| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2008 |
| Session | January |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: tension at specific point |
| Difficulty | Standard +0.3 This is a standard M3 vertical circle problem with routine application of circular motion equations (F=mv²/r) and energy methods. Part (a) involves straightforward force resolution at two points, while part (b) requires Hooke's law and energy calculations but follows predictable patterns. The 'show that' in (b)(i) guides students through the algebra, and part (b)(ii) involves standard energy formulas with algebraic manipulation that, while slightly involved, is typical for M3 level. |
| Spec | 6.02g Hooke's law: T = k*x or T = lambda*x/l6.05c Horizontal circles: conical pendulum, banked tracks6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0.1 + 0.01 \times 9.8 = 0.01 \times \frac{u^2}{0.55}\) | M1 | Using acceleration \(u^2/0.55\) |
| A1 | ||
| Speed is \(3.3 \text{ ms}^{-1}\) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2}m(v^2 - u^2) = mg(2 \times 0.55 - 0.15)\) | M1 | Using conservation of energy |
| \(\frac{1}{2}(v^2 - 3.3^2) = 9.8 \times 0.95\) | A1 | *(ft is \(v^2 = u^2 + 18.62\))* |
| \(v^2 = 29.51\) | ||
| \(R - mg\cos\theta = m\frac{v^2}{a}\) | M1 | Forces and acceleration towards centre |
| \(R - 0.01 \times 9.8 \times \frac{0.4}{0.55} = 0.01 \times \frac{29.51}{0.55}\) | A1 | |
| Normal reaction is \(0.608 \text{ N}\) | A1 | [5] *(ft is \(\frac{u^2 + 22.54}{55}\))* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T = 0.8r\omega^2\) | B1 | |
| \(T = \frac{160}{2}(r-2)\) | B1 | |
| \(\omega^2 = \frac{80(r-2)}{0.8r} = \frac{100(r-2)}{r}\) | E1 | |
| \(\omega^2 = 100 - \frac{200}{r} < 100\), so \(\omega < 10\) | E1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{EE} = \frac{1}{2} \times \frac{160}{2} \times (r-2)^2 = 40(r-2)^2\) | B1 | |
| \(\text{KE} = \frac{1}{2}m(r\omega)^2\) | M1 | Use of \(\frac{1}{2}mv^2\) with \(v = r\omega\) |
| \(= \frac{1}{2} \times 0.8 \times r^2 \times \frac{100(r-2)}{r}\) | ||
| \(= 40r(r-2)\) | A1 | |
| Since \(r > r-2\), \(\quad 40r(r-2) > 40(r-2)^2\) | ||
| i.e. \(\text{KE} > \text{EE}\) | E1 | [4] From fully correct working only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| When \(\omega = 6\): \(\quad 36 = \frac{100(r-2)}{r}\), \(\quad r = 3.125\) | M1 | Obtaining \(r\) |
| \(T = 80(r-2) = 80(3.125 - 2)\) | M1 | |
| Tension is \(90 \text{ N}\) | A1 cao | [3] |
# Question 2:
## Part 2(a)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0.1 + 0.01 \times 9.8 = 0.01 \times \frac{u^2}{0.55}$ | M1 | Using acceleration $u^2/0.55$ |
| | A1 | |
| Speed is $3.3 \text{ ms}^{-1}$ | A1 | **[3]** |
## Part 2(a)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}m(v^2 - u^2) = mg(2 \times 0.55 - 0.15)$ | M1 | Using conservation of energy |
| $\frac{1}{2}(v^2 - 3.3^2) = 9.8 \times 0.95$ | A1 | *(ft is $v^2 = u^2 + 18.62$)* |
| $v^2 = 29.51$ | | |
| $R - mg\cos\theta = m\frac{v^2}{a}$ | M1 | Forces and acceleration towards centre |
| $R - 0.01 \times 9.8 \times \frac{0.4}{0.55} = 0.01 \times \frac{29.51}{0.55}$ | A1 | |
| Normal reaction is $0.608 \text{ N}$ | A1 | **[5]** *(ft is $\frac{u^2 + 22.54}{55}$)* |
## Part 2(b)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T = 0.8r\omega^2$ | B1 | |
| $T = \frac{160}{2}(r-2)$ | B1 | |
| $\omega^2 = \frac{80(r-2)}{0.8r} = \frac{100(r-2)}{r}$ | E1 | |
| $\omega^2 = 100 - \frac{200}{r} < 100$, so $\omega < 10$ | E1 | **[4]** |
## Part 2(b)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{EE} = \frac{1}{2} \times \frac{160}{2} \times (r-2)^2 = 40(r-2)^2$ | B1 | |
| $\text{KE} = \frac{1}{2}m(r\omega)^2$ | M1 | Use of $\frac{1}{2}mv^2$ with $v = r\omega$ |
| $= \frac{1}{2} \times 0.8 \times r^2 \times \frac{100(r-2)}{r}$ | | |
| $= 40r(r-2)$ | A1 | |
| Since $r > r-2$, $\quad 40r(r-2) > 40(r-2)^2$ | | |
| i.e. $\text{KE} > \text{EE}$ | E1 | **[4]** From fully correct working only |
## Part 2(b)(iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| When $\omega = 6$: $\quad 36 = \frac{100(r-2)}{r}$, $\quad r = 3.125$ | M1 | Obtaining $r$ |
| $T = 80(r-2) = 80(3.125 - 2)$ | M1 | |
| Tension is $90 \text{ N}$ | A1 cao | **[3]** |
---2
\begin{enumerate}[label=(\alph*)]
\item A small ball of mass 0.01 kg is moving in a vertical circle of radius 0.55 m on the smooth inside surface of a fixed sphere also of radius 0.55 m . When the ball is at the highest point of the circle, the normal reaction between the surface and the ball is 0.1 N . Modelling the ball as a particle and neglecting air resistance, find
\begin{enumerate}[label=(\roman*)]
\item the speed of the ball when it is at the highest point of the circle,
\item the normal reaction between the surface and the ball when the vertical height of the ball above the lowest point of the circle is 0.15 m .
\end{enumerate}\item A small object Q of mass 0.8 kg moves in a circular path, with centre O and radius $r$ metres, on a smooth horizontal surface. A light elastic string, with natural length 2 m and modulus of elasticity 160 N , has one end attached to Q and the other end attached to O . The object Q has a constant angular speed of $\omega$ rad s $^ { - 1 }$.
\begin{enumerate}[label=(\roman*)]
\item Show that $\omega ^ { 2 } = \frac { 100 ( r - 2 ) } { r }$ and deduce that $\omega < 10$.
\item Find expressions, in terms of $r$ only, for the elastic energy stored in the string, and for the kinetic energy of Q . Show that the kinetic energy of Q is greater than the elastic energy stored in the string.
\item Given that the angular speed of Q is $6 \mathrm { rad } \mathrm { s } ^ { - 1 }$, find the tension in the string.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI M3 2008 Q2 [19]}}