| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2006 |
| 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 | Standard +0.3 This is a straightforward M3 impulse question requiring integration of a polynomial force function and application of the impulse-momentum theorem. The integration is routine (power rule only), and the multi-part structure guides students through standard steps with no novel problem-solving required. Slightly above average difficulty due to being M3 content and requiring careful sign handling, but otherwise a textbook exercise. |
| Spec | 6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| \(I = 1.4 \times 10^3 \int_0^{0.1} (t^2 - 10t^3) dt\) | M1A1 | |
| \(= 1.4 \times 10^3 \left[\frac{1}{3}t^3 - \frac{10}{4}t^4\right]_0^{0.1}\) | m1 | |
| \(= 11.7 \text{ Ns}\) | A1 | AG |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Initial momentum \(= 0.45(-15) = -6.75 \text{ Ns}\) | M1 | |
| Final momentum \(= 11.7 - 6.75 = 4.95 \text{ Ns}\) | M1 | |
| Velocity after impact \(= \frac{4.95}{0.45}\) | m1 | Dependent on both previous M1s |
| \(= 11 \text{ ms}^{-1}\) | A1 | |
| 4 |
| Answer | Marks |
|---|---|
| The ball is not perfectly elastic or \(e \neq 1\) or energy loss | E1 |
| 1 | |
| 9 | Total |
### Part (a)
$I = 1.4 \times 10^3 \int_0^{0.1} (t^2 - 10t^3) dt$ | M1A1 |
$= 1.4 \times 10^3 \left[\frac{1}{3}t^3 - \frac{10}{4}t^4\right]_0^{0.1}$ | m1 |
$= 11.7 \text{ Ns}$ | A1 | AG
| | **4** |
### Part (b)
Initial momentum $= 0.45(-15) = -6.75 \text{ Ns}$ | M1 |
Final momentum $= 11.7 - 6.75 = 4.95 \text{ Ns}$ | M1 |
Velocity after impact $= \frac{4.95}{0.45}$ | m1 | Dependent on both previous M1s
$= 11 \text{ ms}^{-1}$ | A1 |
| | **4** |
### Part (c)
The ball is not perfectly elastic or $e \neq 1$ or energy loss | E1 |
| | **1** |
| | **9** | **Total** |3 A ball of mass 0.45 kg is travelling horizontally with speed $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when it strikes a fixed vertical bat directly and rebounds from it. The ball stays in contact with the bat for 0.1 seconds.
At time $t$ seconds after first coming into contact with the bat, the force exerted on the ball by the bat is $1.4 \times 10 ^ { 5 } \left( t ^ { 2 } - 10 t ^ { 3 } \right)$ newtons, where $0 \leqslant t \leqslant 0.1$.
In this simple model, ignore the weight of the ball and model the ball as a particle.
\begin{enumerate}[label=(\alph*)]
\item Show that the magnitude of the impulse exerted by the bat on the ball is 11.7 Ns , correct to three significant figures.
\item Find, to two significant figures, the speed of the ball immediately after the impact.
\item Give a reason why the speed of the ball immediately after the impact is different from the speed of the ball immediately before the impact.
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2006 Q3 [9]}}