Edexcel M3 2024 January — Question 2 9 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Year2024
SessionJanuary
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHooke's law and elastic energy
TypeElastic string on smooth inclined plane
DifficultyStandard +0.3 This is a standard M3 elastic energy problem requiring conservation of energy (part a) and force resolution (part b). The setup is straightforward with given numerical values, requiring routine application of elastic PE formula (½λx²/l), gravitational PE, and Hooke's law. The algebra is manageable and the problem follows a typical textbook pattern, making it slightly easier than average for M3 level.
Spec3.02d Constant acceleration: SUVAT formulae3.03v Motion on rough surface: including inclined planes6.02d Mechanical energy: KE and PE concepts6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle
2. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{92131234-bfc1-4e0e-87d4-db9335fbf343-04_401_1031_287_516} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} A light elastic spring has natural length \(l\) and modulus of elasticity \(\lambda\) One end of the spring is attached to a point \(A\) on a smooth plane.
The plane is inclined at angle \(\theta\) to the horizontal, where \(\tan \theta = \frac { 5 } { 12 }\) A particle \(P\) of mass \(m\) is attached to the other end of the spring. Initially \(P\) is held at the point \(B\) on the plane, where \(A B\) is a line of greatest slope of the plane. The point \(B\) is lower than \(A\) and \(A B = 2 l\), as shown in Figure 1 .
The particle is released from rest at \(B\) and first comes to instantaneous rest at the point \(C\) on \(A B\), where \(A C = 0.7 l\)
  1. Use the principle of conservation of mechanical energy to show that $$\lambda = \frac { 100 } { 91 } m g$$
  2. Find the acceleration of \(P\) when it is released from rest at \(B\).
Question 2:
Part 2a:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Change in GPE \(= mg \times 1.3l\sin\theta\,(= 0.5lmg)\)M1 Condone sin/cos confusion
\(\text{EPE} = \dfrac{\lambda l^2}{2l}\) or \(\text{EPE} = \dfrac{\lambda(0.3l)^2}{2l}\)B1 One correct term for EPE
Energy equation \(B\) to \(A\)M1 Dimensionally correct with all the required terms. Condone sign errors and sin/cos confusion
\(\dfrac{\lambda l^2}{2l} - \dfrac{\lambda(0.3l)^2}{2l} = 0.5lmg\)A1 Correct unsimplified equation
\(\Rightarrow \lambda = mg\dfrac{1}{1-0.09} = \dfrac{100}{91}mg\)A1* Obtain given answer from correct working. Must see evidence of simplification.
[5]
Part 2b:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Equation of motionM1 Dimensionally correct with all the required terms. Condone sign errors and sin/cos confusion.
\(T - mg\sin\theta = ma\)A1 Correct unsimplified equation
\(\dfrac{\lambda \times l}{l} - mg\sin\theta = ma\) \(\left(\dfrac{100}{91}mg - \dfrac{5}{13}mg = ma\right)\)A1 Correct unsimplified equation with HL used to replace \(T\)
\(a = \dfrac{5}{7}g\)A1 Accept \(0.71g\) or better. If \(g = 9.8\) is used, accept 7.
[4]
(9) total
*Note: If \(g = 9.81\) is used then penalise once per complete question. SHM equations can only be used if the motion is proven to be SHM first.*
# Question 2:

## Part 2a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Change in GPE $= mg \times 1.3l\sin\theta\,(= 0.5lmg)$ | M1 | Condone sin/cos confusion |
| $\text{EPE} = \dfrac{\lambda l^2}{2l}$ or $\text{EPE} = \dfrac{\lambda(0.3l)^2}{2l}$ | B1 | One correct term for EPE |
| Energy equation $B$ to $A$ | M1 | Dimensionally correct with all the required terms. Condone sign errors and sin/cos confusion |
| $\dfrac{\lambda l^2}{2l} - \dfrac{\lambda(0.3l)^2}{2l} = 0.5lmg$ | A1 | Correct unsimplified equation |
| $\Rightarrow \lambda = mg\dfrac{1}{1-0.09} = \dfrac{100}{91}mg$ | A1* | Obtain given answer from correct working. Must see evidence of simplification. |
| | [5] | |

## Part 2b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation of motion | M1 | Dimensionally correct with all the required terms. Condone sign errors and sin/cos confusion. |
| $T - mg\sin\theta = ma$ | A1 | Correct unsimplified equation |
| $\dfrac{\lambda \times l}{l} - mg\sin\theta = ma$ $\left(\dfrac{100}{91}mg - \dfrac{5}{13}mg = ma\right)$ | A1 | Correct unsimplified equation with HL used to replace $T$ |
| $a = \dfrac{5}{7}g$ | A1 | Accept $0.71g$ or better. If $g = 9.8$ is used, accept 7. |
| | [4] | |
| | (9) total | |

*Note: If $g = 9.81$ is used then penalise once per complete question. SHM equations can only be used if the motion is proven to be SHM first.*

---
2.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{92131234-bfc1-4e0e-87d4-db9335fbf343-04_401_1031_287_516}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

A light elastic spring has natural length $l$ and modulus of elasticity $\lambda$ One end of the spring is attached to a point $A$ on a smooth plane.\\
The plane is inclined at angle $\theta$ to the horizontal, where $\tan \theta = \frac { 5 } { 12 }$ A particle $P$ of mass $m$ is attached to the other end of the spring.

Initially $P$ is held at the point $B$ on the plane, where $A B$ is a line of greatest slope of the plane.

The point $B$ is lower than $A$ and $A B = 2 l$, as shown in Figure 1 .\\
The particle is released from rest at $B$ and first comes to instantaneous rest at the point $C$ on $A B$, where $A C = 0.7 l$
\begin{enumerate}[label=(\alph*)]
\item Use the principle of conservation of mechanical energy to show that

$$\lambda = \frac { 100 } { 91 } m g$$
\item Find the acceleration of $P$ when it is released from rest at $B$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel M3 2024 Q2 [9]}}
This paper (7 questions)
View full paper
Q1 6 Q2 9 Q3 8 Q4 9 Q5 16 Q6 14 Q7 13