| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Work done against resistance - penetration into material |
| Difficulty | Moderate -0.3 This is a straightforward two-part energy conservation problem requiring standard application of SUVAT or energy methods. Part (a) uses kinetic + potential energy or kinematics with constant acceleration; part (b) applies work-energy principle with constant resistance. Both are routine A-level mechanics calculations with no conceptual challenges beyond direct formula application. |
| Spec | 3.02h Motion under gravity: vector form6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Using \(v^2 = u^2 + 2as\) with \(s = 0.8,\ u = 6,\ a = 9.8\): \(v^2 = 6^2 + 2 \times 9.8 \times 0.8\) | M1 | Allow for suvat equation(s) used leading to a value for \(v\) or \(v^2\). Allow sign errors |
| \(v = \sqrt{51.68} = 7.19\ \text{ms}^{-1}\) | A1 | Allow even if sign of \(u\) does not match sign of \(s\) and \(a\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Using \(v^2 = u^2 + 2as\) with \(s = 0.03,\ u = \sqrt{51.68},\ v = 0\): \(0 = 51.68 + 2 \times 0.03a\) | M1 | Allow for suvat equation(s) leading to a value for \(a\). Allow \(s = 3\) used. FT their (a). Allow sign errors |
| \(a = -861.3\ldots\ \text{ms}^{-2}\) | A1 | Need not be evaluated |
| N2L for pebble (downwards positive): \(0.04g - R = 0.04a\); \(0.04g - R = -0.04 \times 861.3\) | M1 A1 | Use of N2L allow one error or omission; Fully correct equation; FT their acceleration. Weight must be included |
| \(R = 34.8\ \text{N}\) | A1 | Must be rounded to 3 sf. Accept 34.8 or 34.9 only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Initial \(\text{KE} = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.04 \times 51.68\) | M1 A1 | Attempt to calculate change in KE or GPE |
| \(\text{GPE} = 0.04 \times 9.8 \times 0.03\) | A1 | |
| Work done against \(R\) is \(1.04536\); \(R = \dfrac{1.04536}{0.03} = 34.8\ \text{N}\) | M1 A1 | Allow if GPE is not included |
## Question 9:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Using $v^2 = u^2 + 2as$ with $s = 0.8,\ u = 6,\ a = 9.8$: $v^2 = 6^2 + 2 \times 9.8 \times 0.8$ | M1 | Allow for suvat equation(s) used leading to a value for $v$ or $v^2$. Allow sign errors |
| $v = \sqrt{51.68} = 7.19\ \text{ms}^{-1}$ | A1 | Allow even if sign of $u$ does not match sign of $s$ and $a$ |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Using $v^2 = u^2 + 2as$ with $s = 0.03,\ u = \sqrt{51.68},\ v = 0$: $0 = 51.68 + 2 \times 0.03a$ | M1 | Allow for suvat equation(s) leading to a value for $a$. Allow $s = 3$ used. FT their (a). Allow sign errors |
| $a = -861.3\ldots\ \text{ms}^{-2}$ | A1 | Need not be evaluated |
| N2L for pebble (downwards positive): $0.04g - R = 0.04a$; $0.04g - R = -0.04 \times 861.3$ | M1 A1 | Use of N2L allow one error or omission; Fully correct equation; FT their acceleration. Weight must be included |
| $R = 34.8\ \text{N}$ | A1 | Must be rounded to 3 sf. Accept 34.8 or 34.9 only |
**Further Maths energy method for (b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Initial $\text{KE} = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.04 \times 51.68$ | M1 A1 | Attempt to calculate change in KE or GPE |
| $\text{GPE} = 0.04 \times 9.8 \times 0.03$ | A1 | |
| Work done against $R$ is $1.04536$; $R = \dfrac{1.04536}{0.03} = 34.8\ \text{N}$ | M1 A1 | Allow if GPE is not included |
---9 A child throws a pebble of mass 40 g vertically downwards with a speed of $6 \mathrm {~ms} ^ { - 1 }$ from a point 0.8 m above a sandy beach.
\begin{enumerate}[label=(\alph*)]
\item Calculate the speed at which the pebble hits the beach.
The pebble travels 3 cm through the sand before coming to rest.
\item Find the magnitude of the resistance force of the sand on the pebble, assuming it is constant. Give your answer correct to $\mathbf { 3 }$ significant figures.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2024 Q9 [7]}}