| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Year | 2024 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Energy loss in collision |
| Difficulty | Challenging +1.2 This is a multi-part Further Mechanics collision problem requiring conservation of momentum, coefficient of restitution equations, and energy loss calculations. While it involves several steps and two collision scenarios, the techniques are standard for Further Maths mechanics: setting up simultaneous equations from momentum/restitution, then analyzing conditions for no further collisions. The algebraic manipulation is straightforward, making this moderately above average difficulty but not requiring novel insight. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt at conservation of momentum | M1 | AO 3.3; 4 terms each of correct form; condone 1 error e.g. sign error |
| Attempt at NEL | M1 | AO 1.1; allow 1 error, but not approach and separation speeds reversed |
| \(m_A u_A + 5(-2) = m_A(-3.25) + 5(0.5)\); \(\frac{0.5-(-3.25)}{u_A - (-2)} = 0.75\) | A1 | AO 1.1; both equations correct and consistent; FT their value of \(m_A\) if substituted into COLM equation |
| \(3.75 = 0.75u_A + 1.5 \Rightarrow u_A = \ldots\); \(3m_A - 10 = -3.25m_A + 2.5 \Rightarrow 6.25m_A = 12.5 \Rightarrow m_A = \ldots\) | M1 | AO 1.1; solving to find \(u_A\) or \(m_A\); from their attempt at COLM and/or NEL |
| \(u_A = 3\), \(m_A = 2\) | A1 | AO 1.1; both correct, could be BC; both correct from correct equations implies last M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(KE\ \text{Before} = \frac{1}{2} \times \text{``}2\text{''} \times \text{``}3\text{''}^2 + \frac{1}{2} \times 5 \times 2^2 = 19\) (J) | B1FT | AO 1.1; total KE before (FT their values for \(u_A\) and \(m_A\)); allow one slip |
| \(KE\ \text{After} = \frac{1}{2} \times \text{``}2\text{''} \times 3.25^2 + \frac{1}{2} \times 5 \times 0.5^2 = 11.1875\) (J) | M1 | AO 1.1; allow one slip |
| \(\% KE\ \text{Lost} = \left(1 - \frac{11.1875}{19}\right) \times 100 = 41.11\ldots \approx 41\%\) AG | A1 | AO 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(5(0.5) + 3(-5.5) = 5V_B + 3V_C\) | M1 | AO 1.1; attempt at conservation of momentum; allow one error; NB \(V_B = -3.25\) may be substituted from start |
| \(e = \frac{V_C - V_B}{0.5-(-5.5)}\) | M1 | AO 1.1; attempt at NEL; allow one error, but not approach and separation speeds reversed |
| \(V_C - V_B = 6e\) and \(5V_B + 3V_C = -14\) | A1 | AO 1.1; both equations correct and consistent; may be unsimplified |
| \(3V_C - 3V_B = 18e \Rightarrow 8V_B = -14 - 18e \Rightarrow V_B = \frac{-14-18e}{8} = \frac{-7-9e}{4}\) | M1 | AO 1.1; solving simultaneously to reach expression for \(V_B\) or \(V_C\) in terms of \(e\) only; \(V_C = \frac{-14+30e}{8} = \frac{-7+15e}{4}\); allow 1 error |
| No further collision \(\Rightarrow V_B \geq -3.25\) | B1 | AO 3.1b; correct condition for no further collision; condone \(>\), allow \(=\), but not \(<\) or \(\leq\); must be consistent with previous equations; award for substituting \(V_B = -3.25\) into both equations at start |
| \(\frac{-7-9e}{4} \geq -3.25 \Rightarrow -7 - 9e \geq -13 \Rightarrow 9e \leq 6 \Rightarrow e \leq \frac{2}{3}\) | A1 | AO 1.1; inequality signs must be correct and lower limit of 0 required |
| But \(0 \leq e \leq 1\) so \(0 \leq e \leq \frac{2}{3}\) |
# Question 5:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt at conservation of momentum | M1 | AO 3.3; 4 terms each of correct form; condone 1 error e.g. sign error |
| Attempt at NEL | M1 | AO 1.1; allow 1 error, but not approach and separation speeds reversed |
| $m_A u_A + 5(-2) = m_A(-3.25) + 5(0.5)$; $\frac{0.5-(-3.25)}{u_A - (-2)} = 0.75$ | A1 | AO 1.1; both equations correct and consistent; FT their value of $m_A$ if substituted into COLM equation |
