| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Vector impulse: find deflection angle or impulse magnitude from angle |
| Difficulty | Challenging +1.2 This is a Further Mechanics question requiring vector manipulation, impulse-momentum theorem, and geometric constraints (angle deflection). It involves setting up simultaneous equations from magnitude and angle conditions, then solving a quadratic. More challenging than standard A-level mechanics but routine for FM1 students who have practiced deflection problems. |
| Spec | 1.10d Vector operations: addition and scalar multiplication6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Momentum of \(P\) after impulse \(= a\mathbf{i}\) (or \(a\mathbf{j}\)) | B1 | Correct interpretation of angle of deflection (velocity or momentum a multiple of \(\mathbf{i}\) or \(\mathbf{j}\)) |
| Either: Use of \(\mathbf{I} = m(\mathbf{v} - \mathbf{u})\): \((\mathbf{I} =) 0.5(2a\mathbf{i} - (4\mathbf{i} + 4\mathbf{j})) (= (a-2)\mathbf{i} - 2\mathbf{j})\) | M1 | Form vector triangle or equation for \(\mathbf{v}\) or their \(a\mathbf{i}\) |
| Use of Pythagoras to form equation in \(a\) | M1 | Use trigonometry or Pythagoras' theorem to form equation in \(a\) |
| \(6.25 = 0.25\left((2a-4)^2 + 16\right)\) | A1ft | Unsimplified equation with at most one error. Follow their \(a\mathbf{i}\) |
| \(\left(4a^2 - 16a + 7 = 0\right)\) | A1 | Correct unsimplified equation |
| Or: \(\lambda^2 + \mu^2 = \frac{25}{4}\) | M1 | |
| \(\mathbf{I} = \lambda\mathbf{i} + \mu\mathbf{j} = \frac{1}{2}((x-4)\mathbf{i} - 4\mathbf{j})\) | M1 | |
| \(\mu = -2\) | A1 | Dependent on 2nd M (for impulse) |
| \(\lambda^2 = \frac{9}{4}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Use of \(\mathbf{I} = m(\mathbf{v} - \mathbf{u})\) to form vector triangle | M1 | |
| Form equation in their \(a\) | M1 | |
| \(6.25 = a^2 + 8 - 2a\sqrt{8} \times \frac{1}{\sqrt{2}}\) | A1ft | |
| \(\left(4 \times 6.25 = b^2 + 32 - 2b\sqrt{32} \times \frac{1}{\sqrt{2}} \text{ for velocity } b\mathbf{i}\right)\) | A1 | |
| \(\left(4a^2 - 16a + 7 = 0\right)\) | ||
| \(a = \frac{7}{2}, \frac{1}{2} \Rightarrow \mathbf{I} = \frac{3}{2}\mathbf{i} - 2\mathbf{j}\) (Ns) | M1 | Complete correct method to solve to find a pair of values for \(\lambda\) and \(\mu\) |
| or \(\mathbf{I} = -\frac{3}{2}\mathbf{i} - 2\mathbf{j}\) (Ns) | A1 | Two correct pairs of values for \(\lambda\) and \(\mu\) |
| or \(\mathbf{I} = -2\mathbf{i} - \frac{3}{2}\mathbf{j}\) (Ns) | M1 | Use symmetry in complete correct method to find one of the other pairs of values for \(\lambda\) and \(\mu\) |
| or \(\mathbf{I} = -2\mathbf{i} + \frac{3}{2}\mathbf{j}\) (Ns) | A1 | All four correct pairs (They do not need to write out the impulse in full) |
| Total: 9 marks | (9) |
# Question 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Momentum of $P$ after impulse $= a\mathbf{i}$ (or $a\mathbf{j}$) | B1 | Correct interpretation of angle of deflection (velocity or momentum a multiple of $\mathbf{i}$ or $\mathbf{j}$) |
| **Either:** Use of $\mathbf{I} = m(\mathbf{v} - \mathbf{u})$: $(\mathbf{I} =) 0.5(2a\mathbf{i} - (4\mathbf{i} + 4\mathbf{j})) (= (a-2)\mathbf{i} - 2\mathbf{j})$ | M1 | Form vector triangle or equation for $\mathbf{v}$ or their $a\mathbf{i}$ |
| Use of Pythagoras to form equation in $a$ | M1 | Use trigonometry or Pythagoras' theorem to form equation in $a$ |
| $6.25 = 0.25\left((2a-4)^2 + 16\right)$ | A1ft | Unsimplified equation with at most one error. Follow their $a\mathbf{i}$ |
| $\left(4a^2 - 16a + 7 = 0\right)$ | A1 | Correct unsimplified equation |
| **Or:** $\lambda^2 + \mu^2 = \frac{25}{4}$ | M1 | |
| $\mathbf{I} = \lambda\mathbf{i} + \mu\mathbf{j} = \frac{1}{2}((x-4)\mathbf{i} - 4\mathbf{j})$ | M1 | |
| $\mu = -2$ | A1 | Dependent on 2nd M (for impulse) |
| $\lambda^2 = \frac{9}{4}$ | A1 | |
## Question (Impulse - vector triangle method):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Use of $\mathbf{I} = m(\mathbf{v} - \mathbf{u})$ to form vector triangle | M1 | |
| Form equation in their $a$ | M1 | |
| $6.25 = a^2 + 8 - 2a\sqrt{8} \times \frac{1}{\sqrt{2}}$ | A1ft | |
| $\left(4 \times 6.25 = b^2 + 32 - 2b\sqrt{32} \times \frac{1}{\sqrt{2}} \text{ for velocity } b\mathbf{i}\right)$ | A1 | |
| $\left(4a^2 - 16a + 7 = 0\right)$ | | |
| $a = \frac{7}{2}, \frac{1}{2} \Rightarrow \mathbf{I} = \frac{3}{2}\mathbf{i} - 2\mathbf{j}$ (Ns) | M1 | Complete correct method to solve to find a pair of values for $\lambda$ and $\mu$ |
| or $\mathbf{I} = -\frac{3}{2}\mathbf{i} - 2\mathbf{j}$ (Ns) | A1 | Two correct pairs of values for $\lambda$ and $\mu$ |
| or $\mathbf{I} = -2\mathbf{i} - \frac{3}{2}\mathbf{j}$ (Ns) | M1 | Use symmetry in complete correct method to find one of the other pairs of values for $\lambda$ and $\mu$ |
| or $\mathbf{I} = -2\mathbf{i} + \frac{3}{2}\mathbf{j}$ (Ns) | A1 | All four correct pairs (They do not need to write out the impulse in full) |
| **Total: 9 marks** | **(9)** | |
---\begin{enumerate}
\item A particle $P$, of mass 0.5 kg , is moving with velocity ( $4 \mathbf { i } + 4 \mathbf { j }$ ) $\mathrm { m } \mathrm { s } ^ { - 1 }$ when it receives an impulse I of magnitude 2.5 Ns.
\end{enumerate}
As a result of the impulse, the direction of motion of $P$ is deflected through an angle of $45 ^ { \circ }$ Given that $\mathbf { I } = ( \lambda \mathbf { i } + \mu \mathbf { j } )$ Ns, find all the possible pairs of values of $\lambda$ and $\mu$.
\hfill \mbox{\textit{Edexcel FM1 2019 Q3 [9]}}