| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Year | 2021 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Vector impulse: find deflection angle or impulse magnitude from angle |
| Difficulty | Standard +0.3 This is a standard FM1 impulse-momentum question requiring vector addition and magnitude calculation. Students apply impulse-momentum theorem (Δp = I), find the final velocity vector, use the given speed to solve for λ, then calculate the angle using dot product. While it involves multiple steps and vector manipulation, it follows a predictable template with no novel insight required—slightly easier than average A-level maths. |
| Spec | 1.10d Vector operations: addition and scalar multiplication6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Use of Impulse \(=\) change in momentum | M1 | Must be subtracting two values for momentum, but condone subtraction in the wrong order |
| \(0.5(\mathbf{v} - 8\mathbf{i}) = \lambda(-\mathbf{i}+\mathbf{j})\) \(\mathbf{v} = (-2\lambda+8)\mathbf{i} + 2\lambda\mathbf{j}\) | A1 | Correct unsimplified equation |
| Use of Pythagoras | M1 | Correct use of final speed with their v |
| \(160 = (-2\lambda+8)^2 + (2\lambda)^2\) \(\left(160 = 4\lambda^2 - 32\lambda + 64 + 4\lambda^2\right)\) | A1 | Correct unsimplified equation in one unknown or pair of simultaneous equations |
| Form and solve quadratic in \(\lambda\): \(8\lambda^2 - 32\lambda - 96 = 0\) \(\left(\lambda^2 - 4\lambda - 12 = (\lambda-6)(\lambda+2)=0\right)\) | M1 | Simplify and solve for \(\lambda\) from correct working |
| \(\Rightarrow \lambda = 6\) | A1 | Correct positive solution only |
| Find the required angle: \(180° - \tan^{-1}3\) | M1 | Complete method to solve for \(\theta\) |
| \(\theta = 108°\) | A1 | 108 or better (108.4349…) |
# Question 4:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Use of Impulse $=$ change in momentum | M1 | Must be subtracting two values for momentum, but condone subtraction in the wrong order |
| $0.5(\mathbf{v} - 8\mathbf{i}) = \lambda(-\mathbf{i}+\mathbf{j})$ $\mathbf{v} = (-2\lambda+8)\mathbf{i} + 2\lambda\mathbf{j}$ | A1 | Correct unsimplified equation |
| Use of Pythagoras | M1 | Correct use of final speed with their **v** |
| $160 = (-2\lambda+8)^2 + (2\lambda)^2$ $\left(160 = 4\lambda^2 - 32\lambda + 64 + 4\lambda^2\right)$ | A1 | Correct unsimplified equation in one unknown or pair of simultaneous equations |
| Form and solve quadratic in $\lambda$: $8\lambda^2 - 32\lambda - 96 = 0$ $\left(\lambda^2 - 4\lambda - 12 = (\lambda-6)(\lambda+2)=0\right)$ | M1 | Simplify and solve for $\lambda$ from correct working |
| $\Rightarrow \lambda = 6$ | A1 | Correct positive solution only |
| Find the required angle: $180° - \tan^{-1}3$ | M1 | Complete method to solve for $\theta$ |
| $\theta = 108°$ | A1 | 108 or better (108.4349…) |
---\begin{enumerate}
\item A particle $P$ has mass 0.5 kg . It is moving in the $x y$ plane with velocity $8 \mathbf { i } \mathrm {~ms} ^ { - 1 }$ when it receives an impulse $\lambda ( - \mathbf { i } + \mathbf { j } )$ Ns, where $\lambda$ is a positive constant.
\end{enumerate}
The angle between the direction of motion of $P$ immediately before receiving the impulse and the direction of motion of $P$ immediately after receiving the impulse is $\theta ^ { \circ }$
Immediately after receiving the impulse, $P$ is moving with speed $4 \sqrt { 10 } \mathrm {~ms} ^ { - 1 }$\\
Find (i) the value of $\lambda$\\
(ii) the value of $\theta$
\hfill \mbox{\textit{Edexcel FM1 2021 Q4 [8]}}