| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | June |
| Marks | 6 |
| 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 two-part impulse question requiring (a) basic impulse-momentum theorem with sign consideration, and (b) integration of a given force function and equating to the impulse from part (a). Standard M3 material with no novel insight required, slightly easier than average due to clear setup and routine calculus. |
| Spec | 1.08h Integration by substitution6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Impulse \(= 0.2 \times 32 + 0.2 \times 18\) | M1 | Taking change in momentum, must consider direction (addition of momenta) |
| \(= 10\) N s | A1 | Accept \(10\) kg m s\(^{-1}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Impulse \(= \int_0^{0.09} k(0.9t - 10t^2)\, dt\) | M1 | Setting up integral of force |
| \(= k\left[\frac{0.9t^2}{2} - \frac{10t^3}{3}\right]_0^{0.09}\) | A1 | Correct integration |
| \(= k\left[\frac{0.9 \times 0.0081}{2} - \frac{10 \times 0.000729}{3}\right]\) | ||
| \(= k[0.003645 - 0.00243]\) | ||
| \(= 0.001215k\) | A1 | Correct evaluation |
| \(0.001215k = 10\) | M1 | Setting integral equal to impulse from (a) |
| \(k = \dfrac{10}{0.001215} \approx 8230\) | A1 | Accept \(k = \dfrac{80000}{9.72}\) or equivalent exact form |
# Question 1:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Impulse $= 0.2 \times 32 + 0.2 \times 18$ | M1 | Taking change in momentum, must consider direction (addition of momenta) |
| $= 10$ N s | A1 | Accept $10$ kg m s$^{-1}$ |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Impulse $= \int_0^{0.09} k(0.9t - 10t^2)\, dt$ | M1 | Setting up integral of force |
| $= k\left[\frac{0.9t^2}{2} - \frac{10t^3}{3}\right]_0^{0.09}$ | A1 | Correct integration |
| $= k\left[\frac{0.9 \times 0.0081}{2} - \frac{10 \times 0.000729}{3}\right]$ | | |
| $= k[0.003645 - 0.00243]$ | | |
| $= 0.001215k$ | A1 | Correct evaluation |
| $0.001215k = 10$ | M1 | Setting integral equal to impulse from (a) |
| $k = \dfrac{10}{0.001215} \approx 8230$ | A1 | Accept $k = \dfrac{80000}{9.72}$ or equivalent exact form |
---1 A ball of mass 0.2 kg is hit directly by a bat. Just before the impact, the ball is travelling horizontally with speed $18 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Just after the impact, the ball is travelling horizontally with speed $32 \mathrm {~ms} ^ { - 1 }$ in the opposite direction.
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the impulse exerted on the ball.
\item At time $t$ seconds after the ball first comes into contact with the bat, the force exerted by the bat on the ball is $k \left( 0.9 t - 10 t ^ { 2 } \right)$ newtons, where $k$ is a constant and $0 \leqslant t \leqslant 0.09$. The bat stays in contact with the ball for 0.09 seconds.
Find the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2011 Q1 [6]}}