| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2012 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string on smooth inclined plane |
| Difficulty | Standard +0.8 This is a multi-part M3 question requiring energy methods with elastic strings on an inclined plane. Students must identify the equilibrium position, apply conservation of energy with both gravitational PE and elastic PE, and use calculus or energy principles to find maximum speed (which occurs at the natural length, not equilibrium). The problem requires careful bookkeeping of multiple energy terms and understanding that maximum speed occurs when net force is zero, making it moderately challenging but still a standard M3 exercise. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([0.8mgx/0.78 = mg(5/13)]\) | M1 | For resolving forces and using \(T = \lambda x/L\) at equilibrium position |
| \(x = 0.375\) | A1 | Accept 1.155 for \(e + l\) |
| \(\text{PE} = mg(0.78 + 0.375) \times 5/13\) | B1 FT | FT value of \(x\) |
| \(\text{EE} = 0.8mg \times 0.375^2 \div (2 \times 0.78)\) | B1 FT | FT value of \(x\) |
| \([\frac{1}{2}mv^2 = m(4.353... - 0.7067...)]\) | M1 | For using \(\frac{1}{2}mv^2 = \text{PE loss} - \text{EE gain}\) |
| Maximum speed is 2.70 ms\(^{-1}\) | A1 | |
| [6] | ||
| *Or* at extension \(x\): \(\text{PE} = mg(x+0.78) \times \frac{5}{13}\) | B1 | |
| \(\text{EE} = \frac{0.8mgx^2}{2 \times 0.78}\) | B1 | |
| \(mg(x+0.78)\times\frac{5}{13} = \frac{1}{2}mv^2 + \frac{0.8mgx^2}{2\times0.78}\) | M1 | For using \(\frac{1}{2}mv^2 = \text{PE loss} - \text{EE gain}\) |
| \(v^2 = -10.05x^2 + 7.53x + 5.88\) | ||
| \(v^2 = -10.05(x^2 - 0.749x - 0.585)\) | ||
| For attempting to complete square | M1 | |
| \(v^2 = -10.05((x-0.375)^2 - 0.726)\) | A1 | |
| Max speed is 2.70 ms\(^{-1}\) | A1 | Note: after getting equation for \(v^2\), can differentiate \(v^2\) wrt \(x\) M1; establish max at \(x = 0.375\) A1; Max speed 2.70 ms\(^{-1}\) A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(mg(0.78+x) \times 5/13 = 0.8mgx^2 \div (2 \times 0.78)\) | M1* | For using PE loss = EE gain; or \(mg(x) \times 5/13 = 0.8mg(x-0.78)^2 \div (2\times0.78)\) if \(PO = x\); or \(mg(x+0.78+0.375)\times5/13 = 0.8mg(x+0.375)^2\div(2\times0.78)\) if \(PO = x+0.78+0.375\) |
| \([x^2 - 0.75x - 0.585 = 0\) if \(x\) is extension\(]\) | *M1 | For arranging in quadratic form and attempting to solve; all necessary terms required |
| \(x = 1.2268\) so Distance is 2.01 m | A1 | \([x^2 - 2.31x + 0.6084 = 0\) if \(PO = x]\); \([20x^2 = 14.5125\) if \(PO = x + 0.78 + 0.375]\) |
| [4] | ||
| *Or* put \(v = 0\) in \(v^2\) equation from above and solve to get \(x = 1.23\;(+0.78) = 2.01\) m | M1A1ft, M1A1 |
# Question 5:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[0.8mgx/0.78 = mg(5/13)]$ | M1 | For resolving forces and using $T = \lambda x/L$ at equilibrium position |
| $x = 0.375$ | A1 | Accept 1.155 for $e + l$ |
| $\text{PE} = mg(0.78 + 0.375) \times 5/13$ | B1 FT | FT value of $x$ |
| $\text{EE} = 0.8mg \times 0.375^2 \div (2 \times 0.78)$ | B1 FT | FT value of $x$ |
| $[\frac{1}{2}mv^2 = m(4.353... - 0.7067...)]$ | M1 | For using $\frac{1}{2}mv^2 = \text{PE loss} - \text{EE gain}$ |
| Maximum speed is 2.70 ms$^{-1}$ | A1 | |
| | **[6]** | |
| *Or* at extension $x$: $\text{PE} = mg(x+0.78) \times \frac{5}{13}$ | B1 | |
| $\text{EE} = \frac{0.8mgx^2}{2 \times 0.78}$ | B1 | |
| $mg(x+0.78)\times\frac{5}{13} = \frac{1}{2}mv^2 + \frac{0.8mgx^2}{2\times0.78}$ | M1 | For using $\frac{1}{2}mv^2 = \text{PE loss} - \text{EE gain}$ |
| $v^2 = -10.05x^2 + 7.53x + 5.88$ | | |
| $v^2 = -10.05(x^2 - 0.749x - 0.585)$ | | |
| For attempting to complete square | M1 | |
| $v^2 = -10.05((x-0.375)^2 - 0.726)$ | A1 | |
| Max speed is 2.70 ms$^{-1}$ | A1 | Note: after getting equation for $v^2$, can differentiate $v^2$ wrt $x$ M1; establish max at $x = 0.375$ A1; Max speed 2.70 ms$^{-1}$ A1 |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $mg(0.78+x) \times 5/13 = 0.8mgx^2 \div (2 \times 0.78)$ | M1* | For using PE loss = EE gain; or $mg(x) \times 5/13 = 0.8mg(x-0.78)^2 \div (2\times0.78)$ if $PO = x$; or $mg(x+0.78+0.375)\times5/13 = 0.8mg(x+0.375)^2\div(2\times0.78)$ if $PO = x+0.78+0.375$ |
| $[x^2 - 0.75x - 0.585 = 0$ if $x$ is extension$]$ | *M1 | For arranging in quadratic form and attempting to solve; all necessary terms required |
| $x = 1.2268$ so Distance is 2.01 m | A1 | $[x^2 - 2.31x + 0.6084 = 0$ if $PO = x]$; $[20x^2 = 14.5125$ if $PO = x + 0.78 + 0.375]$ |
| | **[4]** | |
| *Or* put $v = 0$ in $v^2$ equation from above and solve to get $x = 1.23\;(+0.78) = 2.01$ m | M1A1ft, M1A1 | |
---5 One end of a light elastic string, of natural length 0.78 m and modulus of elasticity 0.8 mgN , is attached to a fixed point $O$ on a smooth plane inclined at angle $\alpha$ to the horizontal, where $\sin \alpha = \frac { 5 } { 13 }$. A particle $P$ of mass $m \mathrm {~kg}$ is attached to the other end of the string. $P$ is released from rest at $O$ and moves down the plane without reaching the bottom. Find\\
(i) the maximum speed of $P$ in the subsequent motion,\\
(ii) the distance of $P$ from $O$ when it is at its lowest point.
\hfill \mbox{\textit{OCR M3 2012 Q5 [10]}}