| $3.75 = 0.75u_A + 1.5 \Rightarrow u_A = \ldots$; $3m_A - 10 = -3.25m_A + 2.5 \Rightarrow 6.25m_A = 12.5 \Rightarrow m_A = \ldots$ | M1 | AO 1.1; solving to find $u_A$ or $m_A$; from their attempt at COLM and/or NEL |
| $u_A = 3$, $m_A = 2$ | A1 | AO 1.1; both correct, could be BC; both correct from correct equations implies last M1 |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $KE\ \text{Before} = \frac{1}{2} \times \text{``}2\text{''} \times \text{``}3\text{''}^2 + \frac{1}{2} \times 5 \times 2^2 = 19$ (J) | B1FT | AO 1.1; total KE before (FT their values for $u_A$ and $m_A$); allow one slip |
| $KE\ \text{After} = \frac{1}{2} \times \text{``}2\text{''} \times 3.25^2 + \frac{1}{2} \times 5 \times 0.5^2 = 11.1875$ (J) | M1 | AO 1.1; allow one slip |
| $\% KE\ \text{Lost} = \left(1 - \frac{11.1875}{19}\right) \times 100 = 41.11\ldots \approx 41\%$ **AG** | A1 | AO 2.2a |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $5(0.5) + 3(-5.5) = 5V_B + 3V_C$ | M1 | AO 1.1; attempt at conservation of momentum; allow one error; NB $V_B = -3.25$ may be substituted from start |
| $e = \frac{V_C - V_B}{0.5-(-5.5)}$ | M1 | AO 1.1; attempt at NEL; allow one error, but not approach and separation speeds reversed |
| $V_C - V_B = 6e$ and $5V_B + 3V_C = -14$ | A1 | AO 1.1; both equations correct and consistent; may be unsimplified |
| $3V_C - 3V_B = 18e \Rightarrow 8V_B = -14 - 18e \Rightarrow V_B = \frac{-14-18e}{8} = \frac{-7-9e}{4}$ | M1 | AO 1.1; solving simultaneously to reach expression for $V_B$ or $V_C$ in terms of $e$ only; $V_C = \frac{-14+30e}{8} = \frac{-7+15e}{4}$; allow 1 error |
| No further collision $\Rightarrow V_B \geq -3.25$ | B1 | AO 3.1b; correct condition for no further collision; condone $>$, allow $=$, but not $<$ or $\leq$; must be consistent with previous equations; award for substituting $V_B = -3.25$ into both equations at start |
| $\frac{-7-9e}{4} \geq -3.25 \Rightarrow -7 - 9e \geq -13 \Rightarrow 9e \leq 6 \Rightarrow e \leq \frac{2}{3}$ | A1 | AO 1.1; inequality signs must be correct and lower limit of 0 required |
| But $0 \leq e \leq 1$ so $0 \leq e \leq \frac{2}{3}$ | | |
---5 Two particles, $A$ of mass $m _ { A } \mathrm {~kg}$ and $B$ of mass 5 kg , are moving directly towards each other on a smooth horizontal floor. Before they collide they have speeds $\mathrm { u } _ { \mathrm { A } } \mathrm { m } \mathrm { s } ^ { - 1 }$ and $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ respectively. Immediately after they collide the direction of motion of each particle has been reversed and $A$ and $B$ have speeds $3.25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $0.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ respectively (see diagram). The coefficient of restitution between $A$ and $B$ is 0.75 .
Before:\\
\includegraphics[max width=\textwidth, alt={}, center]{d2156252-71f2-4084-89a2-4d246583eb65-4_218_711_552_283}
After:\\
\includegraphics[max width=\textwidth, alt={}, center]{d2156252-71f2-4084-89a2-4d246583eb65-4_218_707_552_1078}
\begin{enumerate}[label=(\alph*)]
\item Determine the value of $m _ { A }$ and the value of $u _ { A }$.\\[0pt]
[5]
\item Show that approximately $41 \%$ of the kinetic energy of the system is lost in this collision.
After the collision between $A$ and $B$, $B$ goes on to collide directly with a third particle $C$ of mass 3 kg which is travelling towards $B$ with a speed of $5.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The coefficient of restitution between $B$ and $C$ is denoted by $e$.
\item Given that, after $B$ and $C$ collide, there are no further collisions between $A , B$ and $C$ determine the range of possible values of $e$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics AS 2024 Q5 [14]}}