Edexcel M3 — Question 4 10 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Marks10
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)
DifficultyChallenging +1.2 This is a multi-stage mechanics problem requiring energy conservation with elastic strings, coefficient of restitution, and careful tracking through three phases (fall, impact, rebound). While it involves several M3 concepts and requires systematic bookkeeping across stages, the individual techniques are standard and the problem structure is clearly signposted, making it moderately challenging but not requiring novel insight.
Spec6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.03k Newton's experimental law: direct impact

A small stone \(P\) of mass \(m\) kg is attached to one end of a light elastic string of modulus \(3mg\) N and natural length \(2l\) m. The other end of the string is fixed to a point \(O\) at a height \(3l\) m above a horizontal surface. \(P\) is released from rest at \(O\); it hits the surface and rebounds to a height of \(2l\) m. The coefficient of restitution between \(P\) and the surface is \(e\). Calculate the value of \(e\). [9 marks] State one assumption (other than the string being light) that you have used in your solution. [1 mark]

AnswerMarks
(a) \(u = \) speed before impact. Energy: \(3mgl = \frac{3mg}{4\ell}P + \frac{1}{2}mu^2\) (1)M1 A1 A1
\(eu = \) speed after impact. Energy: \(2mgl = \frac{3mg}{4\ell}P + \frac{1}{2}me^2u^2\) (2)B1 M1 A1
(1) gives \(u^2 = \frac{2g\ell}{}, so \)\frac{s}{4} = \frac{9e^2}{4}$M1 A1 A1
\(9e^2 = 5\)M1 A1 A1
\(e = \frac{1}{3}\sqrt{5}\)M1 A1 A1
(b) Assumption: \(P\) is a particle; no air resistanceB1
Total: 10 marks
(a) $u = $ speed before impact. Energy: $3mgl = \frac{3mg}{4\ell}P + \frac{1}{2}mu^2$ (1) | M1 A1 A1 |

$eu = $ speed after impact. Energy: $2mgl = \frac{3mg}{4\ell}P + \frac{1}{2}me^2u^2$ (2) | B1 M1 A1 |

(1) gives $u^2 = \frac{2g\ell}{}, so $\frac{s}{4} = \frac{9e^2}{4}$ | M1 A1 A1 |

$9e^2 = 5$ | M1 A1 A1 |

$e = \frac{1}{3}\sqrt{5}$ | M1 A1 A1 |

(b) Assumption: $P$ is a particle; no air resistance | B1 |

| **Total: 10 marks** |
A small stone $P$ of mass $m$ kg is attached to one end of a light elastic string of modulus $3mg$ N and natural length $2l$ m. The other end of the string is fixed to a point $O$ at a height $3l$ m above a horizontal surface. $P$ is released from rest at $O$; it hits the surface and rebounds to a height of $2l$ m. The coefficient of restitution between $P$ and the surface is $e$.

Calculate the value of $e$. [9 marks]

State one assumption (other than the string being light) that you have used in your solution. [1 mark]

\hfill \mbox{\textit{Edexcel M3  Q4 [10]}}