| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Oblique collision, find velocities/angles |
| Difficulty | Challenging +1.2 This is a standard Further Mechanics 1 oblique collision problem requiring conservation of momentum (parallel and perpendicular to line of centres) and Newton's restitution law. Part (a) is a guided 'show that' with straightforward algebra, and part (b) follows directly by substituting m=eM. While it requires multiple equations and careful component resolution, it follows a well-established method taught explicitly in FM1 with no novel insight required. |
| 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 |
| \(U\sin\alpha\) seen as velocity component of \(S\), perpendicular to line of centres after impact | B1 | \(U\sin\alpha\) or \(U\cos(90-\alpha)\) must be seen in working or velocity diagram |
| CLM along line of centres | M1 | Dimensionally correct, correct no. of terms, condone sin/cos confusion and sign errors |
| \(mU\cos\alpha = mv_1 + Mv_2\) | A1 | Correct equation |
| NEL used along line of centres | M1 | \(e\) appearing on correct side; condone sin/cos confusion if consistent with CLM; must have correct number of terms |
| \(eU\cos\alpha = -v_1 + v_2\) | A1 | Signs of \(v_1\) and \(v_2\) consistent with CLM |
| \(\tan\beta = \frac{U\sin\alpha}{v_1}\) | dM1 | Use of \(S\) moving at \(\beta\) to line of centres; dependent on both previous M marks |
| Eliminate \(v_1\); note \(v_1 = u\cos\alpha\left(\frac{m-eM}{m+M}\right)\) | dM1 | Dependent on first two M marks |
| \(\tan\beta = \frac{(m+M)\tan\alpha}{(m-eM)}\) | A1* | Must match printed answer exactly |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(m = eM \Rightarrow \tan\beta = \infty\) or \(\tan\beta\) undefined so \(\beta = 90°\); or \(m = eM \Rightarrow v_1 = 0\) so velocity component of \(S\) parallel to line of centres is zero | M1 | Use of given condition to deduce \(\beta = 90°\) or velocity component parallel to line of centres is zero |
| After collision, \(S\) moves perpendicular to line of centres and other sphere moves parallel to line of centres, i.e. they move at right angles | A1* | Must refer correctly to direction of both particles; do not accept "horizontally and vertically" since surface is defined as horizontal |
# Question 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $U\sin\alpha$ seen as velocity component of $S$, perpendicular to line of centres after impact | B1 | $U\sin\alpha$ or $U\cos(90-\alpha)$ must be seen in working or velocity diagram |
| CLM along line of centres | M1 | Dimensionally correct, correct no. of terms, condone sin/cos confusion and sign errors |
| $mU\cos\alpha = mv_1 + Mv_2$ | A1 | Correct equation |
| NEL used along line of centres | M1 | $e$ appearing on correct side; condone sin/cos confusion if consistent with CLM; must have correct number of terms |
| $eU\cos\alpha = -v_1 + v_2$ | A1 | Signs of $v_1$ and $v_2$ consistent with CLM |
| $\tan\beta = \frac{U\sin\alpha}{v_1}$ | dM1 | Use of $S$ moving at $\beta$ to line of centres; dependent on both previous M marks |
| Eliminate $v_1$; note $v_1 = u\cos\alpha\left(\frac{m-eM}{m+M}\right)$ | dM1 | Dependent on first two M marks |
| $\tan\beta = \frac{(m+M)\tan\alpha}{(m-eM)}$ | A1* | Must match printed answer exactly |
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# Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $m = eM \Rightarrow \tan\beta = \infty$ or $\tan\beta$ undefined so $\beta = 90°$; or $m = eM \Rightarrow v_1 = 0$ so velocity component of $S$ parallel to line of centres is zero | M1 | Use of given condition to deduce $\beta = 90°$ or velocity component parallel to line of centres is zero |
| After collision, $S$ moves perpendicular to line of centres and other sphere moves parallel to line of centres, i.e. they move at right angles | A1* | Must refer correctly to direction of both particles; do not accept "horizontally and vertically" since surface is defined as horizontal |5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0da9cd5b-6f6f-4607-bd4f-c8ae164466ae-16_758_1399_280_333}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A smooth uniform sphere $S$ of mass $m$ is moving with speed $U$ on a smooth horizontal plane. The sphere $S$ collides obliquely with another uniform sphere of mass $M$ which is at rest on the plane. The two spheres have the same radius.
Immediately before the collision the direction of motion of $S$ makes an angle $\alpha$, where $0 < \alpha < 90 ^ { \circ }$, with the line joining the centres of the spheres.
Immediately after the collision the direction of motion of $S$ makes an angle $\beta$ with the line joining the centres of the spheres, as shown in Figure 1.
The coefficient of restitution between the spheres is $e$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\tan \beta = \frac { ( m + M ) \tan \alpha } { ( m - e M ) }$
Given that $m = e M$,
\item show that the directions of motion of the two spheres immediately after the collision are perpendicular.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM1 2023 Q5 [10]}}