June 2024 Paper 3 Q13
13
The points \(A\) and \(B\) are the lower and upper ends, respectively, of a line of greatest slope on a plane inclined at an angle \(\theta\) to the horizontal, where \(\sin\theta = 0.6\) and \(AB = 1.375\) m (see diagram).
A particle \(P\) is projected up the plane with speed \(6\,\mathrm{m\,s^{-1}}\) from \(A\) towards \(B\).
The plane at \(A\) is fixed to the ground which is horizontal.
The surface of the plane is rough and the coefficient of friction between \(P\) and the plane is 0.25.
The particle leaves the slope at \(B\) and moves freely under gravity.
The particle first lands at a point \(C\) on the horizontal ground. The time taken for \(P\) to travel from \(A\) to \(C\) is \(T\) seconds.
| Scheme | Marks | AO |
|---|---|---|
| \(R(\text{perp. plane}): R = mg\cos 36.86\ldots\) or \(mg \times 0.8\) | B1 | 3.3 |
| \(F = 0.25 \times mg\cos 36.86\ldots\) or \(0.25 \times mg \times 0.8\) | M1* | 3.4 |
| N2L (parallel to plane): \(-0.25 \times mg \times 0.8 - mg \times 0.6 = ma\) or \(-0.25 \times mg \times \cos 36.86\ldots - mg \times \sin 36.86\ldots = ma\) | M1dep* | 3.1b |
| \(a = -7.84\) or \(7.84\) | A1 | 1.1 |
| \(v^2 = 6^2 + 2 \times -7.84 \times 1.375\) | M1dep* | 3.4 |
| \(v^2 = 14.44 \Rightarrow v = 3.8\ (\mathrm{m\,s^{-1}})\) | A1 | 2.2a |
| [6] |
Notes
B1: Allow using an angle of 37 or better
soi (possibly in N2L)
M1*: \(F = 0.25R\) with their \(R\) (with \(R \ne mg\)) – using an angle of 37 or better – allow sin/cos mix
\(R\) must be a component of weight soi (possibly in N2L)
M1dep*: Applying N2L parallel to the plane with correct number of terms and weight term resolved, dimensionally consistent – if down the plane is the positive direction expect \(0.25 \times mg \times 0.8 + mg \times 0.6 = ma\)
Allow sign errors and sin/cos mix, using an angle of 37 or better
A1: \(\pm 0.8g\) - allow awrt \(\pm 7.84\) or awrt \(\pm 7.85\) (using an angle of 37) www
Condone \(\pm 7.8\) or \(\pm 7.9\) www
M1dep*: Use of \(v^2 = u^2 + 2as\) with correct \(u\) and \(s\) and their negative \(a\) (\(\ne -9.8\)) - dependent on both previous M marks – must see the values substituted into the formula (as value for \(v\) is AG) – therefore just seeing \(v^2 = \frac{361}{25}\) or 14.44 is M0
Condone finding a positive value of \(a\) and then using the negative version
A1: AG – do not award this mark if using any non-exact values for \(\sin\theta / \cos\theta\) (so seeing e.g. \(\cos 36.86\ldots\) in their working loses this final A mark)
Must see either \(v^2 = 14.44\) or \(v^2 = 36 - 21.56\) or \(v^2 = 6^2 - 21.56\) or a correct expression for \(v\) before given answer
SC – if \(m\) absent throughout (so implying \(m = 1\)) or \(m\) given a numerical value then at most B0 M1 M1 A1 M1 A0 can be awarded.
