0$ in expression for $v$ to obtain value of $x$. |
| $x = 5.939\ldots \Rightarrow OA = 5.9$ or $5.94\ \text{m}$ | M1A1 (2) | A1: Correct value of $OA$. Allow if $x$ given instead of $OA$. Must be 2 or 3 sf (unless already penalised in (a)). |

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\begin{enumerate}
  \item A particle $P$ of mass $m \mathrm {~kg}$ is initially held at rest at the point $O$ on a smooth inclined plane. The plane is inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = \frac { 2 } { 5 }$
\end{enumerate}

The particle is released from rest and slides down the plane against a force which acts towards $O$. The force has magnitude $\frac { 1 } { 3 } m x ^ { 2 } \mathrm {~N}$, where $x$ metres is the distance of $P$ from $O$.\\
(a) Find the speed of $P$ when $x = 2$

The particle first comes to instantaneous rest at the point $A$.\\
(b) Find the distance $O A$.

\hfill \mbox{\textit{Edexcel M3 2022 Q3 [8]}}
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