Edexcel M3 2003 January — Question 6 12 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Year2003
SessionJanuary
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHooke's law and elastic energy
TypeVertical elastic string: released from rest at natural length or above (string initially slack)
DifficultyStandard +0.8 This M3 question requires energy conservation with elastic strings in two scenarios, including finding when the string becomes slack. It demands understanding of equilibrium positions, extension calculations, and careful application of energy principles across multiple stages of motion. More conceptually demanding than standard M3 questions but follows established methods.
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

6. A light elastic string has natural length 4 m and modulus of elasticity 58.8 N . A particle \(P\) of mass 0.5 kg is attached to one end of the string. The other end of the string is attached to a vertical point \(A\). The particle is released from rest at \(A\) and falls vertically.
  1. Find the distance travelled by \(P\) before it immediately comes to instantaneous rest for the first time. The particle is now held at a point 7 m vertically below \(A\) and released from rest.
  2. Find the speed of the particle when the string first becomes slack.

Question 6:
Part (a):
AnswerMarks Guidance
WorkingMarks Notes
\(\frac{1}{2}\times\frac{58.8}{4}x^2 = 0.5\times 9.8(x+4)\)M1 A1 A1
\(3x^2 - 2x - 8 = 0\)M1 A1
\((3x+4)(x-2)=0,\ x=2\)
Distance fallen \(= 6\) mM1 A1 (7)
Part (b):
AnswerMarks Guidance
WorkingMarks Notes
\(\frac{1}{2}\times 0.5v^2 = \frac{1}{2}\times\frac{58.8}{4}\times 3^2 - 0.5\times 9.8\times 3\)M1 A1 A1
\(v = 14.3\) ms\(^{-1}\)M1 A1 (5)
# Question 6:

## Part (a):
| Working | Marks | Notes |
|---------|-------|-------|
| $\frac{1}{2}\times\frac{58.8}{4}x^2 = 0.5\times 9.8(x+4)$ | M1 A1 A1 | |
| $3x^2 - 2x - 8 = 0$ | M1 A1 | |
| $(3x+4)(x-2)=0,\ x=2$ | | |
| Distance fallen $= 6$ m | M1 A1 | (7) |

## Part (b):
| Working | Marks | Notes |
|---------|-------|-------|
| $\frac{1}{2}\times 0.5v^2 = \frac{1}{2}\times\frac{58.8}{4}\times 3^2 - 0.5\times 9.8\times 3$ | M1 A1 A1 | |
| $v = 14.3$ ms$^{-1}$ | M1 A1 | (5) |

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6. A light elastic string has natural length 4 m and modulus of elasticity 58.8 N . A particle $P$ of mass 0.5 kg is attached to one end of the string. The other end of the string is attached to a vertical point $A$. The particle is released from rest at $A$ and falls vertically.
\begin{enumerate}[label=(\alph*)]
\item Find the distance travelled by $P$ before it immediately comes to instantaneous rest for the first time.

The particle is now held at a point 7 m vertically below $A$ and released from rest.
\item Find the speed of the particle when the string first becomes slack.
\end{enumerate}

\hfill \mbox{\textit{Edexcel M3 2003 Q6 [12]}}