| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Sphere rebounds off fixed wall obliquely |
| Difficulty | Standard +0.8 This is a Further Maths mechanics question requiring decomposition of velocity into components parallel and perpendicular to an oblique wall, application of coefficient of restitution to the normal component only, and impulse calculation. While methodical, it requires careful vector resolution and understanding that restitution applies only to the normal component—more sophisticated than standard A-level mechanics but a standard FM1 exercise. |
| Spec | 1.10d Vector operations: addition and scalar multiplication6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Component parallel to wall: \(\left[\frac{1}{\sqrt{2}}(\mathbf{i}+\mathbf{j}).(8\mathbf{i}+2\mathbf{j})\right] = 5\sqrt{2}\) | M1, A1 | Use of scalar product or equivalent. Allow M1 if not using unit vector. Correct unsimplified expression for component parallel to wall |
| Component perpendicular to wall after impact: \(\frac{1}{3}\left[\frac{1}{\sqrt{2}}(-\mathbf{i}+\mathbf{j}).(8\mathbf{i}+2\mathbf{j})\right] = -\sqrt{2}\) | M1, A1 | Correct use of impact law perpendicular to wall. Condone sign error. Correct unsimplified expression for component perpendicular to wall |
| \(\Rightarrow \mathbf{v} = (5\mathbf{i}+5\mathbf{j})+(-\mathbf{i}+\mathbf{j}) = (4\mathbf{i}+6\mathbf{j})\) | M1, A1* | Complete method to solve for v. Obtain given result from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| If \(\mathbf{v} = a\mathbf{i}+b\mathbf{j}\): \((8\mathbf{i}+2\mathbf{j}).(\mathbf{i}+\mathbf{j}) = (a\mathbf{i}+b\mathbf{j}).(\mathbf{i}+\mathbf{j})\), \((a+b=10)\) | M1, A1 | |
| \(-\frac{1}{3}(8\mathbf{i}+2\mathbf{j}).(-\mathbf{i}+\mathbf{j}) = (a\mathbf{i}+b\mathbf{j}).(-\mathbf{i}+\mathbf{j})\), \((2=-a+b)\) | M1, A1 | |
| \(\Rightarrow \mathbf{v} = (4\mathbf{i}+6\mathbf{j})\) | M1, A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\mathbf{I} = 0.25(4\mathbf{i}+6\mathbf{j}) - 0.25(8\mathbf{i}+2\mathbf{j})\) \(\left(\mathbf{I} = (-\mathbf{i}+\mathbf{j})\right)\) | M1 | Use of \(\mathbf{I} = m\mathbf{v} - m\mathbf{u}\) with velocities or perpendicular components of velocities. Must be subtracting but allow subtraction in either order |
| Use of Pythagoras | M1 | Correct use of Pythagoras to find modulus |
| \( | \mathbf{I} | = \sqrt{2}\) (Ns) |
## Question 7(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Component parallel to wall: $\left[\frac{1}{\sqrt{2}}(\mathbf{i}+\mathbf{j}).(8\mathbf{i}+2\mathbf{j})\right] = 5\sqrt{2}$ | M1, A1 | Use of scalar product or equivalent. Allow M1 if not using unit vector. Correct unsimplified expression for component parallel to wall |
| Component perpendicular to wall after impact: $\frac{1}{3}\left[\frac{1}{\sqrt{2}}(-\mathbf{i}+\mathbf{j}).(8\mathbf{i}+2\mathbf{j})\right] = -\sqrt{2}$ | M1, A1 | Correct use of impact law perpendicular to wall. Condone sign error. Correct unsimplified expression for component perpendicular to wall |
| $\Rightarrow \mathbf{v} = (5\mathbf{i}+5\mathbf{j})+(-\mathbf{i}+\mathbf{j}) = (4\mathbf{i}+6\mathbf{j})$ | M1, A1* | Complete method to solve for **v**. Obtain given result from correct working |
**Alternative method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| If $\mathbf{v} = a\mathbf{i}+b\mathbf{j}$: $(8\mathbf{i}+2\mathbf{j}).(\mathbf{i}+\mathbf{j}) = (a\mathbf{i}+b\mathbf{j}).(\mathbf{i}+\mathbf{j})$, $(a+b=10)$ | M1, A1 | |
| $-\frac{1}{3}(8\mathbf{i}+2\mathbf{j}).(-\mathbf{i}+\mathbf{j}) = (a\mathbf{i}+b\mathbf{j}).(-\mathbf{i}+\mathbf{j})$, $(2=-a+b)$ | M1, A1 | |
| $\Rightarrow \mathbf{v} = (4\mathbf{i}+6\mathbf{j})$ | M1, A1* | |
## Question 7(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\mathbf{I} = 0.25(4\mathbf{i}+6\mathbf{j}) - 0.25(8\mathbf{i}+2\mathbf{j})$ $\left(\mathbf{I} = (-\mathbf{i}+\mathbf{j})\right)$ | M1 | Use of $\mathbf{I} = m\mathbf{v} - m\mathbf{u}$ with velocities or perpendicular components of velocities. Must be subtracting but allow subtraction in either order |
| Use of Pythagoras | M1 | Correct use of Pythagoras to find modulus |
| $|\mathbf{I}| = \sqrt{2}$ (Ns) | A1 | Accept 1.4 Ns or better |
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\item \hspace{0pt} [In this question, $\mathbf { i }$ and $\mathbf { j }$ are perpendicular unit vectors in a horizontal plane.]
\end{enumerate}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{7f077b82-6b39-4cb5-8574-bfa308c88df3-24_543_789_294_639}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 represents the plan view of part of a smooth horizontal floor, where $A B$ is a fixed smooth vertical wall.
The direction of $\overrightarrow { A B }$ is in the direction of the vector $( \mathbf { i } + \mathbf { j } )$\\
A small ball of mass 0.25 kg is moving on the floor when it strikes the wall $A B$.\\
Immediately before its impact with the wall $A B$, the velocity of the ball is $( 8 \mathbf { i } + 2 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$\\
Immediately after its impact with the wall $A B$, the velocity of the ball is $\mathbf { v m s } ^ { - 1 }$\\
The coefficient of restitution between the ball and the wall is $\frac { 1 } { 3 }$\\
By modelling the ball as a particle,\\
(a) show that $\mathbf { v } = 4 \mathbf { i } + 6 \mathbf { j }$\\
(b) Find the magnitude of the impulse received by the ball in the impact.
\hfill \mbox{\textit{Edexcel FM1 2021 Q7 [9]}}