M3 June 2017 Q3
3. A particle \(P\) of mass \(m\) kg is initially held at rest at the point \(O\) on a smooth plane which is inclined at 30\(^\circ\) to the horizontal. The particle is released from rest and slides down the plane against a force of magnitude \(\dfrac{1}{2}mx^2\) newtons acting towards \(O\), where \(x\) metres is the distance of \(P\) from \(O\).
| Scheme | Marks |
|---|---|
| \(mv\dfrac{\mathrm{d}v}{\mathrm{d}x} = mg\sin 30 - \dfrac{1}{2}mx^2\) | M1A1A1 |
| \(\dfrac{1}{2}v^2 = xg\sin 30 - \dfrac{1}{6}x^3\ \ (+c)\) | dM1A1ft |
| \(x = 3 \quad \dfrac{1}{2}v^2 = 3g\sin 30 - \dfrac{9}{2}\) | dM1 |
| \((v = 4.5166\ldots)\) | |
| \(v = 4.5\) or 4.52 \(\left(\text{m s}^{-1}\right)\) | A1cso |
| (7) |
Notes
M1 Attempt NL2 parallel to the plane. Acceleration must be \(v\dfrac{\mathrm{d}v}{\mathrm{d}x}\) and weight must be resolved. (Variable force not resolved.) \(m\) may be cancelled. Integrating \(a\) to obtain \(\dfrac{1}{2}v^2\) gains this mark by implication.
A1 A1 Deduct 1 mark for each error in the equation. Both signs incorrect on RHS is one error.
dM1 Attempt the integration (wrt \(x\)) of both sides of the equation. Depends on the first M mark.
A1ft Correct integration with or without the constant. Follow through their integrand.
dM1 Substitute \(x = 3\) in their integrated equation. Depends on both previous M marks.
A1cso Correct value of \(v\). Must be 2 or 3 sf. CSO: Evidence of a constant of integration must be seen. \(C\) included and then crossed out or disappearing is sufficient evidence.
Definite integration:
M1A1A1 as above
dM1A1ft For the integration - ignore any limits shown
dM1 Use of correct limits. No sub need be shown for 0.
A1 Correct value of \(v\). Must be 2 or 3 sf. CSO: Evidence of a zero lower limit must be seen.
By work-energy:
\(F\) is variable, so if no integral seen score 0/7
\(\dfrac{1}{2}v^2\ (-0) = xg\sin 30 - \displaystyle\int \frac{1}{2}x^2\,\mathrm{d}x\ldots\) M1A1A1
\(\dfrac{1}{2}v^2\ (-0) = xg\sin 30 - \dfrac{1}{6}x^3\) M1A1
For the final A mark, evidence of initial KE being 0 must be seen.
| Scheme | Marks |
|---|---|
| \(v = 0 \Rightarrow x^2 = 6g\sin 30\ \ (x \neq 0)\) | |
| \(x = 5.4\) or 5.42 (m) | M1A1 |
| (2) | |
| (9 marks) |
Notes
M1 Substitute \(v = 0\) in their equation for \(v^2\) (from (a)) and obtain a numerical value of \(x\)
A1 Correct value of \(x\). Must be 2 or 3 sf. Do not penalise missing constant here.