| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2018 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Three-particle sequential collisions |
| Difficulty | Standard +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. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03k Newton's experimental law: direct impact |
| Answer | Marks |
|---|---|
| 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) |
| Answer | Marks |
|---|---|
| \(v = \frac{u}{3}\) | 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 |
| Answer | Marks |
|---|---|
| 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 |
## 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]}}