| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2009 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Calculate impulse from force-time data |
| Difficulty | Standard +0.3 This is a straightforward application of impulse-momentum theorem requiring integration of a polynomial force function and basic algebraic manipulation. The integration is routine (∫(t³+t)dt), and parts (b) and (c) follow directly from the impulse-momentum relationship with no conceptual challenges beyond standard M3 content. |
| Spec | 6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(I = \int_0^4 (t^3 + t)\, dt\) | M1 | |
| \(= \left[\frac{1}{4}t^4 + \frac{1}{2}t^2\right]_0^4\) | m1 | |
| \(= 72\) N s | A1 | |
| Subtotal | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(72 = 0.5v - 0.5(0)\) | M1 | Condone \(-5(0)\) |
| \(v = 144\) | A1F | |
| Subtotal | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\int_0^T (t^3 + t)\, dt = 0.5(12) - 0.5(0)\) | M1 | Condone \(-5(0)\) |
| \(\left[\frac{1}{4}t^4 + \frac{1}{2}t^2\right]_0^T = 6\) | ||
| \(T^4 + 2T^2 - 24 = 0\) | A1 | |
| \(T^2 = \frac{-2 \pm \sqrt{2^2 - 4(1)(-24)}}{2(1)}\) | m1 A1F | |
| or \((T^2 - 4)(T^2 + 6) = 0\) | ||
| \(T^2 = 4\) | ||
| \(T = 2\) | A1F | |
| Subtotal | 5 | |
| Total | 10 |
## Question 4:
### Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $I = \int_0^4 (t^3 + t)\, dt$ | M1 | |
| $= \left[\frac{1}{4}t^4 + \frac{1}{2}t^2\right]_0^4$ | m1 | |
| $= 72$ N s | A1 | |
| **Subtotal** | **3** | |
### Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $72 = 0.5v - 0.5(0)$ | M1 | Condone $-5(0)$ |
| $v = 144$ | A1F | |
| **Subtotal** | **2** | |
### Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\int_0^T (t^3 + t)\, dt = 0.5(12) - 0.5(0)$ | M1 | Condone $-5(0)$ |
| $\left[\frac{1}{4}t^4 + \frac{1}{2}t^2\right]_0^T = 6$ | | |
| $T^4 + 2T^2 - 24 = 0$ | A1 | |
| $T^2 = \frac{-2 \pm \sqrt{2^2 - 4(1)(-24)}}{2(1)}$ | m1 A1F | |
| or $(T^2 - 4)(T^2 + 6) = 0$ | | |
| $T^2 = 4$ | | |
| $T = 2$ | A1F | |
| **Subtotal** | **5** | |
| **Total** | **10** | |4 A particle of mass 0.5 kg is initially at rest. The particle then moves in a straight line under the action of a single force. This force acts in a constant direction and has magnitude $\left( t ^ { 3 } + t \right) \mathrm { N }$, where $t$ is the time, in seconds, for which the force has been acting.
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the impulse exerted by the force on the particle between the times $t = 0$ and $t = 4$.
\item Hence find the speed of the particle when $t = 4$.
\item Find the time taken for the particle to reach a speed of $12 \mathrm {~ms} ^ { - 1 }$.
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2009 Q4 [10]}}