Edexcel FM1 Specimen — Question 7 14 marks

Exam BoardEdexcel
ModuleFM1 (Further Mechanics 1)
SessionSpecimen
Marks14
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Mark schemeDownload PDF ↗
TopicHooke's law and elastic energy
TypeVertical elastic string: general symbolic/proof questions
DifficultyStandard +0.8 This is a multi-part Further Mechanics question requiring equilibrium analysis, Newton's second law with elastic forces, and energy conservation for vertical elastic motion. While the techniques are standard for FM1, the question demands careful handling of multiple stages (equilibrium, initial acceleration, maximum speed at equilibrium position, and energy conservation to find turning point), making it moderately challenging but within expected FM1 scope.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=13.03f Weight: W=mg6.02g 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 particle \(P\) of mass \(m\) is attached to one end of a light elastic string of natural length \(a\) and modulus of elasticity \(3 m g\).
The other end of the string is attached to a fixed point \(O\) on a ceiling.
The particle hangs freely in equilibrium at a distance \(d\) vertically below \(O\).
  1. Show that \(d = \frac { 4 } { 3 } a\). The point \(A\) is vertically below \(O\) such that \(O A = 2 a\).
    The particle is held at rest at \(A\), then released and first comes to instantaneous rest at the point \(B\).
  2. Find, in terms of \(g\), the acceleration of \(P\) immediately after it is released from rest.
  3. Find, in terms of \(g\) and \(a\), the maximum speed attained by \(P\) as it moves from \(A\) to \(B\).
  4. Find, in terms of \(a\), the distance \(O B\).
Question 7:
Part (a):
AnswerMarks Guidance
Working/AnswerMark Guidance
In equilibrium ⟹ no resultant vertical forceM1 Use \(T = \frac{\lambda x}{a}\) to form equation for equilibrium
\(\frac{3mgx}{a} = mg\)A1 Correct unsimplified equation
\(x = \frac{a}{3}\), \(d = \frac{4}{3}a\)A1* Requires sufficient working to justify given answer plus a statement that the required result has been achieved
Part (b):
AnswerMarks Guidance
Working/AnswerMark Guidance
Equation of motionM1 Use \(T = \frac{\lambda x}{a}\) to form equation of motion; need all 3 terms; condone sign errors
\(\frac{3mga}{a} - mg = m\ddot{x}\)A1 Correct unsimplified equation
\(\ddot{x} = 2g\)A1 cao
Part (c):
AnswerMarks Guidance
Working/AnswerMark Guidance
Max speed at equilibrium positionB1 Seen or implied
Work energy & use of \(EPE = \frac{\lambda x^2}{2a}\)M1 Form work-energy equation; all 4 terms needed; condone sign errors
\(\frac{3mga^2}{2a} = \frac{3mg\left(\frac{a}{3}\right)^2}{2a} + \frac{1}{2}mv^2 + mg\frac{2a}{3}\)A1 A1 Correct unsimplified equation A1A1; one error in equation A1A0
\(\frac{1}{2}v^2 = ga\left(\frac{3}{2} - \frac{1}{6} - \frac{2}{3}\right) = \frac{2}{3}ga\), \(v = \sqrt{\frac{4ga}{3}}\)A1 cao
Part (d):
AnswerMarks Guidance
Working/AnswerMark Guidance
At max height: \(KE = 0\), \(EPE\) lost \(= GPE\) gainedM1 Form energy equation
\(\frac{3mga^2}{2a} = mgh\)A1 Correct unsimplified equation
\(OB = \frac{a}{2}\)A1 cao 
## Question 7:

### Part (a):

| Working/Answer | Mark | Guidance |
|---|---|---|
| In equilibrium ⟹ no resultant vertical force | M1 | Use $T = \frac{\lambda x}{a}$ to form equation for equilibrium |
| $\frac{3mgx}{a} = mg$ | A1 | Correct unsimplified equation |
| $x = \frac{a}{3}$, $d = \frac{4}{3}a$ | A1* | Requires sufficient working to justify given answer plus a statement that the required result has been achieved |

### Part (b):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Equation of motion | M1 | Use $T = \frac{\lambda x}{a}$ to form equation of motion; need all 3 terms; condone sign errors |
| $\frac{3mga}{a} - mg = m\ddot{x}$ | A1 | Correct unsimplified equation |
| $\ddot{x} = 2g$ | A1 | cao |

### Part (c):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Max speed at equilibrium position | B1 | Seen or implied |
| Work energy & use of $EPE = \frac{\lambda x^2}{2a}$ | M1 | Form work-energy equation; all 4 terms needed; condone sign errors |
| $\frac{3mga^2}{2a} = \frac{3mg\left(\frac{a}{3}\right)^2}{2a} + \frac{1}{2}mv^2 + mg\frac{2a}{3}$ | A1 A1 | Correct unsimplified equation A1A1; one error in equation A1A0 |
| $\frac{1}{2}v^2 = ga\left(\frac{3}{2} - \frac{1}{6} - \frac{2}{3}\right) = \frac{2}{3}ga$, $v = \sqrt{\frac{4ga}{3}}$ | A1 | cao |

### Part (d):

| Working/Answer | Mark | Guidance |
|---|---|---|
| At max height: $KE = 0$, $EPE$ lost $= GPE$ gained | M1 | Form energy equation |
| $\frac{3mga^2}{2a} = mgh$ | A1 | Correct unsimplified equation |
| $OB = \frac{a}{2}$ | A1 | cao |

---
\begin{enumerate}
  \item A particle $P$ of mass $m$ is attached to one end of a light elastic string of natural length $a$ and modulus of elasticity $3 m g$.
\end{enumerate}

The other end of the string is attached to a fixed point $O$ on a ceiling.\\
The particle hangs freely in equilibrium at a distance $d$ vertically below $O$.\\
(a) Show that $d = \frac { 4 } { 3 } a$.

The point $A$ is vertically below $O$ such that $O A = 2 a$.\\
The particle is held at rest at $A$, then released and first comes to instantaneous rest at the point $B$.\\
(b) Find, in terms of $g$, the acceleration of $P$ immediately after it is released from rest.\\
(c) Find, in terms of $g$ and $a$, the maximum speed attained by $P$ as it moves from $A$ to $B$.\\
(d) Find, in terms of $a$, the distance $O B$.

\hfill \mbox{\textit{Edexcel FM1  Q7 [14]}}
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