| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Year | 2023 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Direct collision with given impulse |
| Difficulty | Standard +0.3 This is a straightforward application of conservation of momentum and impulse definition in a two-particle system. The setup is clear, requiring (a) a single momentum equation to find B's speed, and (b) direct use of impulse = change in momentum. While it's Further Mechanics, the question involves standard textbook methods with no conceptual subtlety or multi-step reasoning beyond basic algebra. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2.4m + 0(3m) = (m + 3m)v\) | M1 | Conservation of momentum with \(A\) and \(B\) having some post-collision velocity and initial momentum \(> 0\). Treat consistent calculation using \(m_B = m\) (possibly combined with \(m_A = 3m\)) as MR. Can be awarded if seen in (b) |
| \(4v = 2.4 \Rightarrow v = 0.6\), so speed of \(B\) is \(0.6\ \text{ms}^{-1}\) | A1 | \(m_B = m_A = m \Rightarrow v = 1.2\); \(m_B = m,\ m_A = 3m \Rightarrow v = 1.8\) |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Impulse on \(B\) = change in \(B\)'s momentum \(= 3m \times 0.6 - 0(3m)\) | M1 | Using \(I = \Delta mv\) |
| So magnitude of impulse is \(1.8m\) (Ns) | A1 | Do not allow \(-1.8m\). \(m_B = m_A = m \Rightarrow I = 1.2m\); \(m_B = m,\ m_A = 3m \Rightarrow I = 1.8m\) |
| [2] |
## Question 1:
**Part (a)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2.4m + 0(3m) = (m + 3m)v$ | M1 | Conservation of momentum with $A$ and $B$ having some post-collision velocity and initial momentum $> 0$. Treat consistent calculation using $m_B = m$ (possibly combined with $m_A = 3m$) as MR. Can be awarded if seen in (b) |
| $4v = 2.4 \Rightarrow v = 0.6$, so speed of $B$ is $0.6\ \text{ms}^{-1}$ | A1 | $m_B = m_A = m \Rightarrow v = 1.2$; $m_B = m,\ m_A = 3m \Rightarrow v = 1.8$ |
| **[2]** | | |
**Part (b)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Impulse on $B$ = change in $B$'s momentum $= 3m \times 0.6 - 0(3m)$ | M1 | Using $I = \Delta mv$ |
| So magnitude of impulse is $1.8m$ (Ns) | A1 | Do not allow $-1.8m$. $m_B = m_A = m \Rightarrow I = 1.2m$; $m_B = m,\ m_A = 3m \Rightarrow I = 1.8m$ |
| **[2]** | | |
---1 Two particles $A$, of mass $m \mathrm {~kg}$, and $B$, of mass $3 m \mathrm {~kg}$, are connected by a light inextensible string and placed together at rest on a smooth horizontal surface with the string slack. $A$ is projected along the surface, directly away from $B$, with a speed of $2.4 \mathrm {~ms} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the speed of $B$ immediately after the string becomes taut.
\item Find, in terms of $m$, the magnitude of the impulse exerted on $B$ as a result of the string becoming taut.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics AS 2023 Q1 [4]}}