AQA M3 2006 June — Question 2 12 marks

Exam BoardAQA
ModuleM3 (Mechanics 3)
Year2006
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicOblique and successive collisions
TypeSuccessive collisions, three particles in line
DifficultyStandard +0.3 This is a standard two-collision momentum problem requiring application of conservation of momentum and Newton's restitution law twice. Part (a) is routine bookwork with equal masses, and part (b) involves algebraic manipulation to find e from given information. The setup is straightforward with no geometric complications or novel insights required.
Spec6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact

2 Three smooth spheres \(A , B\) and \(C\) of equal radii and masses \(m , m\) and \(2 m\) respectively lie at rest on a smooth horizontal table. The centres of the spheres lie in a straight line with \(B\) between \(A\) and \(C\). The coefficient of restitution between any two spheres is \(e\). The sphere \(A\) is projected directly towards \(B\) with speed \(u\) and collides with \(B\).
  1. Find, in terms of \(u\) and \(e\), the speed of \(B\) immediately after the impact between \(A\) and \(B\).
  2. The sphere \(B\) subsequently collides with \(C\). The speed of \(C\) immediately after this collision is \(\frac { 3 } { 8 } u\). Find the value of \(e\).

Part (a)
AnswerMarks Guidance
Conservation of momentum: \(mu = mv_A + mv_B\)M1
\(u = v_A + v_B\)A1
Restitution: \(eu = v_B - v_A\)M1A1 OE
\(v_B = \frac{1}{2}u(1 + e)\)A1F OE
5
Part (b)
AnswerMarks Guidance
\(mv_B = mw_B + 2m\frac{3u}{8}\)M1A1 OE
\(ev_B = \frac{3u}{8} - w_B\)M1A1 OE
Elimination of \(w_B\)m1 Dependent on both M1s
\(4e^2 + 8e - 5 = 0\)A1F Simplified quadratic equation in \(e\) only
\(e = \frac{1}{2}\)A1F Stated as the only value \((0 < e < 1\) for follow through\()\)
7
12Total
### Part (a)
Conservation of momentum: $mu = mv_A + mv_B$ | M1 |
$u = v_A + v_B$ | A1 |
Restitution: $eu = v_B - v_A$ | M1A1 | OE
$v_B = \frac{1}{2}u(1 + e)$ | A1F | OE
| | **5** |

### Part (b)
$mv_B = mw_B + 2m\frac{3u}{8}$ | M1A1 | OE
$ev_B = \frac{3u}{8} - w_B$ | M1A1 | OE
Elimination of $w_B$ | m1 | Dependent on both M1s
$4e^2 + 8e - 5 = 0$ | A1F | Simplified quadratic equation in $e$ only
$e = \frac{1}{2}$ | A1F | Stated as the only value $(0 < e < 1$ for follow through$)$
| | **7** |
| | **12** | **Total** |
2 Three smooth spheres $A , B$ and $C$ of equal radii and masses $m , m$ and $2 m$ respectively lie at rest on a smooth horizontal table. The centres of the spheres lie in a straight line with $B$ between $A$ and $C$. The coefficient of restitution between any two spheres is $e$.

The sphere $A$ is projected directly towards $B$ with speed $u$ and collides with $B$.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $u$ and $e$, the speed of $B$ immediately after the impact between $A$ and $B$.
\item The sphere $B$ subsequently collides with $C$. The speed of $C$ immediately after this collision is $\frac { 3 } { 8 } u$. Find the value of $e$.
\end{enumerate}

\hfill \mbox{\textit{AQA M3 2006 Q2 [12]}}