June 2022 Paper 2 Q19
19 In this question use \(g\) = 9.8 m s−2
A rough wooden ramp is 10 metres long and is inclined at an angle of 25° above the horizontal.
The bottom of the ramp is at the point \(O\).
A crate of mass 20 kg is at rest at the point \(A\) on the ramp.
The crate is pulled up the ramp using a rope attached to the crate.
Once in motion, the rope remains taut and parallel to the line of greatest slope of the ramp.
(a) The tension in the rope is 230 N
The crate accelerates up the ramp at 1.2 m s−2
Find the coefficient of friction between the crate and the ramp. [7 marks]
(b)
(i) The crate takes 3.8 seconds to reach the top of the ramp.
Find the distance \(OA\). [3 marks]
(ii) Other than air resistance, state one assumption you have made about the crate in answering part (b)(i). [1 mark]
| Scheme | Marks | AO |
|---|---|---|
| States \(F = \mu R\) seen anywhere PI by use of \(\mu R\) in their 4-term equation of motion or on diagram | B1 | 3.3 |
| Resolves the weight parallel to the slope to obtain \(mg\sin 25\) or better | B1 | 1.1b |
| Resolves perpendicular to the slope to obtain \(R = mg\cos 25\) or better | B1 | 1.1b |
| Uses F = ma to form a four-term equation with consistent signs. eg \(T - \text{weight} - \text{Friction} = ma\) Condone omission of \(g\) in weight and friction component | M1 | 3.3 |
| Substitutes \(T\) = 230 and \(F = \mu mg\cos 25\) into their four term \(F = ma\) equation with consistent signs. Condone ‘\(mga\)’ in \(F = ma\) for this mark | M1 | 1.1a |
| Obtains single correct equation with all numerical values substituted. eg \(230 - 196\sin 25 - 196\cos 25\,\mu = 24\) Scores B1B1B1M1M1A1 | A1 | 1.1b |
| Obtains \(\mu = 0.69\) CAO | A1 | 3.2a |
| (7) |
Typical solution
\[T - \text{weight} - \text{Friction} = ma\]\[T - 20g\sin 25 - F = m \times 1.2\]\[T - 20g\sin 25 - 20g\cos 25\,\mu = m \times 1.2\]\[230 - 196\sin 25 - 196\cos 25\,\mu = 24\]\[\mu = 0.69\]| Scheme | Marks | AO |
|---|---|---|
| (i) Substitutes \(u\) = 0, \(a\) = 1.2 and \(t\) = 3.8 into \(s = ut + \dfrac{1}{2}at^2\) Or uses appropriate constant acceleration equations that forms a complete method to obtain \(s\) | M1 | 1.1a |
| Obtains AWRT 8.7 | A1 | 1.1b |
| Obtains 10 − their 8.7 AWRT 1.3 or FT their 8.7 provided it is less than 10 Condone missing units | A1F | 1.1b |
| (3) | ||
| (ii) States the crate has been modelled as a particle OE | E1 | 3.5b |
| (1) | ||
| (11 marks) |
Typical solution
(i)
\[s = ut + \frac{1}{2}at^2\]\[s = \frac{1}{2} \times 1.2 \times 3.8^2 = 8.664\]\[OA = 10 - 8.664 = 1.3 \text{ m}\](ii)
The crate is a particle