| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2015 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: complete revolution conditions |
| Difficulty | Standard +0.3 This is a standard M3 circular motion problem requiring energy conservation and Newton's second law for circular motion. Part (i) is routine derivation combining these principles, part (ii) is the classic 'complete circle' condition (T=0 at top), and part (iii) applies the same methods with given values. All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration |
| Answer | Marks |
|---|---|
| M1 | must have the right 3 terms; allow sign error/ sin for cos for M1 |
| A1 | |
| M1 | |
| A1 | |
| A1 [5] | (AG) with no errors and no gaps in argument |
| Answer | Marks |
|---|---|
| M1 | allow inequalities for M1A1 |
| A1 | |
| A1 [3] | |
| B1 | allow inequalities for B1M1 |
| M1 | |
| A1 |
| Answer | Marks |
|---|---|
| M1 | |
| A1 | might see \(\theta = 99.6°\) or 1.74 radians |
| M1 | accept use of their \(\theta\) |
| A1 [4] | |
| M1 | |
| A1 |
### Part (i)
**Answer:** using $F = ma$
$T - 0.2g\cos\theta = 0.2v^2/0.5$
by energy
$\frac{1}{2} \times 0.2u^2 = \frac{1}{2}0.2v^2 + 0.2g \times 0.5(1 - \cos\theta)$
$T = 5.88\cos\theta + 0.4u^2 - 3.92$
| M1 | must have the right 3 terms; allow sign error/ sin for cos for M1 |
| A1 | |
| M1 | |
| A1 | |
| A1 [5] | (AG) with no errors and no gaps in argument |
**Guidance:** $v^2 = u^2 - 9.8(1 - \cos\theta)$
### Part (ii)
**Answer:** when $\theta = 180°$, $5.88\cos\theta + 0.4u^2 - 3.92 = 0$
$-5.88 + 0.4u^2 - 3.92 = 0$
min $u$ is 4.95 (m s$^{-1}$)
OR, at top, $mg = \frac{mu^2}{r}$, so $v^2 = 0.5g$
by energy $\frac{1}{2}0.2u^2 = \frac{1}{2}0.2 \times 0.5g + 0.2g$
min $u$ is 4.95 (m s$^{-1}$)
| M1 | allow inequalities for M1A1 |
| A1 | |
| A1 [3] | |
| | |
| B1 | allow inequalities for B1M1 |
| M1 | |
| A1 | |
**Guidance:** $\frac{7}{2}\sqrt{2}$; 4.9497... Not > 4.95
### Part (iii)
**Answer:** $5.88\cos\theta + 0.4 \times 12.25 - 3.92 = 0$
$\cos\theta = (3.92 - 4.9)/5.88$ (= $-1/6$)
use energy eq$^n$ from (i)
$\frac{1}{2}0.2 \times 3.5^2 = \frac{1}{2}0.2v^2 + 0.2g \times 0.5(1 - \cos\theta)$
$v = 0.904$ m s$^{-1}$
OR use $T$ equation from (i)
$0 - 0.2g(-1/6) = 0.2v^2/0.5$
$v = 0.904$ m s$^{-1}$
| M1 | |
| A1 | might see $\theta = 99.6°$ or 1.74 radians |
| M1 | accept use of their $\theta$ |
| A1 [4] | |
| | |
| M1 | |
| A1 | |
**Guidance:** 99.49406...°, 1.73824...rads; 0.903696...\includegraphics{figure_7}
One end of a light inextensible string of length $0.5$ m is attached to a fixed point $O$. A particle $P$ of mass $0.2$ kg is attached to the other end of the string. $P$ is projected horizontally from the point $0.5$ m below $O$ with speed $u\text{ ms}^{-1}$. When the string makes an angle of $\theta$ with the downward vertical the particle has speed $v\text{ ms}^{-1}$ (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Show that, while the string is taut, the tension, $T$ N, in the string is given by
$$T = 5.88\cos \theta + 0.4u^2 - 3.92.$$ [5]
\item Find the least value of $u$ for which the particle will move in a complete circle. [3]
\item If in fact $u = 3.5\text{ ms}^{-1}$, find the speed of the particle at the point where the string first becomes slack. [4]
\end{enumerate}
END OF QUESTION PAPER
\hfill \mbox{\textit{OCR M3 2015 Q7 [12]}}