| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/3 (Pre-U Mathematics Paper 3) |
| Year | 2016 |
| Session | June |
| Marks | 8 |
| Topic | Variable acceleration (1D) |
| Type | Velocity from acceleration by integration |
| Difficulty | Standard +0.8 This is a two-part mechanics problem requiring application of SUVAT equations with resistance forces. Part (i) needs students to find acceleration from kinematic data then resolve forces; part (ii) requires finding terminal velocity concepts and solving motion with resistance during descent. While methodical, it demands careful force analysis in both directions and multi-step reasoning beyond standard projectile motion, placing it moderately above average difficulty. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration6.02a Work done: concept and definition |
| Answer | Marks | Guidance |
|---|---|---|
| \(0^2 = 165^2 + 2 \times a \times 1237.5\) | M1 | Use of appropriate 'suvat' equation. |
| \(\therefore a = -11\) (ms\(^{-2}\)) | A1 | Correct outcome. Must be either negative or suitably qualified. |
| \(0.01 \times (-11) = -0.01 \times 10 - R\) | M1 | Use of N2; signs correct. |
| \(\therefore R = 0.01\) (N) | A1 | Ft c's \(a\) provided it is negative. |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.01 \times a = 0.01 \times 10 - 0.01\) | M1 | Use of N2; signs adjusted correctly. |
| \(\therefore a = 9\) (ms\(^{-2}\)) | A1 | Ft c's \(R\). |
| \(1237.5 = 0 + \frac{9}{2}t^2\) | M1 | Depends on the previous M mark. Use of appropriate 'suvat' equation. |
| \(\therefore t = \sqrt{275} = 16.583\ldots \approx 16.6\) (s) | A1 | Correct outcome. c.a.o. |
### (i)
$0^2 = 165^2 + 2 \times a \times 1237.5$ | M1 | Use of appropriate 'suvat' equation.
$\therefore a = -11$ (ms$^{-2}$) | A1 | Correct outcome. Must be either negative or suitably qualified.
$0.01 \times (-11) = -0.01 \times 10 - R$ | M1 | Use of N2; signs correct.
$\therefore R = 0.01$ (N) | A1 | Ft c's $a$ provided it is negative. | [4]
### (ii)
$0.01 \times a = 0.01 \times 10 - 0.01$ | M1 | Use of N2; signs adjusted correctly.
$\therefore a = 9$ (ms$^{-2}$) | A1 | Ft c's $R$.
$1237.5 = 0 + \frac{9}{2}t^2$ | M1 | Depends on the previous M mark. Use of appropriate 'suvat' equation.
$\therefore t = \sqrt{275} = 16.583\ldots \approx 16.6$ (s) | A1 | Correct outcome. c.a.o. | [4]
---A particle of mass $0.01$ kg is projected vertically upwards from a point $G$ at ground level with speed $165 \text{ m s}^{-1}$ and reaches a maximum height of $1237.5$ m. Throughout its motion it experiences a constant resistance.
\begin{enumerate}[label=(\roman*)]
\item Find the acceleration of the particle as it ascends and hence the magnitude of the resistance. [4]
\item During its descent back to $G$ the particle experiences the same constant resistance. Find the time taken for the descent. [4]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2016 Q9 [8]}}