October 2021 Paper 1 Q12
12 A box of mass \(m\) kg slides down a rough slope inclined at \(15^\circ\) to the horizontal. The coefficient of friction between the box and the slope is 0.4. The box has an initial velocity of \(1.2\,\text{m}\,\text{s}^{-1}\) down the slope.
Calculate the distance the box travels before coming to rest. [7]
| Scheme | Marks | AO |
|---|---|---|
| Normal reaction \(mg\cos 15^\circ\) | B1 | 3.1b |
| Max friction \(\mu N = 0.4mg\cos 15^\circ\) | M1 | 1.1b |
| Resolve down the slope \(mg\sin 15^\circ - F = ma\) | B1 | 1.1b |
| \(mg\sin 15^\circ - 0.4mg\cos 15^\circ = ma\) giving \(a = -1.25\,\text{m}\,\text{s}^{-2}\) | M1 A1 | 3.1b 1.1b |
| Using \(v^2 = u^2 + 2as\) \(0^2 = 1.2^2 + 2 \times (-1.25)s\) | M1 | 3.1b |
| giving \(s = 0.576\) m | A1 | 1.1b |
| [7] |
Notes
B1: Correct normal reaction
M1: Attempt to evaluate friction FT their normal reaction.
Only allow \(0.4mg\) if it is clear that \(mg\) is their normal reaction and not just weight
B1: Correct component of weight seen \((2.536m)\)
M1: All terms present; allow sign errors, sin/cos interchange for weight and their \(F\)
A1: Correct equation (\(a\) need not be explicitly evaluated here)
M1: Use of suvat equation(s) leading to a value for \(s\) using \(v = 0\)
A1: FT their negative \(a\)