Edexcel FM1 2023 June — Question 4 15 marks

Exam BoardEdexcel
ModuleFM1 (Further Mechanics 1)
Year2023
SessionJune
Marks15
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHooke's law and elastic energy
TypeElastic string equilibrium and statics
DifficultyStandard +0.8 This is a multi-part Further Mechanics question requiring equilibrium analysis, energy methods with resistance forces, and finding maximum speed conditions. Part (a) is routine equilibrium, but parts (b) and (c) require careful energy accounting with resistance and understanding that maximum speed occurs when net force is zero. The combination of elastic potential energy, gravitational potential energy, work against resistance, and force analysis across three distinct scenarios elevates this above standard A-level mechanics.
Spec6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle
  1. A light elastic string has natural length \(2 a\) and modulus of elasticity \(4 m g\). One end of the elastic string is attached to a fixed point \(O\). A particle \(P\) of mass \(m\) is attached to the other end of the elastic string.
    The particle \(P\) hangs freely in equilibrium at the point \(E\), which is vertically below \(O\)
    1. Find the length \(O E\).
    Particle \(P\) is now pulled vertically downwards to the point \(A\), where \(O A = 4 a\), and released from rest. The resistance to the motion of \(P\) is a constant force of magnitude \(\frac { 1 } { 4 } m g\).
  2. Find, in terms of \(a\) and \(g\), the speed of \(P\) after it has moved a distance \(a\). Particle \(P\) is now held at \(O\) Particle \(P\) is released from rest and reaches its maximum speed at the point \(B\). The resistance to the motion of \(P\) is again a constant force of magnitude \(\frac { 1 } { 4 } m g\).
  3. Find the distance \(O B\).
Question 4(a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(T = \frac{4mge}{2a}\)B1
\(T = mg\)M1
\(e = \frac{1}{2}a\)A1
\(OE = \frac{5a}{2}\)A1
Question 4(b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
GPE term \(\pm mga\)B1
Work done against resistance \(\pm\frac{1}{4}mga\)B1
Use of EPE formula onceM1
\(\pm\frac{4mg}{2\times 2a}\{(2a)^2 - a^2\}\)A1
Work energy equationM1
\(\frac{1}{4}mga = \frac{4mg}{2\times 2a}\{(2a)^2-a^2\} - mga - \frac{1}{2}mv^2\)A1
\(v = \sqrt{\frac{7ag}{2}}\) oeA1
Question 4(c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(mg - T - \frac{1}{4}mg = 0\)M1
\(mg - \frac{4mgx}{2a} - \frac{1}{4}mg = 0\)A1
\(x = \frac{3a}{8}\)A1
\(OB = \frac{19a}{8}\) oeA1
Question 4(b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
GPE term seenB1 Ignore sign
Work term \(\frac{mga}{4}\) seenB1 Ignore sign. Allow B1 for WD = \(\frac{5mga}{4}\) where resistance included within term
Use of EPE formula, EPE in form \(\frac{\lambda x^2}{ka}\)M1
Difference between two correct EPE terms seen, unsimplifiedA1
Work-energy equation formed with all relevant terms: KE, GPE, 2EPE, WD, condone sign errorsM1 M0 if WD = \(\frac{5mga}{4}\) and a GPE term present (weight counted twice)
Correct unsimplified equationA1
\(v = \sqrt{\frac{7ag}{2}}\), \(v = \frac{1}{2}\sqrt{14ag}\)A1 Correct answer in terms of \(a\) and \(g\); do not allow 9.8 for \(g\)
Question 4(c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Vertical equilibrium or equation of motion with \(a=0\), correct number of termsM1 All 3 forces included; \(\left(mg \pm \frac{mg}{4}\right)\) may be simplified; Hooke's Law need not be substituted but M0 if equilibrium position from (a) used
Correct equation in one unknownA1
caoA1
cao; if extension is \((OB - 2a)\), \(OB\) found directly, both A marks earned togetherA1
Question 4(c) Alt 1 (Differentiation with work-energy from point of release):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\frac{1}{2}mv^2 = mgh - \frac{4mg(h-2a)^2}{2(2a)} - \frac{mgh}{4}\)M1 All relevant terms included, correct form, no extra terms
Correct equation for \(v^2\) and \(h\)A1 May use different letter
\(\frac{d}{dh}(v^2) = 0 \rightarrow \frac{3g}{2} = \frac{4g(h-2a)}{a}\)A1 Correct equation after differentiating and setting to zero
\(OB = \frac{19a}{8}\)A1 Correct answer 
# Question 4(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $T = \frac{4mge}{2a}$ | B1 | |
| $T = mg$ | M1 | |
| $e = \frac{1}{2}a$ | A1 | |
| $OE = \frac{5a}{2}$ | A1 | |

