| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2014 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string equilibrium and statics |
| Difficulty | Standard +0.8 This M3 question requires multiple sophisticated techniques: 3D geometry to find string extensions, equilibrium with elastic strings on inclined planes, and energy methods with variable tension. The multi-stage problem with changing forces (string slack at B) and the need to carefully track geometry throughout makes this significantly harder than standard mechanics questions, though the individual components are A-level accessible. |
| Spec | 3.03u Static equilibrium: on rough surfaces6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(T\cos\beta = mg\sin\alpha\) | M1 | Resolving parallel to slope; \(\alpha\) is angle of slope, \(\beta = C\hat{A}B\); \(\alpha = 16.26°\), \(\beta = 53.13° - \alpha = 36.87°\) |
| \(0.8T = (9)(9.8)(0.28)\) | A1 | Accept \(\cos 36.9°\) etc |
| OR \(T\sin\gamma + R\cos\alpha = mg\) | M1 | Resolving vertically and horizontally; \(\gamma\) is between string and horizontal, \(\gamma = 53.13°\) (\(R\) is normal reaction) |
| \(T\cos\gamma = R\sin\alpha\) | ||
| \(0.8T + 0.96R = 9 \times 9.8\) | A1 | Both equations correct |
| \(0.6T = 0.28R\) | ||
| \(T = 30.87\) | A1 | Accept anything rounding to 31; Dep on M1A1 (may be implied) |
| \(T = \dfrac{\lambda(4.0 - 2.2)}{2.2}\) | B1 | Correct equation linking \(T\) and \(\lambda\) |
| Modulus of elasticity is \(37.73\) N | E1 [5] | Working must lead to \(37.73\) to 4 sf |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Resultant force is \(18\) N (up the slope) | M1 | Or \(18 + T\cos\beta - mg\sin\alpha\); accept positive direction indicated clearly on diagram |
| Acceleration is \(2\) ms\(^{-2}\) in direction AB | A1 [2] | Just '\(2\)' implies M1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| At B, tension is \(\dfrac{37.73 \times (2.5-2.2)}{2.2}\) \((= 5.145)\) | B1 | |
| \(18 + T_B\sin\alpha - mg\sin\alpha = ma\) | M1 | Equation of motion; at least two forces required for M1 |
| \(18 + 5.145 \times 0.28 - 9 \times 9.8 \times 0.28 = 9a\) | A1 | FT for wrong tension |
| Acceleration is \(0.584\) ms\(^{-2}\) in direction BA (3 sf) | A1 [4] | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| WD by force is \(18 \times 2.5\) \((= 45)\) | B1 | |
| EE at A is \(\dfrac{37.73 \times 1.8^2}{2 \times 2.2}\) \((= 27.783)\) | B1 | For either of these |
| EE at B is \(\dfrac{37.73 \times 0.3^2}{2 \times 2.2}\) \((= 0.77175)\) | ||
| Change in PE is \(9 \times 9.8 \times 0.7\) \((= 61.74)\) | B1 | |
| \(45 + 27.783 = 0.77175 + 61.74 + \frac{1}{2}(9)v^2\) | M1, A1 | Equation involving KE and at least two of WD, EE, PE; FT from any B0 above, but all 5 terms must be non-zero and all signs correct |
| Speed is \(1.51\) ms\(^{-1}\) (3 sf) | A1 [6] | CAO; dependent on previous 5 marks |
# Question 3:
## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $T\cos\beta = mg\sin\alpha$ | M1 | Resolving parallel to slope; $\alpha$ is angle of slope, $\beta = C\hat{A}B$; $\alpha = 16.26°$, $\beta = 53.13° - \alpha = 36.87°$ |
| $0.8T = (9)(9.8)(0.28)$ | A1 | Accept $\cos 36.9°$ etc |
| **OR** $T\sin\gamma + R\cos\alpha = mg$ | M1 | Resolving vertically and horizontally; $\gamma$ is between string and horizontal, $\gamma = 53.13°$ ($R$ is normal reaction) |
| $T\cos\gamma = R\sin\alpha$ | | |
| $0.8T + 0.96R = 9 \times 9.8$ | A1 | Both equations correct |
| $0.6T = 0.28R$ | | |
| $T = 30.87$ | A1 | Accept anything rounding to 31; Dep on M1A1 (may be implied) |
| $T = \dfrac{\lambda(4.0 - 2.2)}{2.2}$ | B1 | Correct equation linking $T$ and $\lambda$ |
| Modulus of elasticity is $37.73$ N | E1 [5] | Working must lead to $37.73$ to 4 sf |
## Part (ii)(A):
| Answer | Mark | Guidance |
|--------|------|----------|
| Resultant force is $18$ N (up the slope) | M1 | Or $18 + T\cos\beta - mg\sin\alpha$; accept positive direction indicated clearly on diagram |
| Acceleration is $2$ ms$^{-2}$ in direction AB | A1 [2] | Just '$2$' implies M1A0 |
## Part (ii)(B):
| Answer | Mark | Guidance |
|--------|------|----------|
| At B, tension is $\dfrac{37.73 \times (2.5-2.2)}{2.2}$ $(= 5.145)$ | B1 | |
| $18 + T_B\sin\alpha - mg\sin\alpha = ma$ | M1 | Equation of motion; at least two forces required for M1 |
| $18 + 5.145 \times 0.28 - 9 \times 9.8 \times 0.28 = 9a$ | A1 | FT for wrong tension |
| Acceleration is $0.584$ ms$^{-2}$ in direction BA (3 sf) | A1 [4] | CAO |
## Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| WD by force is $18 \times 2.5$ $(= 45)$ | B1 | |
| EE at A is $\dfrac{37.73 \times 1.8^2}{2 \times 2.2}$ $(= 27.783)$ | B1 | For either of these |
| EE at B is $\dfrac{37.73 \times 0.3^2}{2 \times 2.2}$ $(= 0.77175)$ | | |
| Change in PE is $9 \times 9.8 \times 0.7$ $(= 61.74)$ | B1 | |
| $45 + 27.783 = 0.77175 + 61.74 + \frac{1}{2}(9)v^2$ | M1, A1 | Equation involving KE and at least two of WD, EE, PE; FT from any B0 above, but all 5 terms must be non-zero and all signs correct |
| Speed is $1.51$ ms$^{-1}$ (3 sf) | A1 [6] | CAO; dependent on previous 5 marks |
---3 The fixed points A and B lie on a line of greatest slope of a smooth inclined plane, with B higher than A . The horizontal distance from A to B is 2.4 m and the vertical distance is 0.7 m . The fixed point C is 2.5 m vertically above B . A light elastic string of natural length 2.2 m has one end attached to C and the other end attached to a small block of mass 9 kg which is in contact with the plane. The block is in equilibrium when it is at A, as shown in Fig. 3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5a0df44f-f8f0-44d4-b2f6-70a5314706f9-4_712_641_488_687}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Show that the modulus of elasticity of the string is 37.73 N .
The block starts at A and is at rest. A constant force of 18 N , acting in the direction AB , is then applied to the block so that it slides along the line AB .
\item Find the magnitude and direction of the acceleration of the block\\
(A) when it leaves the point A ,\\
(B) when it reaches the point B .
\item Find the speed of the block when it reaches the point B .
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M3 2014 Q3 [17]}}