| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Mechanics (Further AS Paper 2 Mechanics) |
| Year | 2018 |
| 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 question requiring: (a) basic impulse calculation using change in momentum with sign consideration; (b)(i) integrating a quadratic force function and equating to impulse; (b)(ii) differentiating to find maximum force. All techniques are standard Further Maths mechanics with no novel problem-solving required, making it slightly easier than average A-level. |
| Spec | 6.02b Calculate work: constant force, resolved component6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(I = mv - mu = 250(1.2) - 250(-1.8)\) | M1 | Uses correct formula for impulse with one velocity negative |
| \(I = 750\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(750 = \int_0^{0.8} kt(4-5t)\,dt\) | M1 | Forms equation involving appropriate integral using their value from part (a) to find \(k\) |
| \(\int_0^{0.8} t(4-5t)\,dt = \frac{32}{75}\) | B1 | Evaluates definite integral correctly |
| \(k = \frac{750}{\frac{32}{75}} = \frac{28125}{16} = 1800\) (2sf) | A1F | Solves equation to find value of \(k\); follow through their '750' |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Maximum occurs at \(t = 0.4\); maximum value is \(\frac{28125}{16} \times 0.4 \times [4 - 5(0.4)] = \frac{5625}{4}\) N | M1 | Deduces when maximum value of force occurs and substitutes correct value of \(t\) into formula for force |
| \(= 1400\) (2sf) | A1F | Obtains correct value for their maximum force; follow through incorrect \(k\) |
## Question 6(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I = mv - mu = 250(1.2) - 250(-1.8)$ | M1 | Uses correct formula for impulse with one velocity negative |
| $I = 750$ | A1 | |
## Question 6(b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $750 = \int_0^{0.8} kt(4-5t)\,dt$ | M1 | Forms equation involving appropriate integral using their value from part (a) to find $k$ |
| $\int_0^{0.8} t(4-5t)\,dt = \frac{32}{75}$ | B1 | Evaluates definite integral correctly |
| $k = \frac{750}{\frac{32}{75}} = \frac{28125}{16} = 1800$ (2sf) | A1F | Solves equation to find value of $k$; follow through their '750' |
## Question 6(b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Maximum occurs at $t = 0.4$; maximum value is $\frac{28125}{16} \times 0.4 \times [4 - 5(0.4)] = \frac{5625}{4}$ N | M1 | Deduces when maximum value of force occurs and substitutes correct value of $t$ into formula for force |
| $= 1400$ (2sf) | A1F | Obtains correct value for their maximum force; follow through incorrect $k$ |
---6 At a fairground a dodgem car is moving in a straight horizontal line towards a side wall that is perpendicular to the velocity of the car.
The speed of the car is $1.8 \mathrm {~ms} ^ { - 1 }$\\
It collides with the side wall and rebounds along its original path with a speed of $1.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
The total mass of the dodgem car and the passengers is 250 kg\\
6
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the impulse on the car during the collision with the side wall.\\
6
\item A possible model for the magnitude of the force, $F$ newtons, acting on the dodgem car due to its collision with the side wall is given by
$$F = k t ( 4 - 5 t ) \quad \text { for } 0 \leq t \leq 0.8$$
6 (b) (i) Find the value of $k$.\\
(b) (ii) Determine the maximum magnitude of the force predicted by the model.
6 (b) (ii) Determine the maximum magnitude of the fored bed bed at
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Mechanics 2018 Q6 [7]}}