---

# Question 4(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| GPE term $\pm mga$ | B1 | |
| Work done against resistance $\pm\frac{1}{4}mga$ | B1 | |
| Use of EPE formula once | M1 | |
| $\pm\frac{4mg}{2\times 2a}\{(2a)^2 - a^2\}$ | A1 | |
| Work energy equation | M1 | |
| $\frac{1}{4}mga = \frac{4mg}{2\times 2a}\{(2a)^2-a^2\} - mga - \frac{1}{2}mv^2$ | A1 | |
| $v = \sqrt{\frac{7ag}{2}}$ oe | A1 | |

---

# Question 4(c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $mg - T - \frac{1}{4}mg = 0$ | M1 | |
| $mg - \frac{4mgx}{2a} - \frac{1}{4}mg = 0$ | A1 | |
| $x = \frac{3a}{8}$ | A1 | |
| $OB = \frac{19a}{8}$ oe | A1 | |

# Question 4(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| GPE term seen | B1 | Ignore sign |
| Work term $\frac{mga}{4}$ seen | B1 | Ignore sign. Allow B1 for WD = $\frac{5mga}{4}$ where resistance included within term |
| Use of EPE formula, EPE in form $\frac{\lambda x^2}{ka}$ | M1 | |
| Difference between two correct EPE terms seen, unsimplified | A1 | |
| Work-energy equation formed with all relevant terms: KE, GPE, 2EPE, WD, condone sign errors | M1 | M0 if WD = $\frac{5mga}{4}$ **and** a GPE term present (weight counted twice) |
| Correct unsimplified equation | A1 | |
| $v = \sqrt{\frac{7ag}{2}}$, $v = \frac{1}{2}\sqrt{14ag}$ | A1 | Correct answer in terms of $a$ and $g$; do not allow 9.8 for $g$ |

---

# Question 4(c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Vertical equilibrium or equation of motion with $a=0$, correct number of terms | M1 | All 3 forces included; $\left(mg \pm \frac{mg}{4}\right)$ may be simplified; Hooke's Law need not be substituted but M0 if equilibrium position from (a) used |
| Correct equation in one unknown | A1 | |
| cao | A1 | |
| cao; if extension is $(OB - 2a)$, $OB$ found directly, both A marks earned together | A1 | |

---

# Question 4(c) Alt 1 (Differentiation with work-energy from point of release):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}mv^2 = mgh - \frac{4mg(h-2a)^2}{2(2a)} - \frac{mgh}{4}$ | M1 | All relevant terms included, correct form, no extra terms |
| Correct equation for $v^2$ and $h$ | A1 | May use different letter |
| $\frac{d}{dh}(v^2) = 0 \rightarrow \frac{3g}{2} = \frac{4g(h-2a)}{a}$ | A1 | Correct equation after differentiating and setting to zero |
| $OB = \frac{19a}{8}$ | A1 | Correct answer |

---
\begin{enumerate}
  \item A light elastic string has natural length $2 a$ and modulus of elasticity $4 m g$. One end of the elastic string is attached to a fixed point $O$. A particle $P$ of mass $m$ is attached to the other end of the elastic string.\\
The particle $P$ hangs freely in equilibrium at the point $E$, which is vertically below $O$\\
(a) Find the length $O E$.
\end{enumerate}

Particle $P$ is now pulled vertically downwards to the point $A$, where $O A = 4 a$, and released from rest. The resistance to the motion of $P$ is a constant force of magnitude $\frac { 1 } { 4 } m g$.\\
(b) Find, in terms of $a$ and $g$, the speed of $P$ after it has moved a distance $a$.

Particle $P$ is now held at $O$ Particle $P$ is released from rest and reaches its maximum speed at the point $B$. The resistance to the motion of $P$ is again a constant force of magnitude $\frac { 1 } { 4 } m g$.\\
(c) Find the distance $O B$.

\hfill \mbox{\textit{Edexcel FM1 2023 Q4 [15]}}
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