| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2016 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Collision/impulse during SHM |
| Difficulty | Challenging +1.2 This is a substantial M3 question requiring multiple mechanics techniques (elastic strings, conservation of momentum, SHM identification and analysis), but follows a standard template with guided steps. The 'show that' parts provide checkpoints, and while part (iv) requires careful integration of energy/SHM concepts over multiple stages, the individual techniques are all standard M3 material without requiring novel insight or particularly complex algebra. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings6.03b Conservation of momentum: 1D two particles |
| Answer | Marks | Guidance |
|---|---|---|
| In equilibrium, \(mg = \frac{24mge}{0.6}\) | M1 | Working essential, with 24mg and 0.6 used |
| \((e = 0.025\) m); OP = 0.625 m | A1 2 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(V = 2.1\) (ms⁻¹) | M1, A1, A1 3 | Or \(\frac{1}{2}mv^2 = mga × 0.4\). Accept \(\sqrt{(0.8g)}\). Valid method shown – no slips. AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(4mg - \frac{24mg(x - 0.6)}{0.6} = 4ma\) | M1 | Or \(4mg - \frac{24mg(x + 0.1)}{0.6} = 4m\ddot{x}\). Allow sign error, \(m\) for 4m. \(4mg - \frac{24mgx}{0.6} = 4m\ddot{x}\) \(-10g(x - 0.1) = \ddot{x}\) |
| Answer | Marks | Guidance |
|---|---|---|
| amplitude = 0.225 (m) | A1, B1, M1, A1 5 | Accept \(a\) for \(\ddot{x}\); \(-10gx = \ddot{x}\). Allow their \(ω\); any \(x\) |
| Answer | Marks | Guidance |
|---|---|---|
| 0 = 1.99 – 9.8t | M1, M1, M1, M1, M1, M1, A1 7 | Falling dist 0.4 m: 0.285714 s. Must be their 2.8. Travel 0.075m to centre: 0.0343287 s. Half period: 0.31734878 s. Travel 0.1m above centre: 0.046523 s. Speed when string becomes slack 1.995307 m s⁻¹. Allow their values; allow cos. Allow for whole or part period. Allow their values; allow sin. Allow their values. Dep M6 |
| Time = 0.888 s |
## (i)
In equilibrium, $mg = \frac{24mge}{0.6}$ | M1 | Working essential, with 24mg and 0.6 used
$(e = 0.025$ m); OP = 0.625 m | A1 2 | AG
## (ii)
For O falling 0.4m
$v^2 = 2g × 0.4$
$v = 2.8$ (m s⁻¹)
For combined particle
$3mx2.8 = 4mV$
$V = 2.1$ (ms⁻¹) | M1, A1, A1 3 | Or $\frac{1}{2}mv^2 = mga × 0.4$. Accept $\sqrt{(0.8g)}$. Valid method shown – no slips. AG
## (iii)
When a distance $x$ below O
$4mg - \frac{24mg(x - 0.6)}{0.6} = 4ma$ | M1 | Or $4mg - \frac{24mg(x + 0.1)}{0.6} = 4m\ddot{x}$. Allow sign error, $m$ for 4m. $4mg - \frac{24mgx}{0.6} = 4m\ddot{x}$ $-10g(x - 0.1) = \ddot{x}$
$-10ga(x - 0.7) = \ddot{x}$
(SHM about) a point 0.7 m below O
2.1² = 10g(a² - 0.075²)
amplitude = 0.225 (m) | A1, B1, M1, A1 5 | Accept $a$ for $\ddot{x}$; $-10gx = \ddot{x}$. Allow their $ω$; any $x$ | Use of $v = a\omega$ gains M1
## (iv)
0.4 = $\frac{1}{2} × 9.8t^2$; or 2.8 = 9.8t
0.075 = 0.225 sin($\sqrt{98}$t)
$\frac{1}{2}T = \frac{π}{\sqrt{98}}$
0.1 = 0.225 sin($\sqrt{98}$t)
$v = (0.225\sqrt{98}\cos(\sqrt{98} × 0.046523)) (= 1.995)$
or $v^2 = 98(0.225^2 - 0.1^2)$
0 = 1.99 – 9.8t | M1, M1, M1, M1, M1, M1, A1 7 | Falling dist 0.4 m: 0.285714 s. Must be their 2.8. Travel 0.075m to centre: 0.0343287 s. Half period: 0.31734878 s. Travel 0.1m above centre: 0.046523 s. Speed when string becomes slack 1.995307 m s⁻¹. Allow their values; allow cos. Allow for whole or part period. Allow their values; allow sin. Allow their values. Dep M6
Time = 0.888 s |
---A particle $P$ of mass $m$ kg is attached to one end of a light elastic string of modulus of elasticity $24mg$ N and natural length $0.6$ m. The other end of the string is attached to a fixed point $O$; the particle $P$ rests in equilibrium at a point $A$ with the string vertical.
\begin{enumerate}[label=(\roman*)]
\item Show that the distance $OA$ is $0.625$ m. [2]
\end{enumerate}
Another particle $Q$, of mass $3m$ kg, is released from rest from a point $0.4$ m above $P$ and falls onto $P$. The two particles coalesce.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Show that the combined particle initially moves with speed $2.1$ m s$^{-1}$. [3]
\item Show that the combined particle initially performs simple harmonic motion, and find the centre of this motion and its amplitude. [5]
\item Find the time that elapses between $Q$ being released from rest and the combined particle first reaching the highest point of its subsequent motion. [7]
\end{enumerate}
\hfill \mbox{\textit{OCR M3 2016 Q7 [17]}}