| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2012 |
| Session | June |
| Marks | 7 |
| 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 impulse-momentum problem requiring integration of a given force function and application of the impulse-momentum theorem. The force function is polynomial (easy to integrate), and the problem clearly guides students through (a) finding impulse via integration, then (b) applying conservation of momentum. While it involves calculus and momentum concepts, it's a standard M3 textbook exercise with no conceptual surprises or problem-solving insight required—slightly easier than average A-level. |
| Spec | 1.08d Evaluate definite integrals: between limits6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Impulse = \(\int_0^{0.5} 4 \times 10^4 t^2(1-2t) \, dt\) | M1 | Attempt to integrate force with respect to time |
| \(= 4 \times 10^4 \left[\frac{t^3}{3} - \frac{t^4}{2}\right]_0^{0.5}\) | A1 | Correct integration |
| \(= 4 \times 10^4 \left(\frac{(0.5)^3}{3} - \frac{(0.5)^4}{2}\right)\) | M1 | Substituting limits |
| \(= 4 \times 10^4 \times \frac{1}{96} = \frac{4000}{96} \approx 41.7\) N s | A1 | \(\frac{125}{3}\) or \(41.\overline{6}\) N s |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Initial momentum \(= 60 \times 5 = 300\) N s (towards wall) | B1 | |
| Impulse-momentum: \(\frac{125}{3} = 60v + 300\) or \(\frac{125}{3} = 60v - (-300)\) | M1 | Applying impulse = change in momentum, signs must be consistent |
| \(v = \frac{\frac{125}{3} - 300}{60}\) giving speed \(= \frac{175}{60} \approx 2.92\) m s\(^{-1}\) | A1 | Accept \(\frac{35}{12}\) or \(2.9\overline{16}\) m s\(^{-1}\) |
# Question 1:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Impulse = $\int_0^{0.5} 4 \times 10^4 t^2(1-2t) \, dt$ | M1 | Attempt to integrate force with respect to time |
| $= 4 \times 10^4 \left[\frac{t^3}{3} - \frac{t^4}{2}\right]_0^{0.5}$ | A1 | Correct integration |
| $= 4 \times 10^4 \left(\frac{(0.5)^3}{3} - \frac{(0.5)^4}{2}\right)$ | M1 | Substituting limits |
| $= 4 \times 10^4 \times \frac{1}{96} = \frac{4000}{96} \approx 41.7$ N s | A1 | $\frac{125}{3}$ or $41.\overline{6}$ N s |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Initial momentum $= 60 \times 5 = 300$ N s (towards wall) | B1 | |
| Impulse-momentum: $\frac{125}{3} = 60v + 300$ or $\frac{125}{3} = 60v - (-300)$ | M1 | Applying impulse = change in momentum, signs must be consistent |
| $v = \frac{\frac{125}{3} - 300}{60}$ giving speed $= \frac{175}{60} \approx 2.92$ m s$^{-1}$ | A1 | Accept $\frac{35}{12}$ or $2.9\overline{16}$ m s$^{-1}$ |
---1 An ice-hockey player has mass 60 kg . He slides in a straight line at a constant speed of $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ on the horizontal smooth surface of an ice rink towards the vertical perimeter wall of the rink, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{a90a2de3-5cc0-4e87-b29a-2562f86eee17-02_476_594_769_715}
The player collides directly with the wall, and remains in contact with the wall for 0.5 seconds.
At time $t$ seconds after coming into contact with the wall, the force exerted by the wall on the player is $4 \times 10 ^ { 4 } t ^ { 2 } ( 1 - 2 t )$ newtons, where $0 \leqslant t \leqslant 0.5$.
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the impulse exerted by the wall on the player.
\item The player rebounds from the wall. Find the player's speed immediately after the collision.
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2012 Q1 [7]}}