Alternative for final two marks
| Scheme | Marks |
|---|---|
| \(1.375 = 6t + \frac{1}{2} \times -7.84 \times t^2\) \(t = \frac{55}{196}\) (or \(t = 1.25\)) \(v = 6 + (-7.84) \times \frac{55}{196}\) | M1dep* |
| \(v = 3.8\ (\mathrm{m\,s^{-1}})\) | A1 |
M1dep*: Use of \(s = ut + \frac{1}{2}at^2\) with correct \(u\) and \(s\) and their negative \(a\) (\(\ne -9.8\)), and selects the smaller of the two positive \(t\) values and uses this value correctly in \(v = u + at\) with correct \(u\) and their negative \(a\) - dependent on both previous M marks
So, using the second positive \(t\) (if correct this is 1.25) is M0
A1: AG – do not award this mark if using any non-exact values for \(\sin\theta / \cos\theta\)
Must see a correct expression for \(v\) before given answer
Alternative energy approach to 13(a) (Appendix)
| Scheme | Marks |
|---|---|
| \(R(\text{perp. plane}): R = mg\cos 36.86\ldots\) or \(mg \times 0.8\) | B1 |
| \(F = 0.25 \times mg\cos 36.86\ldots\) or \(0.25 \times mg \times 0.8\) | M1 |
| Work done by gravity/PE term \(= \pm mg \times (1.375 \times 0.6)\) or \(\pm mg \times (1.375 \times \sin 36.86\ldots)\) OR Work done by friction \(\pm 1.375 \times (0.25 \times mg \times 0.8)\) or \(\pm 1.375 \times (0.25 \times mg \times \cos 36.86\ldots)\) | B1 |
| \(\frac{1}{2}mv^2 - \frac{1}{2}m \times 6^2 = -mg(1.375 \times 0.6) - 1.375 \times (0.25 \times mg \times 0.8)\) | M1 |
| Correct W-E equation | A1 |
| \(v^2 = 14.44 \Rightarrow v = 3.8\ (\mathrm{m\,s^{-1}})\) | A1 |
B1: Allow using an angle of 37 or better
soi (possibly in Work-Energy equation)
M1: Use of \(F = 0.25R\) with their \(R\) (with \(R \ne mg\)) – using an angle of 37 or better – allow sin/cos mix
\(R\) must be a component of weight not mass
soi (possibly in W-E equation)
B1: Correct expression for either WD by gravity or friction – using an angle of 37 or better
soi (possibly in W-E equation)
M1: Attempt at W-E principle – equation with the correct number of relevant terms, dimensionally correct, allow sign errors and sin/cos mix,
M0 if PE term is \(\pm 1.375mg\)
A1: Correct W-E equation
A1: AG – do not award this mark if using any non-exact values for \(\sin\theta / \cos\theta\)
Must see either \(v^2 = 14.44\) or \(v^2 = 36 - 21.56\) or \(v^2 = 6^2 - 21.56\) or a correct expression for \(v\) before given answer
SC – if \(m\) absent throughout (so implying \(m = 1\)) or \(m\) given a numerical value then at most B0 M1 B1 M1 A1 A0 can be awarded.
| Scheme | Marks | AO |
|---|---|---|
| Time from \(A\) to \(B\) is \(t_1\) where \(3.8 = 6 + (-7.84)t_1\) | M1 | 3.4 |
| \([t_1 =]\ 0.281\) (s) | A1 | 1.1 |
| [Height of \(P\) at \(B\) is] 0.825 (m) | B1 | 3.1b |
| Equation for time \(t_2\) (from \(B\) to \(C\)) | M1 | 3.1b |
| \(-0.825 = (3.8 \times 0.6)t_2 + 0.5(-g)t_2^{\,2}\) or \(-0.825 = (3.8 \times \sin 36.86\ldots)t_2 + 0.5(-g)t_2^{\,2}\) \(\left(4.9t_2^{\,2} - 2.28t_2 - 0.825 = 0\right)\) | A1 | 1.1 |
| \(T\,[= 0.280\ldots + 0.704\ldots] = 0.985\) | A1 | 2.2a |
| [6] |
Notes
M1: Use of \(v = u + at\) with correct \(u\) and \(v\), and their negative \(a\) from (a) (or any other complete method e.g. \(1.375 = \left(\dfrac{6 + 3.8}{2}\right)t_1\))
Give BOD if \(6 = 3.8 + 7.84t_1\) seen but if \(v = u + at\) stated before then M0
A1: awrt 0.281 or accept 0.28 www. If exact then expect \(\frac{55}{196}\)
0.2806122...
B1: Must be exact e.g. \(\frac{33}{40}\) – allow un-simplified e.g. \(1.375 \times 0.6\)
Allow negative
M1: M1 for an equation for time \(t_2\) (from \(B\) to \(C\)) with awrt \(\pm 0.83\) for \(s\), \(a = \pm g\) and either \(u = 3.8 \times \sin\theta\) or \(u = 3.8 \times \cos\theta\) with \(\theta = 37\) (or better) substituted (or corresponding exact value) – condone sign errors
A1: A1 for a correct (un-simplified) equation for \(t_2\) - must be using 0.825 but allow \(\theta = 37\) or better
For reference: \(t_2 = 0.704346599\ldots\)
A1: awrt 0.985 only – dependent on all previous marks in this part
0.984958…