Edexcel M2 2018 June — Question 7 14 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2018
SessionJune
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions 1
TypeThree-particle sequential collisions
DifficultyStandard +0.8 This is a challenging M2 mechanics problem requiring sequential collision analysis with restitution coefficients, conservation of momentum, and kinetic energy relationships. It demands careful algebraic manipulation across multiple collision stages and proof that a second collision occurs, going beyond routine collision exercises.
Spec6.03b Conservation of momentum: 1D two particles6.03k Newton's experimental law: direct impact

7. Three particles \(A\), \(B\) and \(C\) have masses \(2 m , 3 m\) and \(4 m\) respectively. The particles lie at rest in a straight line on a smooth horizontal surface, with \(B\) between \(A\) and \(C\). Particle \(A\) is projected towards \(B\) with speed \(u\) and collides directly with \(B\). The coefficient of restitution between \(A\) and \(B\) is \(e\). The kinetic energy of \(A\) immediately after the collision is one ninth of the kinetic energy of \(A\) immediately before the collision. Given that the direction of motion of \(A\) is unchanged by the collision,
  1. find the value of \(e\). After the collision between \(A\) and \(B\) there is a direct collision between \(B\) and \(C\). The coefficient of restitution between \(B\) and \(C\) is \(f\), where \(f < \frac { 3 } { 4 }\). The speed of \(B\) immediately after the collision with \(C\) is \(V\).
    1. Express \(V\) in terms of \(f\) and \(u\).
    2. Hence show that there will be a second collision between \(A\) and \(B\).

7a.
AnswerMarks
CLM: \(2mu = 2mv + 3mw\)M1 A1
Impact law: \(eu = w - v\)M1 A1
Eliminate w: \(5v = u(2 - 3e)\) oeM1
\(v = \frac{u}{3}\)M1
\(e = \frac{1}{9}, 0.11\) or betterA1
OR: \(v = \frac{u}{3}\) is used in CLM and NIL equations (See Alternative below)
ALTERNATIVE:
AnswerMarks
\(v = \frac{u}{3}\)M1 (KE)
\(2mu = 2mv + 3mw\) (\(2mu = 2m\frac{u}{3} + 3mw\))M1 A1 (CLM)
\(\Rightarrow w = \frac{4u}{9}\)M1 (eliminate w)
\(e = \frac{w - v}{u} = \left(e = \frac{\frac{4u}{9} - \frac{u}{3}}{u}\right)\)M1 A1 (Impact Law)
\(e = \frac{1}{9}\)A1
7b(i).
AnswerMarks
Substitute \(e\left(= \frac{1}{9}\right): v = \frac{u}{3}, w = \frac{4}{9}u\)M1
CLM & Impact equations: \(3m \times w = 3mV + 4mW\) (\(\frac{4}{3}u = 3V + 4W\))M1
\(w f = W - V\) (\(\frac{16u}{9}f = 4W - 4V\))A1
Follow their w
Solve for \(V\): \(V = \frac{4u}{21} - \frac{16uf}{63}\)M1
N.B. answer must be positiveA1
since \(f \geq 0\), \(V \leq \frac{4u}{21} < \frac{u}{3}\) and hence A collides with BM1 A1
7b(ii).
\((ii)\)
Fourth M1 (only available if they have a correct expression for \(V\)) for a complete strategy to show that there is a 2\(^{nd}\) collision between A and B, using \(f \geq 0\) (Allow M1 A0 if they use \(f = 0\), see above)
Third A1 for a fully correct justification. N.B. GIVEN ANSWER
(ii) Alternative:
Fourth M1 (only available if they have a correct expression for \(V\)) for a complete strategy to show that there is a 2\(^{nd}\) collision between A and B:
2\(^{nd}\) collision if \(V < \frac{u}{3}\)
i.e. if \(\frac{4u}{21} - \frac{16uf}{63} < \frac{u}{3}\)
i.e. if \(\frac{-9}{16} < f\).
Since \(f \geq 0\) (given), there will be a 2\(^{nd}\) collision.
Third A1 for a fully correct justification. N.B. GIVEN ANSWER.
Notes and General Information:
- Follow through (ft) marks are indicated where appropriate
- DM indicates a dependent mark that requires a previous mark to have been earned
- A marks can be 'follow through' (A ft) when they depend on previous working
- CAO indicates "correct answer only"
- CSO indicates "correct solution only" with no errors
- Allow indicates acceptable variations
## 7a.
CLM: $2mu = 2mv + 3mw$ | M1 A1 |
Impact law: $eu = w - v$ | M1 A1 |
Eliminate w: $5v = u(2 - 3e)$ oe | M1 |
$v = \frac{u}{3}$ | M1 |
$e = \frac{1}{9}, 0.11$ or better | A1 |

