| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Year | 2024 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Oblique collision, direction deflected given angle |
| Difficulty | Challenging +1.8 This is a challenging Further Mechanics oblique collision problem requiring systematic application of momentum conservation (in two perpendicular directions) and the restitution equation, followed by two inequality proofs involving physical constraints. While the techniques are standard for FM1, the three-part structure with 'show that' proofs, the perpendicularity condition creating a constraint, and the need to derive inequalities from physical requirements (e > 0 and kinetic energy considerations) make this significantly harder than routine collision exercises. |
| Spec | 6.03c Momentum in 2D: vector form6.03d Conservation in 2D: vector momentum6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| CLM along line of centres | M1 | 3.1b – Use CLM with mass and velocities paired correctly. Correct no. of terms, condone sin/cos confusion and sign errors on velocities. Allow consistent missing \(m\)'s and/or consistent extra \(g\)'s. |
| \(mU\cos\alpha = -mv_A + 3mv_B\) or \(mU\cos\alpha = -mV\sin\alpha + 3mv_B\) | A1 | 1.1b – Correct unsimplified equation |
| Impact Law used along line of centres | M1 | 3.3 – Correct no. of terms, condone sin/cos confusion and sign errors but \(e\) must be on correct side of equation |
| \(eU\cos\alpha = v_A + v_B\) or \(eU\cos\alpha = V\sin\alpha + v_B\) | A1 | 1.1b – Correct unsimplified equation (signs consistent with CLM equation) |
| Solve for \(v_B\) | DM1 | 2.1 – Dependent on both previous M's |
| \(v_B = \frac{1}{4}(1+e)U\cos\alpha\) | A1* | 2.2a – Given answer correctly obtained and exactly as printed. Working should include an equation in \(v_B\) only before reaching given answer. |
| (6 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Unsimplified expression for \(v_A\) or \(V\sin\alpha\) seen | M1 | 2.1 – Use given expression for \(v_B\) to find unsimplified expression for \(v_A\) or \(V\sin\alpha\). If seen in (a) must be used in (b) to achieve the mark. |
| \(= \frac{1}{4}(3e-1)U\cos\alpha\) oe seen | A1 | 1.1b – Correct unsimplified expression for \(v_A\) or \(V\sin\alpha\) seen |
| Solve \(v_A > 0\) to find an inequality for \(e\), or solve \(V\sin\alpha > 0\) to find an inequality for \(e\) | M1 | 3.1b |
| Correct and complete reasoning leading to \(e > \frac{1}{3}\). Must include: correct inequality for \(v_A\), \(0° < \alpha < 90°\) or \(\alpha\) is acute, \(\cos\alpha > 0\) | A1* | 2.2a – Given answer complete and correctly obtained with no incorrect statements. Complete explanation must include: correct inequality for \(v_A\); \(0°<\alpha<90°\) or \(\alpha\) is acute; \(\cos\alpha > 0\); leading to \(e > \frac{1}{3}\) |
| (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Velocity component of \(A\), perpendicular to line of centres, after collision \(= U\sin\alpha\) | B1 | 3.3 – May be seen labelled on diagram or written as an expression early in working. Does not need to be used here. |
| Use of 90° deflection: \(\tan\alpha = \dfrac{\frac{1}{4}(3e-1)U\cos\alpha}{U\sin\alpha}\) | M1 | 2.1 – Use of perpendicular deflection to form an equation in \(\alpha\) and \(e\) (and \(U\)). Allow reciprocal. Alternative methods: scalar product of vectors \(\binom{U\cos\alpha}{U\sin\alpha}\cdot\binom{-v_A}{U\sin\alpha}=0\) with \(v_A\) substituted; or \(\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\) with \(\alpha+\beta=90°\Rightarrow 1-\tan\alpha\tan\beta=0\Rightarrow\cdots\) |
| \(\tan^2\alpha = \frac{1}{4}(3e-1)\) oe | A1 | 1.1b – Correct equation in \(e\) and \(\tan\alpha\) only, e.g. \(e = \dfrac{4\tan^2\alpha+1}{3}\) oe |
| \(e \leq 1 \Rightarrow \tan^2\alpha \leq \frac{1}{2}\) | DM1 | 3.1b – Dependent on previous M. Clear explanation using \(e\leq 1\) to form an inequality in \(\tan\alpha\). Condone use of max \(e=1\Rightarrow\) max \(\tan^2\alpha = \ldots\) |
| \(0 < \tan\alpha \leq \dfrac{1}{\sqrt{2}}\) since \(0° < \alpha \leq 90°\) | A1* | 2.2a – Given answer obtained from correct and complete working with no incorrect statements. Complete explanation must justify both sides of inequality using both \(e\leq 1\) and \(0°<\alpha<90°\) (allow \(\alpha\) is acute) |
| (5 marks) | ||
| Total: 15 marks |
