| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | String through hole – lower particle also moves in horizontal circle (conical pendulum below) |
| Difficulty | Challenging +1.2 This is a multi-step circular motion problem requiring force analysis on two connected particles, but follows standard Further Mechanics techniques. Part (a) involves resolving forces and applying circular motion equations (T = mrω²) to both particles, with some trigonometry. Part (b) requires understanding relative angular motion. While it has 7 marks total and requires careful setup, it's a fairly typical Further Mechanics question without requiring novel insight—students who've practiced conical pendulum and connected particle problems will recognize the approach. |
| Spec | 3.03k Connected particles: pulleys and equilibrium6.05a Angular velocity: definitions6.05b Circular motion: v=r*omega and a=v^2/r |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (a) | For A: T 1 .2 5 0 .9 2A = |
| Answer | Marks | Guidance |
|---|---|---|
| For B: ↕ Tcos = 2g | B1 | 3.4 |
| vertical for B | T = 24.5 = 2.5g |
| Answer | Marks | Guidance |
|---|---|---|
| B | B1 | 2.2a |
| the radius of B’s circle used | 0.36sin or 0.216 |
| Answer | Marks | Guidance |
|---|---|---|
| B B | M1 | 3.4 |
| Answer | Marks |
|---|---|
| B, | T 𝑠𝑖𝑛𝜃=1.5g |
| Answer | Marks | Guidance |
|---|---|---|
| A | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | for strings to realign need | |
| (35/6)t – (14/3)t = k | M1FT | 3.1b |
| Answer | Marks |
|---|---|
| difference of angular speed | Do not award for partially complete |
| Answer | Marks | Guidance |
|---|---|---|
| 7t/6 = => t = 6 /7 so time is awrt 2.69 s | A1 | 1.1 |
Question 7:
7 | (a) | For A: T 1 .2 5 0 .9 2A = | M1 | 3.3 | NII for A using a = r2 or v2/r or
v and correct values for m and
r
For B: ