| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2010 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | SHM on inclined plane |
| Difficulty | Challenging +1.2 This is a structured SHM question with clear guidance through each part. While it requires understanding of elastic strings on an inclined plane and involves multiple steps (finding equilibrium, deriving tensions, proving SHM, finding period and specific motion), the question provides scaffolding at each stage. The mathematical manipulation is straightforward once the setup is understood, making it moderately above average but not requiring exceptional insight. |
| Spec | 3.03v Motion on rough surface: including inclined planes4.10f Simple harmonic motion: x'' = -omega^2 x6.02g Hooke's law: T = k*x or T = lambda*x/l6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T_{AP} = \dfrac{7.35}{1.5}\times0.05 \; (= 0.245)\) | M1, A1 | Using Hooke's law; or \(\dfrac{7.35}{1.5}(AP - 1.5)\) |
| \(T_{BP} = \dfrac{7.35}{2.5}\times0.5 \; (= 1.47)\) | A1 | or \(\dfrac{7.35}{2.5}(2.05 - AP)\) |
| Resultant force up the plane: \(T_{BP} - T_{AP} - mg\sin30°\) \(= 1.47 - 0.245 - 0.25\times9.8\sin30°\) \(= 1.47 - 0.245 - 1.225 = 0\) | M1 | |
| Hence there is no acceleration | E1 | Correctly shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T_{AP} = \dfrac{7.35}{1.5}(0.05 + x) \; (= 0.245 + 4.9x)\) | B1 | |
| \(T_{BP} = \dfrac{7.35}{2.5}(4.55 - 1.55 - x - 2.5) = 2.94(0.5 - x)\) | M1 | |
| \(= 1.47 - 2.94x\) | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T_{BP} - T_{AP} - mg\sin30° = m\dfrac{d^2x}{dt^2}\) | M1 | Equation of motion parallel to plane |
| \((1.47 - 2.94x) - (0.245 + 4.9x) - 1.225 = 0.25\dfrac{d^2x}{dt^2}\) | A2 | Give A1 for an equation correct apart from sign errors |
| \(\dfrac{d^2x}{dt^2} = -31.36x\) | ||
| Hence the motion is simple harmonic | E1 | Must state conclusion. Working must be fully correct (cao); If \(a\) is used for accn down plane, then \(a = 31.36x\) can earn M1A2; but E1 requires comment about directions |
| Period is \(\dfrac{2\pi}{\sqrt{31.36}} = \dfrac{2\pi}{5.6}\); Period is \(1.12\) s (3 sf) | B1 cao | Accept \(\dfrac{5\pi}{14}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = -0.05\cos5.6t\) | M1, A1 | For \(A\sin\omega t\) or \(A\cos\omega t\); Allow \(\pm0.05\sin/\cos5.6t\); Implied by \(v = \pm0.28\sin/\cos5.6t\) |
| \(v = 0.28\sin5.6t\) | ||
| \(-0.2 = 0.28\sin5.6t\) OR \(0.2^2 = 31.36(0.05^2 - x^2)\), \(x = (\pm)0.035\) | M1 | Using \(v = \pm0.2\) to obtain an equation for \(t\) |
| \(0.035 = -0.05\cos5.6t\) | M1 | |
| \(5.6t = \pi + 0.7956\) | M1 | Fully correct strategy for finding the required time |
| Time is \(0.703\) s (3 sf) | A1 cao |
# Question 4:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T_{AP} = \dfrac{7.35}{1.5}\times0.05 \; (= 0.245)$ | M1, A1 | Using Hooke's law; or $\dfrac{7.35}{1.5}(AP - 1.5)$ |
| $T_{BP} = \dfrac{7.35}{2.5}\times0.5 \; (= 1.47)$ | A1 | or $\dfrac{7.35}{2.5}(2.05 - AP)$ |
| Resultant force up the plane: $T_{BP} - T_{AP} - mg\sin30°$ $= 1.47 - 0.245 - 0.25\times9.8\sin30°$ $= 1.47 - 0.245 - 1.225 = 0$ | M1 | |
| Hence there is no acceleration | E1 | Correctly shown |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T_{AP} = \dfrac{7.35}{1.5}(0.05 + x) \; (= 0.245 + 4.9x)$ | B1 | |
| $T_{BP} = \dfrac{7.35}{2.5}(4.55 - 1.55 - x - 2.5) = 2.94(0.5 - x)$ | M1 | |
| $= 1.47 - 2.94x$ | E1 | |
## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T_{BP} - T_{AP} - mg\sin30° = m\dfrac{d^2x}{dt^2}$ | M1 | Equation of motion parallel to plane |
| $(1.47 - 2.94x) - (0.245 + 4.9x) - 1.225 = 0.25\dfrac{d^2x}{dt^2}$ | A2 | Give A1 for an equation correct apart from sign errors |
| $\dfrac{d^2x}{dt^2} = -31.36x$ | | |
| Hence the motion is simple harmonic | E1 | Must state conclusion. Working must be fully correct (cao); If $a$ is used for accn down plane, then $a = 31.36x$ can earn M1A2; but E1 requires comment about directions |
| Period is $\dfrac{2\pi}{\sqrt{31.36}} = \dfrac{2\pi}{5.6}$; Period is $1.12$ s (3 sf) | B1 cao | Accept $\dfrac{5\pi}{14}$ |
## Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = -0.05\cos5.6t$ | M1, A1 | For $A\sin\omega t$ or $A\cos\omega t$; Allow $\pm0.05\sin/\cos5.6t$; Implied by $v = \pm0.28\sin/\cos5.6t$ |
| $v = 0.28\sin5.6t$ | | |
| $-0.2 = 0.28\sin5.6t$ OR $0.2^2 = 31.36(0.05^2 - x^2)$, $x = (\pm)0.035$ | M1 | Using $v = \pm0.2$ to obtain an equation for $t$ |
| $0.035 = -0.05\cos5.6t$ | M1 | |
| $5.6t = \pi + 0.7956$ | M1 | Fully correct strategy for finding the required time |
| Time is $0.703$ s (3 sf) | A1 cao | |4 Fig. 4 shows a smooth plane inclined at an angle of $30 ^ { \circ }$ to the horizontal. Two fixed points A and B on the plane are 4.55 m apart with B higher than A on a line of greatest slope. A particle P of mass 0.25 kg is in contact with the plane and is connected to A and to B by two light elastic strings. The string AP has natural length 1.5 m and modulus of elasticity 7.35 N ; the string BP has natural length 2.5 m and modulus of elasticity 7.35 N . The particle P moves along part of the line AB , with both strings taut throughout the motion.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{023afdfb-21b6-40fe-9a09-e6769667ee7b-4_598_1006_568_571}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
(i) Show that, when $\mathrm { AP } = 1.55 \mathrm {~m}$, the acceleration of P is zero.\\
(ii) Taking $\mathrm { AP } = ( 1.55 + x ) \mathrm { m }$, write down the tension in the string AP , in terms of $x$, and show that the tension in the string BP is $( 1.47 - 2.94 x ) \mathrm { N }$.\\
(iii) Show that the motion of P is simple harmonic, and find its period.
The particle P is released from rest with $\mathrm { AP } = 1.5 \mathrm {~m}$.\\
(iv) Find the time after release when P is first moving down the plane with speed $0.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\hfill \mbox{\textit{OCR MEI M3 2010 Q4 [18]}}