| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2007 |
| 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 time-varying force. Part (a) requires integrating F(t) = 4t + 5 from 0 to 3, part (b) uses impulse = change in momentum with initial velocity zero, and part (c) extends the same method. All steps are standard M3 techniques with no conceptual challenges—slightly easier than average due to the simple polynomial force function and direct application of formulas. |
| Spec | 1.08h Integration by substitution3.02d Constant acceleration: SUVAT formulae3.03c Newton's second law: F=ma one dimension6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(I = \int_0^3 (4t+5)\,dt\) | M1 | |
| \(= \left[2t^2 + 5t\right]_0^3\) | m1 | Or evaluation of constant |
| \(= 33\) Ns | A1 | Total: 3 marks |
| Alternative: \(I =\) Area under Force-Time graph | (M1) | |
| \(= \frac{17+5}{2} \times 3 = 33\) Ns | (m1)(A1) | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(I = mv - mu\) | ||
| \(33 = 2v - 2(0)\) | M1 | |
| \(v = 16.5 \text{ ms}^{-1}\) | A1F | Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(I = \int_0^t (4t+5)\,dt = 2(37.5) - 2(0)\) | M1 | |
| \(2t^2 + 5t - 75 = 0\) | A1 | |
| \(t = \frac{-5 \pm \sqrt{25 + 8 \times 75}}{4}\) | m1 | |
| \(t = 5\) | A1F | For one value of \(t\) identified only; Total: 4 marks |
## Question 3:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I = \int_0^3 (4t+5)\,dt$ | M1 | |
| $= \left[2t^2 + 5t\right]_0^3$ | m1 | Or evaluation of constant |
| $= 33$ Ns | A1 | Total: 3 marks |
| **Alternative:** $I =$ Area under Force-Time graph | (M1) | |
| $= \frac{17+5}{2} \times 3 = 33$ Ns | (m1)(A1) | (3) |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I = mv - mu$ | | |
| $33 = 2v - 2(0)$ | M1 | |
| $v = 16.5 \text{ ms}^{-1}$ | A1F | Total: 2 marks |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I = \int_0^t (4t+5)\,dt = 2(37.5) - 2(0)$ | M1 | |
| $2t^2 + 5t - 75 = 0$ | A1 | |
| $t = \frac{-5 \pm \sqrt{25 + 8 \times 75}}{4}$ | m1 | |
| $t = 5$ | A1F | For one value of $t$ identified only; Total: 4 marks |
---3 A particle $P$, of mass 2 kg , is initially at rest at a point $O$ on a smooth horizontal surface. The particle moves along a straight line, $O A$, under the action of a horizontal force. When the force has been acting for $t$ seconds, it has magnitude $( 4 t + 5 ) \mathrm { N }$.
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the impulse exerted by the force on $P$ between the times $t = 0$ and $t = 3$.
\item Find the speed of $P$ when $t = 3$.
\item The speed of $P$ at $A$ is $37.5 \mathrm {~ms} ^ { - 1 }$. Find the time taken for the particle to reach $A$.
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2007 Q3 [9]}}