AS June 2025 Q3
3. A plane is inclined to the horizontal at an angle \(\alpha\), where \(\tan\alpha = \dfrac{4}{3}\). A small block of mass \(m\) is held at a point \(A\) on the plane and released from rest.
Initially the block is modelled as a particle, air resistance is modelled as being negligible and the plane is modelled as being smooth.
In a refined model, the block is again modelled as a particle and air resistance is modelled as being negligible, but the plane is modelled as being rough with the coefficient of friction between the block and the plane being \(\dfrac{2}{3}\)
Using the refined model,
| Scheme | Marks | AO |
|---|---|---|
| Use of conservation of energy principle: \(\dfrac{1}{2}m \times 7^2 = mgh\) | M1 | 3.4 |
| \(\dfrac{1}{2}m \times 7^2 = mgd\sin\alpha\) | A1 | 1.1b |
| \(d = 3.1\) or 3.13 (m) | A1 | 1.1b |
| (3) |
Notes
N.B. Only penalise overaccuracy in answer to (a) or fractional answers to (a) and (c) ONCE for the question.
Consistent use of \(\sin\alpha = \dfrac{3}{5}\) and \(\cos\alpha = \dfrac{4}{5}\) could be treated as a MR of \(\tan\alpha = \dfrac{4}{3}\) for \(\tan\alpha = \dfrac{3}{4}\), if there is no evidence to the contrary e.g. a correct triangle, and leads to (a) \(d = 4.17\) or 4.2 (m) (b) \(\dfrac{8mg}{15}\) (c) \(x = 37.5\) or 38 (m). It can score MAX (a) M1A1A0 (b) M1M1A0 (c) B1M1A1A1
M1: Correct no. of terms, dimensionally correct, condone sin/cos confusion and sign errors, allow consistent missing \(m\)’s.
N.B. \(h\) does not need to be substituted.
A1: A correct equation in (\(m\)), \(d\) and \(\alpha\), seen or implied.
N.B. Could be e.g. \(2.5 = d\sin\alpha\) if they find \(h\) first.
A1: Either answer.
25/8 is A0 as is \(245/8g\)
| Scheme | Marks | AO |
|---|---|---|
| Resolve perpendicular to the plane: \(R = mg\cos\alpha\) | M1 | 3.1b |
| Use of \(F = \dfrac{2}{3}R\) | M1 | 1.2 |
| \(F = \dfrac{2}{5}mg\) or \(\dfrac{6}{15}mg\) or \(0.4mg\) (must be in terms of \(m\) and \(g\)) | A1 | 1.1b |
| (3) |
Notes
N.B. Only penalise overaccuracy in answer to (a) or fractional answers to (a) and (c) ONCE for the question.
Consistent use of \(\sin\alpha = \dfrac{3}{5}\) and \(\cos\alpha = \dfrac{4}{5}\) could be treated as a MR of \(\tan\alpha = \dfrac{4}{3}\) for \(\tan\alpha = \dfrac{3}{4}\), if there is no evidence to the contrary e.g. a correct triangle, and leads to (a) \(d = 4.17\) or 4.2 (m) (b) \(\dfrac{8mg}{15}\) (c) \(x = 37.5\) or 38 (m). It can score MAX (a) M1A1A0 (b) M1M1A0 (c) B1M1A1A1
M1: Correct no. of terms, dimensionally correct, condone sin/cos confusion.
M1: Use of \(F = \dfrac{2}{3}R\)
A1: cao
| Scheme | Marks | AO |
|---|---|---|
| WD against friction \(= Fx\) or \(F\left(\dfrac{h}{\sin\alpha}\right)\) | B1 | 3.4 |
| Use of work-energy principle: \(mgh - \dfrac{1}{2}m \times 7^2 = \dfrac{2}{5}mgx\) | M1 | 3.1b |
| \(mgx\sin\alpha - \dfrac{1}{2}m \times 7^2 = \dfrac{2}{5}mgx\) | A1 | 1.1b |
| \(x = 6.3\) or 6.25 (m) | A1 | 1.1b |
| (4) | ||
| (10 marks) |
Notes
N.B. Only penalise overaccuracy in answer to (a) or fractional answers to (a) and (c) ONCE for the question.
Consistent use of \(\sin\alpha = \dfrac{3}{5}\) and \(\cos\alpha = \dfrac{4}{5}\) could be treated as a MR of \(\tan\alpha = \dfrac{4}{3}\) for \(\tan\alpha = \dfrac{3}{4}\), if there is no evidence to the contrary e.g. a correct triangle, and leads to (a) \(d = 4.17\) or 4.2 (m) (b) \(\dfrac{8mg}{15}\) (c) \(x = 37.5\) or 38 (m). It can score MAX (a) M1A1A0 (b) M1M1A0 (c) B1M1A1A1
B1: Seen or implied, \(F\) does not need to be substituted but must be \(F\) not \(R\).
M1: Correct no. of terms, dimensionally correct, condone sin/cos confusion and sign errors, allow consistent missing \(m\)’s.
N.B. Allow if they clearly make a slip and use \(Rd\) instead of \(Fd\) for WD against friction.
N.B. M0 if they use \(h\) from part (a) or any other numerical value for \(h\).
A1: Correct equation in (\(m\)), \(x\) and \(\alpha\), seen or implied.
N.B. Could be e.g. \(5 = x\sin\alpha\) if they find \(h\) first.
A1: Either answer.
25/4 is A0