| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Year | 2021 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Oblique collision, vector velocity form |
| Difficulty | Challenging +1.2 This is a standard Further Mechanics 1 oblique collision question requiring conservation of momentum (perpendicular and parallel components), Newton's experimental law, and the key insight that perpendicular components are unchanged. While it involves multiple steps and vector components, the techniques are routine for FM1 students and the 90° deflection provides a helpful constraint that simplifies the algebra. |
| Spec | 1.10d Vector operations: addition and scalar multiplication6.03c Momentum in 2D: vector form6.03d Conservation in 2D: vector momentum6.03i Coefficient of restitution: e |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Component of \(P\) in j direction after collision \(= 2\) | B1 | Correct only. Check the diagram |
| Deflected through 90° so velocity after \(= \left(-\frac{4}{u}\mathbf{i} + 2\mathbf{j}\right)\) ms\(^{-1}\) | B1 | Correct only. Seen or implied |
| CLM parallel to line of centres: \(0.3\left(u + \frac{4}{u}\right) = 0.5(4+w)\) | M1, A1ft | Correct use of CLM. Need all terms. Condone sign errors. Follow their components of velocity of \(P\) |
| Impact law parallel to line of centres: \(w + \frac{4}{u} = \frac{3}{5}(u+4)\) | M1, A1ft | Correct use of impact law. Condone sign errors. Follow their components of velocity of \(P\) |
| \(\begin{cases} 3u + \frac{12}{u} = 20 + 5w \\ \frac{20}{u} + 5w = 3u + 12 \end{cases}\) | M1 | Solve their correctly formed simultaneous equations to obtain value of \(u\) |
| \(\Rightarrow \frac{32}{u} = 32,\quad u = 1\) | A1 | Correct only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| For \(Q\) after: \(w = -1\) | B1 | Correct only |
| \(\mathbf{v} = -\mathbf{i} + 3\mathbf{j}\) | B1ft | Follow their \(w\) |
| Find relevant angle between directions | M1 | Correct method to find a relevant angle between the directions |
| \(\alpha° = \tan^{-1}3 - \tan^{-1}\frac{3}{4}\) or \(\alpha° = \cos^{-1}\left(\frac{4+9}{5\times\sqrt{10}}\right)\) | A1ft | Correct unsimplified expression. Follow their v |
| \(\alpha = 34.7°\quad(35°)\) | A1 | 35 or better (34.695…) 0.61 radians |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| The line of centres is parallel to the surface the spheres are moving on, so the impulse acts parallel to the surface | B1 | Or equivalent that explains that the line of centres is parallel to the surface |
# Question 3:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Component of $P$ in **j** direction after collision $= 2$ | B1 | Correct only. Check the diagram |
| Deflected through 90° so velocity after $= \left(-\frac{4}{u}\mathbf{i} + 2\mathbf{j}\right)$ ms$^{-1}$ | B1 | Correct only. Seen or implied |
| CLM parallel to line of centres: $0.3\left(u + \frac{4}{u}\right) = 0.5(4+w)$ | M1, A1ft | Correct use of CLM. Need all terms. Condone sign errors. Follow their components of velocity of $P$ |
| Impact law parallel to line of centres: $w + \frac{4}{u} = \frac{3}{5}(u+4)$ | M1, A1ft | Correct use of impact law. Condone sign errors. Follow their components of velocity of $P$ |
| $\begin{cases} 3u + \frac{12}{u} = 20 + 5w \\ \frac{20}{u} + 5w = 3u + 12 \end{cases}$ | M1 | Solve their correctly formed simultaneous equations to obtain value of $u$ |
| $\Rightarrow \frac{32}{u} = 32,\quad u = 1$ | A1 | Correct only |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| For $Q$ after: $w = -1$ | B1 | Correct only |
| $\mathbf{v} = -\mathbf{i} + 3\mathbf{j}$ | B1ft | Follow their $w$ |
| Find relevant angle between directions | M1 | Correct method to find a relevant angle between the directions |
| $\alpha° = \tan^{-1}3 - \tan^{-1}\frac{3}{4}$ or $\alpha° = \cos^{-1}\left(\frac{4+9}{5\times\sqrt{10}}\right)$ | A1ft | Correct unsimplified expression. Follow their **v** |
| $\alpha = 34.7°\quad(35°)$ | A1 | 35 or better (34.695…) 0.61 radians |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| The line of centres is parallel to the surface the spheres are moving on, so the impulse acts parallel to the surface | B1 | Or equivalent that explains that the line of centres is parallel to the surface |
---\begin{enumerate}
\item \hspace{0pt} [In this question, $\mathbf { i }$ and $\mathbf { j }$ are perpendicular unit vectors in a horizontal plane.]
\end{enumerate}
A smooth uniform sphere $P$ has mass 0.3 kg . Another smooth uniform sphere $Q$, with the same radius as $P$, has mass 0.5 kg .
The spheres are moving on a smooth horizontal surface when they collide obliquely. Immediately before the collision the velocity of $P$ is $( u \mathbf { i } + 2 \mathbf { j } ) \mathrm { ms } ^ { - 1 }$, where $u$ is a positive constant, and the velocity of $Q$ is $( - 4 \mathbf { i } + 3 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$
At the instant when the spheres collide, the line joining their centres is parallel to $\mathbf { i }$.\\
The coefficient of restitution between $P$ and $Q$ is $\frac { 3 } { 5 }$\\
As a result of the collision, the direction of motion of $P$ is deflected through an angle of $90 ^ { \circ }$ and the direction of motion of $Q$ is deflected through an angle of $\alpha ^ { \circ }$\\
(a) Find the value of $u$\\
(b) Find the value of $\alpha$\\
(c) State how you have used the fact that $P$ and $Q$ have equal radii.
\hfill \mbox{\textit{Edexcel FM1 2021 Q3 [14]}}