June 2024 Paper 1 Q9
9 A child throws a pebble of mass 40 g vertically downwards with a speed of \(6\ \text{m s}^{-1}\) from a point 0.8 m above a sandy beach.
The pebble travels 3 cm through the sand before coming to rest.
| Scheme | Marks | AO |
|---|---|---|
| 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 | 1.1a |
| \(v = \sqrt{51.68} = 7.19\ \text{m s}^{-1}\) | A1 | 1.1 |
| [2] |
Notes
M1: Allow for suvat equation(s) used leading to a value for \(v\) or \(v^2\)
Allow sign errors
A1: Allow even if the sign of \(u\) does not match the sign of \(s\) and \(a\)
| Scheme | Marks | AO |
|---|---|---|
| Using \(v^2 = u^2 + 2as\) with \(s = 0.03, u = \sqrt{51.68}, v = 0\) \(0^2 = 51.68 + 2 \times 0.03a\) | M1 | 3.1b |
| \(a = -861.3\ldots\ \text{m s}^{-2}\) | A1 | 1.1 |
| N2L for pebble (downwards positive) \(0.04g - R = 0.04a\) \(0.04g - R = -0.04 \times 861.3\) | M1 A1 | 3.1b 1.1 |
| \(R = 34.8\ \text{N}\) | A1 | 1.1 |
| [5] |
Notes
M1: Allow for suvat equation(s) used leading to a value for \(a\)
Allow for \(s = 3\) used.
FT their (a) Allow sign errors
A1: Need not be evaluated
M1: Use of N2L allow one error or omission
A1: Fully correct equation
FT their acceleration. Weight must be included.
A1: Must be rounded to 3 sf. Accept 34.8 or 34.9 only
Further Maths students may attempt an energy method for (b)
| Scheme | Marks | AO |
|---|---|---|
| Initial KE \(= \tfrac{1}{2}mv^2 = \tfrac{1}{2} \times 0.04 \times 51.68\) GPE \(= 0.04 \times 9.8 \times 0.03\) | M1 A1 A1 | |
| Work done against R is 1.04536 \(R = \dfrac{1.04536}{0.03} = 34.8\ \text{N}\) | M1 A1 |
M1: Attempt to calculate change in KE or GPE
M1: Allow if GPE is not included