**OR:** $v = \frac{u}{3}$ is used in CLM and NIL equations (See Alternative below) | |

**ALTERNATIVE:**
$v = \frac{u}{3}$ | M1 (KE) |
$2mu = 2mv + 3mw$ ($2mu = 2m\frac{u}{3} + 3mw$) | M1 A1 (CLM) |
$\Rightarrow w = \frac{4u}{9}$ | M1 (eliminate w) |
$e = \frac{w - v}{u} = \left(e = \frac{\frac{4u}{9} - \frac{u}{3}}{u}\right)$ | M1 A1 (Impact Law) |
$e = \frac{1}{9}$ | A1 |

## 7b(i).
Substitute $e\left(= \frac{1}{9}\right): v = \frac{u}{3}, w = \frac{4}{9}u$ | M1 |
CLM & Impact equations: $3m \times w = 3mV + 4mW$ ($\frac{4}{3}u = 3V + 4W$) | M1 |
$w f = W - V$ ($\frac{16u}{9}f = 4W - 4V$) | A1 |
Follow their w | |
Solve for $V$: $V = \frac{4u}{21} - \frac{16uf}{63}$ | M1 |
N.B. answer must be positive | A1 |

**since $f \geq 0$, $V \leq \frac{4u}{21} < \frac{u}{3}$ and hence A collides with B** | M1 A1 |

## 7b(ii).
$(ii)$
Fourth M1 (only available if they have a correct expression for $V$) for a complete strategy to show that there is a 2$^{nd}$ collision between A and B, using $f \geq 0$ (Allow M1 A0 if they use $f = 0$, see above)
Third A1 for a fully correct justification. N.B. GIVEN ANSWER

**(ii) Alternative:**
Fourth M1 (only available if they have a correct expression for $V$) for a complete strategy to show that there is a 2$^{nd}$ collision between A and B:

2$^{nd}$ collision if $V < \frac{u}{3}$

i.e. if $\frac{4u}{21} - \frac{16uf}{63} < \frac{u}{3}$

i.e. if $\frac{-9}{16} < f$.

Since $f \geq 0$ (given), there will be a 2$^{nd}$ collision.
Third A1 for a fully correct justification. N.B. GIVEN ANSWER.

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# Notes and General Information:

- Follow through (ft) marks are indicated where appropriate
- DM indicates a dependent mark that requires a previous mark to have been earned
- A marks can be 'follow through' (A ft) when they depend on previous working
- CAO indicates "correct answer only"
- CSO indicates "correct solution only" with no errors
- Allow indicates acceptable variations
7. Three particles $A$, $B$ and $C$ have masses $2 m , 3 m$ and $4 m$ respectively. The particles lie at rest in a straight line on a smooth horizontal surface, with $B$ between $A$ and $C$. Particle $A$ is projected towards $B$ with speed $u$ and collides directly with $B$. The coefficient of restitution between $A$ and $B$ is $e$. The kinetic energy of $A$ immediately after the collision is one ninth of the kinetic energy of $A$ immediately before the collision.

Given that the direction of motion of $A$ is unchanged by the collision,
\begin{enumerate}[label=(\alph*)]
\item find the value of $e$.

After the collision between $A$ and $B$ there is a direct collision between $B$ and $C$. The coefficient of restitution between $B$ and $C$ is $f$, where $f < \frac { 3 } { 4 }$. The speed of $B$ immediately after the collision with $C$ is $V$.
\item \begin{enumerate}[label=(\roman*)]
\item Express $V$ in terms of $f$ and $u$.
\item Hence show that there will be a second collision between $A$ and $B$.\\

\begin{center}

\end{center}
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{Edexcel M2 2018 Q7 [14]}}