| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Impulse from variable force (then find velocity) |
| Difficulty | Moderate -0.3 This is a straightforward application of impulse-momentum theorem with a variable force. Part (a) requires integrating a simple linear function, part (b) applies impulse = change in momentum directly, and part (c) involves solving a quadratic equation. All steps are standard M3 techniques with no conceptual challenges or novel problem-solving required, making it slightly easier than average. |
| Spec | 1.08d Evaluate definite integrals: between limits3.03c Newton's second law: F=ma one dimension6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Impulse \(= \displaystyle\int_0^3 (3t+1)\,dt\) | M1 | Integrating force with respect to time |
| \(= \left[\dfrac{3t^2}{2} + t\right]_0^3 = \dfrac{27}{2} + 3 = 16.5\) N s | M1 A1 | Correct integration and evaluation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Impulse \(=\) change in momentum: \(16.5 = 0.5(v - 4)\) | M1 | Applying impulse–momentum theorem |
| \(v = 37\ \text{m s}^{-1}\) | A1 | Correct velocity |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.5\,\dfrac{dv}{dt} = 3t + 1\) | M1 | Newton's second law |
| \(v = \dfrac{1}{0.5}\left(\dfrac{3t^2}{2} + t\right) + C = 3t^2 + 2t + C\) | M1 | Integrating and including constant |
| At \(t=0\), \(v=4\): \(C = 4\), so \(v = 3t^2 + 2t + 4\) | A1 | Correct expression for \(v\) |
| \(20 = 3t^2 + 2t + 4 \Rightarrow 3t^2 + 2t - 16 = 0\) | M1 | Setting \(v = 20\) |
| \((3t + 8)(t - 2) = 0 \Rightarrow t = 2\) | A1 | \(t = 2\) (reject negative) |
# Question 3:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Impulse $= \displaystyle\int_0^3 (3t+1)\,dt$ | M1 | Integrating force with respect to time |
| $= \left[\dfrac{3t^2}{2} + t\right]_0^3 = \dfrac{27}{2} + 3 = 16.5$ N s | M1 A1 | Correct integration and evaluation |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Impulse $=$ change in momentum: $16.5 = 0.5(v - 4)$ | M1 | Applying impulse–momentum theorem |
| $v = 37\ \text{m s}^{-1}$ | A1 | Correct velocity |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.5\,\dfrac{dv}{dt} = 3t + 1$ | M1 | Newton's second law |
| $v = \dfrac{1}{0.5}\left(\dfrac{3t^2}{2} + t\right) + C = 3t^2 + 2t + C$ | M1 | Integrating and including constant |
| At $t=0$, $v=4$: $C = 4$, so $v = 3t^2 + 2t + 4$ | A1 | Correct expression for $v$ |
| $20 = 3t^2 + 2t + 4 \Rightarrow 3t^2 + 2t - 16 = 0$ | M1 | Setting $v = 20$ |
| $(3t + 8)(t - 2) = 0 \Rightarrow t = 2$ | A1 | $t = 2$ (reject negative) |
I can see this is an exam paper (P/Jun14/MM03) showing Question 4 about two boats A and B with position vectors. However, the images shown are **answer space pages** (blank lined pages for students to write their answers), not mark scheme pages.
There is **no mark scheme content visible** in these images — they contain only:
- The question text for Q4 (parts a, b, c, d)
- Blank answer spaces for Q3 and Q4
To extract mark scheme content, you would need to provide images of the **actual mark scheme document** for this paper, which is a separate publication.
If you'd like, I can instead:
- **Work through the solutions** to Q4(a)–(d) showing full working
- **Suggest likely mark allocations** based on the question structure and standard M1/A1 marking conventions
Would either of those be helpful?3 A particle of mass 0.5 kg is moving in a straight line on a smooth horizontal surface.\\
The particle is then acted on by a horizontal force for 3 seconds. This force acts in the direction of motion of the particle and at time $t$ seconds has magnitude $( 3 t + 1 ) \mathrm { N }$.
When $t = 0$, the velocity of the particle is $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the impulse of the force on the particle between the times $t = 0$ and $t = 3$.
\item Hence find the velocity of the particle when $t = 3$.
\item Find the value of $t$ when the velocity of the particle is $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2014 Q3 [9]}}