| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Oblique collision, vector velocity form |
| Difficulty | Standard +0.8 This is a standard Further Mechanics 1 oblique collision problem requiring conservation of momentum, Newton's law of restitution, and vector decomposition. While it involves multiple steps (finding velocities in both directions, applying two key principles, then calculating deflection angle), these are well-practiced techniques for FM1 students with no novel insight required. The oblique nature and vector notation add moderate complexity above basic mechanics, placing it somewhat above average difficulty. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors6.03c Momentum in 2D: vector form6.03d Conservation in 2D: vector momentum6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Overall strategy to find \(\mathbf{V}_A\) | M1 | Correct overall strategy to form sufficient equations and solve for \(\mathbf{V}_A\) |
| Velocity of \(A\) perpendicular to line of centres after collision \(= 3\mathbf{j}\ \text{(m s}^{-1}\text{)}\) | B1 | Use the model to find the component of \(\mathbf{V}_A\) perpendicular to the line of centres |
| CLM parallel to loc: \(2m\times 3 - 3m\times 5 = 3mw - 2mv\ \ (-9 = 3w - 2v)\) | M1, A1 | Use CLM to form equation in \(v\) and \(w\); need all 4 terms, dimensionally correct; correct unsimplified |
| Correct use of impact law: \(v + w = \frac{1}{4}(3+5)\ (=2)\) | M1, A1 | Must be used the right way round; correct unsimplified |
| Solve: \(3w - 2v = -9\), \(2v + 2w = 4\) \(\Rightarrow \mathbf{v}_B = -\mathbf{i} + 2\mathbf{j}\ \text{(m s}^{-1}\text{)}\) | A1ft | \(v_B\) correct; follow their \(2\mathbf{j}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\cos\theta = \dfrac{(-5\mathbf{i}+2\mathbf{j}).(-\mathbf{i}+2\mathbf{j})}{\sqrt{29}\sqrt{5}}\) | M1 | Complete method for finding the required angle; follow their \(v_B\) |
| \(\theta = 41.63...° = 42°\) (nearest degree) | A1 | cao |
| Alternative method: \(\tan^{-1}2 - \tan^{-1}\frac{2}{5} = 41.63...° = 42°\) (nearest degree) |
## Question 6:
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Overall strategy to find $\mathbf{V}_A$ | M1 | Correct overall strategy to form sufficient equations and solve for $\mathbf{V}_A$ |
| Velocity of $A$ perpendicular to line of centres after collision $= 3\mathbf{j}\ \text{(m s}^{-1}\text{)}$ | B1 | Use the model to find the component of $\mathbf{V}_A$ perpendicular to the line of centres |
| CLM parallel to loc: $2m\times 3 - 3m\times 5 = 3mw - 2mv\ \ (-9 = 3w - 2v)$ | M1, A1 | Use CLM to form equation in $v$ and $w$; need all 4 terms, dimensionally correct; correct unsimplified |
| Correct use of impact law: $v + w = \frac{1}{4}(3+5)\ (=2)$ | M1, A1 | Must be used the right way round; correct unsimplified |
| Solve: $3w - 2v = -9$, $2v + 2w = 4$ $\Rightarrow \mathbf{v}_B = -\mathbf{i} + 2\mathbf{j}\ \text{(m s}^{-1}\text{)}$ | A1ft | $v_B$ correct; follow their $2\mathbf{j}$ |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos\theta = \dfrac{(-5\mathbf{i}+2\mathbf{j}).(-\mathbf{i}+2\mathbf{j})}{\sqrt{29}\sqrt{5}}$ | M1 | Complete method for finding the required angle; follow their $v_B$ |
| $\theta = 41.63...° = 42°$ (nearest degree) | A1 | cao |
| Alternative method: $\tan^{-1}2 - \tan^{-1}\frac{2}{5} = 41.63...° = 42°$ (nearest degree) | | |\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 $A$ has mass $2 m \mathrm {~kg}$ and another smooth uniform sphere $B$, with the same radius as $A$, has mass $3 m \mathrm {~kg}$.
The spheres are moving on a smooth horizontal plane when they collide obliquely.\\
Immediately before the collision the velocity of $A$ is $( 3 \mathbf { i } + 3 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$ and the velocity of $B$ is $( - 5 \mathbf { i } + 2 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$.
At the instant of collision, the line joining the centres of the spheres is parallel to $\mathbf { i }$.\\
The coefficient of restitution between the spheres is $\frac { 1 } { 4 }$\\
(a) Find the velocity of $B$ immediately after the collision.\\
(b) Find, to the nearest degree, the size of the angle through which the direction of motion of $B$ is deflected as a result of the collision.
\hfill \mbox{\textit{Edexcel FM1 Q6 [9]}}