| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Sphere rebounds off fixed wall obliquely |
| Difficulty | Standard +0.3 This is a standard Further Mechanics 1 collision problem requiring application of the coefficient of restitution formula in vector form. Part (a) involves identifying the wall direction and applying e = (separation speed)/(approach speed) for the perpendicular component. Part (b) requires finding velocity components after the second collision and calculating the deflection angle using trigonometry. While it involves two collisions and vector decomposition, these are routine FM1 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Complete method to find direction of normal (perpendicular to wall): \((\pm m)(({\bf i}+3{\bf j})-(4{\bf i}-{\bf j}))\) | M1 | Complete method to find direction of normal. Eg find direction of wall first by equating parallel components then find perpendicular direction using scalar product or otherwise. Eg \((4{\bf i}-{\bf j})\cdot(x{\bf i}+y{\bf j}) = ({\bf i}+3{\bf j})\cdot(x{\bf i}+y{\bf j}) \Rightarrow 3x=4y\). Wall direction \(\parallel (4{\bf i}+3{\bf j}) \Rightarrow\) Impulse direction \(\parallel (-3{\bf i}+4{\bf j})\) |
| \((-3{\bf i}+4{\bf j})\) or any parallel vector | A1 | Any correct vector parallel to impulse \((-3{\bf i}+4{\bf j})\) |
| Resolve both velocities parallel to impulse | M1 | Complete method to resolve both velocities parallel to impulse. Using scalar product (main scheme) or using trig from diagram. \(m\) is not required. |
| Approach: \((4{\bf i}-{\bf j})\cdot\left(\frac{1}{5}\right)(-3{\bf i}+4{\bf j}) = -\frac{16}{5}\) N.B. \(\frac{1}{5}\) is not required | A1 | Both correct velocity components parallel to impulse, allow positive or negative values in both cases. |
| Separation: \(({\bf i}+3{\bf j})\cdot\left(\frac{1}{5}\right)(-3{\bf i}+4{\bf j}) = \frac{9}{5}\) N.B. \(\frac{1}{5}\) is not required | ||
| \(e =\) separation speed \(\div\) approach speed | DM1 | Dependent on previous M. Use of Impact Law parallel to impulse. Ratio of speeds correct way up (separation \(\div\) approach). |
| \(e = \frac{9}{16}\) (0.56 or better) | A1 | Correct answer (0.56 or better) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Complete method to find both components of rebound velocity: \(\parallel\) to wall (i), unchanged; Perp (j), \(\frac{1}{3}\times 3\) | M1 | Method to find velocity after impact with second wall: i component unchanged, j component: \(\frac{1}{3}(3)\). Condone sign errors. |
| Both velocity components correct, may be given on diagram. \({\bf w} = ({\bf i}-{\bf j})\) | A1 | Correct velocity components after impact, may be given if seen on a diagram. |
| Complete method to find angle of deflection \(\alpha°\): e.g. \(\alpha° = \theta + \phi = \tan^{-1}\left(\frac{3}{1}\right) + \tan^{-1}\left(\frac{1}{1}\right)\) or using scalar product \(\alpha° = \cos^{-1}\left(\frac{({\bf i}+3{\bf j})\cdot({\bf i}-{\bf j})}{\sqrt{1^2+3^2}\sqrt{1^2+1^2}}\right) = \cos^{-1}\left(\frac{-2}{\sqrt{10}\sqrt{2}}\right)\) | DM1 | Complete method to find angle of deflection for velocity after second impact. Dependent on previous M. |
| \(\alpha = 117°\) (nearest whole number) | A1 | Correct answer, must be rounded to nearest degree. |
# Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete method to find direction of normal (perpendicular to wall): $(\pm m)(({\bf i}+3{\bf j})-(4{\bf i}-{\bf j}))$ | M1 | Complete method to find direction of normal. Eg find direction of wall first by equating parallel components then find perpendicular direction using scalar product or otherwise. Eg $(4{\bf i}-{\bf j})\cdot(x{\bf i}+y{\bf j}) = ({\bf i}+3{\bf j})\cdot(x{\bf i}+y{\bf j}) \Rightarrow 3x=4y$. Wall direction $\parallel (4{\bf i}+3{\bf j}) \Rightarrow$ Impulse direction $\parallel (-3{\bf i}+4{\bf j})$ |
| $(-3{\bf i}+4{\bf j})$ or any parallel vector | A1 | Any correct vector parallel to impulse $(-3{\bf i}+4{\bf j})$ |
| Resolve both velocities parallel to impulse | M1 | Complete method to resolve both velocities parallel to impulse. Using scalar product (main scheme) or using trig from diagram. $m$ is not required. |
| Approach: $(4{\bf i}-{\bf j})\cdot\left(\frac{1}{5}\right)(-3{\bf i}+4{\bf j}) = -\frac{16}{5}$ **N.B.** $\frac{1}{5}$ is not required | A1 | Both correct velocity components parallel to impulse, allow positive or negative values in both cases. |
| Separation: $({\bf i}+3{\bf j})\cdot\left(\frac{1}{5}\right)(-3{\bf i}+4{\bf j}) = \frac{9}{5}$ **N.B.** $\frac{1}{5}$ is not required | | |
| $e =$ separation speed $\div$ approach speed | DM1 | Dependent on previous M. Use of Impact Law parallel to impulse. Ratio of speeds correct way up (separation $\div$ approach). |
| $e = \frac{9}{16}$ (0.56 or better) | A1 | Correct answer (0.56 or better) |
---
# Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete method to find **both** components of rebound velocity: $\parallel$ to wall (i), unchanged; Perp (j), $\frac{1}{3}\times 3$ | M1 | Method to find velocity after impact with second wall: **i** component unchanged, **j** component: $\frac{1}{3}(3)$. Condone sign errors. |
| Both velocity components correct, may be given on diagram. ${\bf w} = ({\bf i}-{\bf j})$ | A1 | Correct velocity components after impact, may be given if seen on a diagram. |
| Complete method to find angle of deflection $\alpha°$: e.g. $\alpha° = \theta + \phi = \tan^{-1}\left(\frac{3}{1}\right) + \tan^{-1}\left(\frac{1}{1}\right)$ or using scalar product $\alpha° = \cos^{-1}\left(\frac{({\bf i}+3{\bf j})\cdot({\bf i}-{\bf j})}{\sqrt{1^2+3^2}\sqrt{1^2+1^2}}\right) = \cos^{-1}\left(\frac{-2}{\sqrt{10}\sqrt{2}}\right)$ | DM1 | Complete method to find angle of deflection for velocity after second impact. Dependent on previous M. |
| $\alpha = 117°$ (nearest whole number) | A1 | Correct answer, must be rounded to nearest degree. |\begin{enumerate}
\item \hspace{0pt} [In this question, $\mathbf { i }$ and $\mathbf { j }$ are horizontal perpendicular unit vectors.]
\end{enumerate}
A particle $P$ is moving with velocity ( $4 \mathbf { i } - \mathbf { j }$ ) $\mathrm { m } \mathrm { s } ^ { - 1 }$ on a smooth horizontal plane. The particle collides with a smooth vertical wall and rebounds with velocity $( \mathbf { i } + 3 \mathbf { j } ) \mathrm { ms } ^ { - 1 }$ The coefficient of restitution between $P$ and the wall is $e$.\\
(a) Find the value of $e$.
After the collision, $P$ goes on to hit a second smooth vertical wall, which is parallel to $\mathbf { i }$.\\
The coefficient of restitution between $P$ and this second wall is $\frac { 1 } { 3 }$\\
The angle through which the direction of motion of $P$ has been deflected by its collision with this second wall is $\alpha ^ { \circ }$.\\
(b) Find the value of $\alpha$, giving your answer to the nearest whole number.
\hfill \mbox{\textit{Edexcel FM1 2024 Q6 [10]}}