| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Vertical elastic string: released from rest at natural length or above (string initially slack) |
| Difficulty | Standard +0.8 This is a challenging M3 question requiring energy conservation with elastic strings and then force analysis at a specific position. Part (i) needs careful setup of EPE and GPE with the 'just reaches O' condition. Part (ii) requires determining whether the string is slack or taut at 1.3m below release point, then applying F=ma with correct force directions. The multi-stage reasoning and potential for sign errors elevates this above standard mechanics questions. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03c Newton's second law: F=ma one dimension6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| By energy: \(\frac{30(d-0.6)^2}{2 \times 0.6} = 48 \times d\) | M1*, A1 | Attempt at elastic energy; allow M1 for \(\frac{30y^2}{(2)\times 0.6} = kd\); \(\frac{30x^2}{2\times 0.6} = 48(x+0.6)\) |
| \(25d^2 - 78d + 9 = 0\) or \(30d^2 - 93.6d + 10.8 = 0\) | *M1 | Get 3 term quadratic and attempt to solve; allow 1 slip or \(25x^2 - 48x - 28.8 = 0\) |
| \((d=)\ 3\ \text{(m)}\) | A1 [4] | Ignore \(d = 0.12\) unless given as answer; \((x=)\ 2.4\) leading to \((d=)\ 3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use \(F = ma\): \(48 - \frac{30\times(3-0.6-1.3)}{0.6} = (\pm)\frac{48}{g}a\) | M1, A1ft | ft their '3'; allow missing \(g\), allow \(1.3\) or \(0.6\) to be omitted |
| \((a=)(+/-)\ 1.43\) | A1 | \(1.4291666\) |
| upwards | A1 [4] | Depends on \(a\) being right; Using energy: \(a = v\frac{dv}{dx} = \frac{g}{48}(50x-72)\) M1A1 |
## Question 2:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| By energy: $\frac{30(d-0.6)^2}{2 \times 0.6} = 48 \times d$ | M1*, A1 | Attempt at elastic energy; allow M1 for $\frac{30y^2}{(2)\times 0.6} = kd$; $\frac{30x^2}{2\times 0.6} = 48(x+0.6)$ |
| $25d^2 - 78d + 9 = 0$ or $30d^2 - 93.6d + 10.8 = 0$ | *M1 | Get 3 term quadratic and attempt to solve; allow 1 slip or $25x^2 - 48x - 28.8 = 0$ |
| $(d=)\ 3\ \text{(m)}$ | A1 [4] | Ignore $d = 0.12$ unless given as answer; $(x=)\ 2.4$ leading to $(d=)\ 3$ |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use $F = ma$: $48 - \frac{30\times(3-0.6-1.3)}{0.6} = (\pm)\frac{48}{g}a$ | M1, A1ft | ft their '3'; allow missing $g$, allow $1.3$ or $0.6$ to be omitted |
| $(a=)(+/-)\ 1.43$ | A1 | $1.4291666$ |
| upwards | A1 [4] | Depends on $a$ being right; Using energy: $a = v\frac{dv}{dx} = \frac{g}{48}(50x-72)$ M1A1 |
---2 One end of a light elastic string, of natural length 0.6 m and modulus of elasticity 30 N , is attached to a fixed point $O$. A particle $P$ of weight 48 N is attached to the other end of the string. $P$ is released from rest at a point $d \mathrm {~m}$ vertically below $O$. Subsequently $P$ just reaches $O$.\\
(i) Find $d$.\\
(ii) Find the magnitude and direction of the acceleration of $P$ when it has travelled 1.3 m from its point of release.
\hfill \mbox{\textit{OCR M3 2014 Q2 [8]}}