Edexcel M2 — Question 6 14 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Marks14
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TopicMomentum and Collisions 1
TypeThree-particle sequential collisions
DifficultyStandard +0.3 This is a standard M2 sequential collision problem with straightforward application of conservation of momentum and Newton's restitution law. Part (a) is routine calculation, and part (b) requires logical reasoning about velocity conditions for a second collision, but follows a well-established problem type with no novel insights needed.
Spec6.03b Conservation of momentum: 1D two particles6.03j Perfectly elastic/inelastic: collisions

6. Three uniform spheres \(A , B\) and \(C\) of equal radius have masses \(3 m , 2 m\) and \(2 m\) respectively. Initially, the spheres are at rest on a smooth horizontal table with their centres in a straight line and with \(B\) between \(A\) and \(C\). Sphere \(A\) is projected directly towards \(B\) with speed \(u\). Given that the coefficient of restitution between \(A\) and \(B\) is \(\frac { 2 } { 3 }\),
  1. show that the speeds of \(A\) and \(B\) after the collision are \(\frac { 1 } { 3 } u\) and \(u\) respectively.
    (6 marks)
    The coefficient of restitution between \(B\) and \(C\) is \(e\). Given that \(A\) and \(B\) collide again,
  2. show that \(e > \frac { 1 } { 3 }\).
    (8 marks)

Part (a):
AnswerMarks
cons. of mom: \(3mu + 0 = 3mv_1 + 2mv_2\)M1
\(\therefore 3v_1 + 2v_2 = 3u\)A1
\(\frac{v_2 - v_1}{u} = \frac{2}{3}\) \(\therefore 3v_2 - 3v_1 = 2u\)M1 A1
solve simul. giving \(v_1 = \frac{1}{5}u\) and \(v_2 = u\)M1 A1
Part (b):
AnswerMarks Guidance
cons. of mom: \(2mu + 0 = 2mw_1 + 2mw_2\)M1
\(w_1 + w_2 = u\)A1
\(\frac{w_2 - w_1}{u} = e\); \(w_2 - w_1 = eu\)A1
solve simul. giving \(w_1 = \frac{1}{2}u(1-e)\)M1 A1
A and B collide again so speed of \(B <\) speed of AM1
\(\frac{1}{2}u(1-e) < \frac{1}{5}u\) so \(\frac{1}{2}e > \frac{1}{2} - \frac{1}{5}\); \(\therefore e > \frac{1}{5}\)M1 A1 (14)
**Part (a):**
cons. of mom: $3mu + 0 = 3mv_1 + 2mv_2$ | M1 |

$\therefore 3v_1 + 2v_2 = 3u$ | A1 |

$\frac{v_2 - v_1}{u} = \frac{2}{3}$ $\therefore 3v_2 - 3v_1 = 2u$ | M1 A1 |

solve simul. giving $v_1 = \frac{1}{5}u$ and $v_2 = u$ | M1 A1 |

**Part (b):**
cons. of mom: $2mu + 0 = 2mw_1 + 2mw_2$ | M1 |

$w_1 + w_2 = u$ | A1 |

$\frac{w_2 - w_1}{u} = e$; $w_2 - w_1 = eu$ | A1 |

solve simul. giving $w_1 = \frac{1}{2}u(1-e)$ | M1 A1 |

A and B collide again so speed of $B <$ speed of A | M1 |

$\frac{1}{2}u(1-e) < \frac{1}{5}u$ so $\frac{1}{2}e > \frac{1}{2} - \frac{1}{5}$; $\therefore e > \frac{1}{5}$ | M1 A1 | (14)

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6. Three uniform spheres $A , B$ and $C$ of equal radius have masses $3 m , 2 m$ and $2 m$ respectively. Initially, the spheres are at rest on a smooth horizontal table with their centres in a straight line and with $B$ between $A$ and $C$. Sphere $A$ is projected directly towards $B$ with speed $u$.

Given that the coefficient of restitution between $A$ and $B$ is $\frac { 2 } { 3 }$,
\begin{enumerate}[label=(\alph*)]
\item show that the speeds of $A$ and $B$ after the collision are $\frac { 1 } { 3 } u$ and $u$ respectively.\\
(6 marks)\\
The coefficient of restitution between $B$ and $C$ is $e$. Given that $A$ and $B$ collide again,
\item show that $e > \frac { 1 } { 3 }$.\\
(8 marks)
\end{enumerate}

\hfill \mbox{\textit{Edexcel M2  Q6 [14]}}