| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2013 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Particle at midpoint of string between two horizontal fixed points: vertical motion |
| Difficulty | Standard +0.8 This is a substantial multi-part mechanics problem requiring: geometry to find string extensions, Hooke's law for tensions, resolving forces in 2D at equilibrium, energy methods for maximum speed, and Newton's second law for acceleration. The combination of elastic strings, 2D geometry, and energy conservation makes this significantly harder than average A-level mechanics, though the individual techniques are standard for M3. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Length of each string is 7.8 m | B1 | |
| \(T = \frac{728}{6.4}(7.8 - 6.4)\) | M1 | Using Hooke's law. Must use extension |
| Tension is 159.25 N | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(2T\cos\theta = mg\) | M1 | Resolving vertically |
| \(2 \times 159.25 \times \frac{5}{13} = m \times 9.8\) | A1 | FT |
| \(m = \frac{122.5}{9.8} = 12.5\) kg | E1 | Working must lead to 12.5 to 3 sf |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| New length of each string is 7.5 m | ||
| \(T = \frac{728}{6.4}(7.5 - 6.4) \, (= 125.125)\) | M1, A1 | Hooke's law with new extension. FT for incorrect \(T\) |
| \(mg - 2T\cos\theta = ma\) | M1, A1 | Vertical equation of motion (3 terms). FT for incorrect \(T\) |
| \(12.5 \times 9.8 - 2 \times 125.125 \times 0.28 = 12.5a\) | ||
| Acceleration is 4.19 ms\(^{-2}\) downwards (3 sf) | A1 | Some indication of downwards required |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| At maximum speed, acceleration is zero. Acceleration is zero in equilibrium position | B1 | Mention of zero acceleration. Reference to \(v^2 = \omega^2(A^2 - x^2)\), SHM, etc, will usually be B0 |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Change of PE is \(12.5 \times 9.8 \times 3 \, (= 367.5)\) | B1 | |
| Initial EE is \(2 \times \frac{728 \times 0.8^2}{2 \times 6.4} \, (= 72.8)\) | B1 | Allow one string (36.4) |
| Final EE is \(2 \times \frac{728 \times 1.4^2}{2 \times 6.4} \, (= 222.95)\) | B1 | Allow one string (111.475) |
| \(\frac{1}{2}(12.5)v^2 - 367.5 + 222.95 = 72.8\) | M1, A1 | Equation involving KE, PE and EE. FT from any B0 above. All signs must be correct. All terms must be non-zero |
| Maximum speed is 5.90 ms\(^{-1}\) (3 sf) | A1 | CAO |
| [6] |
### Part (i)
| Length of each string is 7.8 m | B1 | |
|---|---|---|
| $T = \frac{728}{6.4}(7.8 - 6.4)$ | M1 | Using Hooke's law. Must use extension |
| Tension is 159.25 N | A1 | |
| | [3] | |
|---|---|---|
### Part (ii)
| $2T\cos\theta = mg$ | M1 | Resolving vertically |
|---|---|---|
| $2 \times 159.25 \times \frac{5}{13} = m \times 9.8$ | A1 | FT |
| $m = \frac{122.5}{9.8} = 12.5$ kg | E1 | Working must lead to 12.5 to 3 sf |
| | [3] | |
|---|---|---|
### Part (iii)
| New length of each string is 7.5 m | | |
|---|---|---|
| $T = \frac{728}{6.4}(7.5 - 6.4) \, (= 125.125)$ | M1, A1 | Hooke's law with new extension. FT for incorrect $T$ |
|---|---|---|
| $mg - 2T\cos\theta = ma$ | M1, A1 | Vertical equation of motion (3 terms). FT for incorrect $T$ |
|---|---|---|
| $12.5 \times 9.8 - 2 \times 125.125 \times 0.28 = 12.5a$ | | |
| Acceleration is 4.19 ms$^{-2}$ downwards (3 sf) | A1 | Some indication of downwards required |
|---|---|---|
| | [5] | |
|---|---|---|
### Part (iv)
| At maximum speed, acceleration is zero. Acceleration is zero in equilibrium position | B1 | Mention of zero acceleration. Reference to $v^2 = \omega^2(A^2 - x^2)$, SHM, etc, will usually be B0 |
|---|---|---|
| | [1] | |
|---|---|---|
### Part (v)
| Change of PE is $12.5 \times 9.8 \times 3 \, (= 367.5)$ | B1 | |
|---|---|---|
| Initial EE is $2 \times \frac{728 \times 0.8^2}{2 \times 6.4} \, (= 72.8)$ | B1 | Allow one string (36.4) |
|---|---|---|
| Final EE is $2 \times \frac{728 \times 1.4^2}{2 \times 6.4} \, (= 222.95)$ | B1 | Allow one string (111.475) |
|---|---|---|
| $\frac{1}{2}(12.5)v^2 - 367.5 + 222.95 = 72.8$ | M1, A1 | Equation involving KE, PE and EE. FT from any B0 above. All signs must be correct. All terms must be non-zero |
|---|---|---|
| Maximum speed is 5.90 ms$^{-1}$ (3 sf) | A1 | CAO |
|---|---|---|
| | [6] | |
|---|---|---|3 Two fixed points X and Y are 14.4 m apart and XY is horizontal. The midpoint of XY is M . A particle P is connected to X and to Y by two light elastic strings. Each string has natural length 6.4 m and modulus of elasticity 728 N . The particle P is in equilibrium when it is 3 m vertically below M, as shown in Fig. 3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3f674569-7e99-4ba8-84f1-a1eb438e30ed-3_284_878_404_580}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
(i) Find the tension in each string when P is in the equilibrium position.\\
(ii) Show that the mass of P is 12.5 kg .
The particle P is released from rest at M , and moves in a vertical line.\\
(iii) Find the acceleration of P when it is 2.1 m vertically below M .\\
(iv) Explain why the maximum speed of P occurs at the equilibrium position.\\
(v) Find the maximum speed of P .
\hfill \mbox{\textit{OCR MEI M3 2013 Q3 [18]}}