| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2015 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Multiple particles on string |
| Difficulty | Challenging +1.2 Part (a) is a standard vertical circular motion derivation using energy conservation and Newton's second law—routine for M3 students. Part (b) involves a more complex 3D conical pendulum system with multiple particles, requiring careful force resolution and geometry, but follows systematic methods without requiring novel insight. The multi-step nature and 3D geometry elevate it slightly above average difficulty. |
| Spec | 6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(T_1 + mg = m\frac{v_1^2}{a}\) | B1 | Condone \(r\) for \(a\) |
| \(\frac{1}{2}mv_2^2 = \frac{1}{2}mv_1^2 + mga(1-\cos\theta)\) | M1 A1 | Equation involving KE and PE |
| \(T_2 + mg\cos\theta = m\frac{v_2^2}{a}\) | B1 | Condone same \(v\) for B1B1 |
| \(T_2 - T_1 = \frac{mv_2^2}{a} - \frac{mv_1^2}{a} + mg(1-\cos\theta)\) | ||
| \(T_2 - T_1 = 2mg(1-\cos\theta) + mg(1-\cos\theta)\) | M1 | Eliminating \(v_1\) and \(v_2\); Dependent on previous M1 |
| \(T_2 = T_1 + 3mg(1-\cos\theta)\) | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(T_{QR}\cos 60° = 1.5 \times 9.8\) | M1 | Resolving vertically for R |
| Tension in QR is \(29.4\) N | A1 | |
| \(T_{AQ}\cos\theta = 0.9 \times 9.8 + T_{QR}\cos 60°\) | M1 | Resolving vertically for Q; \(\theta\) is the angle CAQ |
| \(\frac{12}{13}T_{AQ} = 8.82 + 14.7\) | A1 | FT e.g. \(T\cos 22.6 = 0.9\times9.8 + 29.4\cos 60°\) |
| Tension in AQ is \(25.48\) N | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(T_{QR}\sin 60° = 1.5(0.5 + 1.8\sin 60°)\omega^2\) | M1 A1 | Using \(1.5r\omega^2\); Allow \(1.5\frac{v^2}{r}\) provided \(v = r\omega\) also seen |
| Angular speed is \(2.87 \text{ rad s}^{-1}\) (3 sf) | A1 | FT is \(\sqrt{0.2804\, T_{QR}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(T_{CQ} + T_{AQ}\sin\theta - T_{QR}\sin 60° = (0.9)(0.5)\omega^2\) | B1B1 | For LHS and RHS (CAO); At most one error (FT including in B1's) and no missing terms |
| \(T_{CQ} + 25.48 \times \frac{5}{13} - 29.4\sin 60° = 0.45 \times 2.871^2\) | M1 | Numerical equation for \(T_{CQ}\) |
| Tension in CQ is \(19.4\) N (3 sf) | A1 | CAO |
# Question 2:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T_1 + mg = m\frac{v_1^2}{a}$ | B1 | Condone $r$ for $a$ |
| $\frac{1}{2}mv_2^2 = \frac{1}{2}mv_1^2 + mga(1-\cos\theta)$ | M1 A1 | Equation involving KE and PE |
| $T_2 + mg\cos\theta = m\frac{v_2^2}{a}$ | B1 | Condone same $v$ for B1B1 |
| $T_2 - T_1 = \frac{mv_2^2}{a} - \frac{mv_1^2}{a} + mg(1-\cos\theta)$ | | |
| $T_2 - T_1 = 2mg(1-\cos\theta) + mg(1-\cos\theta)$ | M1 | Eliminating $v_1$ and $v_2$; Dependent on previous M1 |
| $T_2 = T_1 + 3mg(1-\cos\theta)$ | E1 | |
## Part (b)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T_{QR}\cos 60° = 1.5 \times 9.8$ | M1 | Resolving vertically for R |
| Tension in QR is $29.4$ N | A1 | |
| $T_{AQ}\cos\theta = 0.9 \times 9.8 + T_{QR}\cos 60°$ | M1 | Resolving vertically for Q; $\theta$ is the angle CAQ |
| $\frac{12}{13}T_{AQ} = 8.82 + 14.7$ | A1 | FT e.g. $T\cos 22.6 = 0.9\times9.8 + 29.4\cos 60°$ |
| Tension in AQ is $25.48$ N | A1 | CAO |
## Part (b)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T_{QR}\sin 60° = 1.5(0.5 + 1.8\sin 60°)\omega^2$ | M1 A1 | Using $1.5r\omega^2$; Allow $1.5\frac{v^2}{r}$ provided $v = r\omega$ also seen |
| Angular speed is $2.87 \text{ rad s}^{-1}$ (3 sf) | A1 | FT is $\sqrt{0.2804\, T_{QR}}$ |
## Part (b)(iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T_{CQ} + T_{AQ}\sin\theta - T_{QR}\sin 60° = (0.9)(0.5)\omega^2$ | B1B1 | For LHS and RHS (CAO); At most one error (FT including in B1's) and no missing terms |
| $T_{CQ} + 25.48 \times \frac{5}{13} - 29.4\sin 60° = 0.45 \times 2.871^2$ | M1 | Numerical equation for $T_{CQ}$ |
| Tension in CQ is $19.4$ N (3 sf) | A1 | CAO |
---2
\begin{enumerate}[label=(\alph*)]
\item A particle P of mass $m$ is attached to a fixed point O by a light inextensible string of length $a$. P is moving without resistance in a complete vertical circle with centre O and radius $a$. When P is at the highest point of the circle, the tension in the string is $T _ { 1 }$. When OP makes an angle $\theta$ with the upward vertical, the tension in the string is $T _ { 2 }$. Show that
$$T _ { 2 } = T _ { 1 } + 3 m g ( 1 - \cos \theta ) .$$
\item The fixed point A is 1.2 m vertically above the fixed point C . A particle Q of mass 0.9 kg is joined to A , to C , and to a particle R of mass 1.5 kg , by three light inextensible strings of lengths $1.3 \mathrm {~m} , 0.5 \mathrm {~m}$ and 1.8 m respectively. The particle Q moves in a horizontal circle with centre C , and R moves in a horizontal circle at the same constant angular speed as Q , in such a way that $\mathrm { A } , \mathrm { C } , \mathrm { Q }$ and R are always coplanar. The string QR makes an angle of $60 ^ { \circ }$ with the downward vertical. This situation is shown in Fig. 2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{70a2c3ce-7bdb-4ddd-92fc-f7dcbdfdcfaf-3_579_1191_881_406}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Find the tensions in the strings QR and AQ .
\item Find the angular speed of the system.
\item Find the tension in the string CQ .
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI M3 2015 Q2 [18]}}