## Question 7:
### Part 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| CLM along line of centres | M1 | 3.1b – Use CLM with mass and velocities paired correctly. Correct no. of terms, condone sin/cos confusion and sign errors on velocities. Allow consistent missing $m$'s and/or consistent extra $g$'s. |
| $mU\cos\alpha = -mv_A + 3mv_B$ or $mU\cos\alpha = -mV\sin\alpha + 3mv_B$ | A1 | 1.1b – Correct unsimplified equation |
| Impact Law used along line of centres | M1 | 3.3 – Correct no. of terms, condone sin/cos confusion and sign errors but $e$ must be on correct side of equation |
| $eU\cos\alpha = v_A + v_B$ or $eU\cos\alpha = V\sin\alpha + v_B$ | A1 | 1.1b – Correct unsimplified equation (signs consistent with CLM equation) |
| Solve for $v_B$ | DM1 | 2.1 – Dependent on both previous M's |
| $v_B = \frac{1}{4}(1+e)U\cos\alpha$ | A1* | 2.2a – Given answer correctly obtained and exactly as printed. Working should include an equation in $v_B$ only before reaching given answer. |
| **(6 marks)** | | |
### Part 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Unsimplified expression for $v_A$ or $V\sin\alpha$ seen | M1 | 2.1 – Use given expression for $v_B$ to find unsimplified expression for $v_A$ or $V\sin\alpha$. If seen in (a) must be used in (b) to achieve the mark. |
| $= \frac{1}{4}(3e-1)U\cos\alpha$ oe seen | A1 | 1.1b – Correct unsimplified expression for $v_A$ or $V\sin\alpha$ seen |
| Solve $v_A > 0$ to find an inequality for $e$, or solve $V\sin\alpha > 0$ to find an inequality for $e$ | M1 | 3.1b |
| Correct and complete reasoning leading to $e > \frac{1}{3}$. Must include: correct inequality for $v_A$, $0° < \alpha < 90°$ or $\alpha$ is acute, $\cos\alpha > 0$ | A1* | 2.2a – Given answer complete and correctly obtained with no incorrect statements. Complete explanation must include: correct inequality for $v_A$; $0°<\alpha<90°$ or $\alpha$ is acute; $\cos\alpha > 0$; leading to $e > \frac{1}{3}$ |
| **(4 marks)** | | |
### Part 7(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Velocity component of $A$, perpendicular to line of centres, after collision $= U\sin\alpha$ | B1 | 3.3 – May be seen labelled on diagram or written as an expression early in working. Does not need to be used here. |
| Use of 90° deflection: $\tan\alpha = \dfrac{\frac{1}{4}(3e-1)U\cos\alpha}{U\sin\alpha}$ | M1 | 2.1 – Use of perpendicular deflection to form an equation in $\alpha$ and $e$ (and $U$). Allow reciprocal. Alternative methods: scalar product of vectors $\binom{U\cos\alpha}{U\sin\alpha}\cdot\binom{-v_A}{U\sin\alpha}=0$ with $v_A$ substituted; or $\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$ with $\alpha+\beta=90°\Rightarrow 1-\tan\alpha\tan\beta=0\Rightarrow\cdots$ |
| $\tan^2\alpha = \frac{1}{4}(3e-1)$ oe | A1 | 1.1b – Correct equation in $e$ and $\tan\alpha$ only, e.g. $e = \dfrac{4\tan^2\alpha+1}{3}$ oe |
| $e \leq 1 \Rightarrow \tan^2\alpha \leq \frac{1}{2}$ | DM1 | 3.1b – Dependent on previous M. Clear explanation using $e\leq 1$ to form an inequality in $\tan\alpha$. Condone use of max $e=1\Rightarrow$ max $\tan^2\alpha = \ldots$ |
| $0 < \tan\alpha \leq \dfrac{1}{\sqrt{2}}$ since $0° < \alpha \leq 90°$ | A1* | 2.2a – Given answer obtained from correct and complete working with no incorrect statements. Complete explanation must justify both sides of inequality using both $e\leq 1$ and $0°<\alpha<90°$ (allow $\alpha$ is acute) |
| **(5 marks)** | | |
| **Total: 15 marks** | | |7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{58a33c19-77c6-4b08-ac09-ce6aa1e641df-20_501_703_251_680}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A smooth uniform sphere $A$ of mass $m$ is moving with speed $U$ on a smooth horizontal plane. The sphere $A$ collides obliquely with a smooth uniform sphere $B$ of mass $3 m$ which is at rest on the plane. The two spheres have the same radius.
Immediately before the collision, the direction of motion of $A$ makes an angle $\alpha$, where $0 ^ { \circ } < \alpha < 90 ^ { \circ }$, with the line joining the centres of the spheres.
Immediately after the collision, the direction of motion of $A$ is perpendicular to its original direction, as shown in Figure 1.
The coefficient of restitution between the spheres is $e$.
\begin{enumerate}[label=(\alph*)]
\item Show that the speed of $B$ immediately after the collision is
$$\frac { 1 } { 4 } ( 1 + e ) U \cos \alpha$$
\item Show that $e > \frac { 1 } { 3 }$
\item Show that $0 < \tan \alpha \leqslant \frac { 1 } { \sqrt { 2 } }$
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM1 2024 Q7 [15]}}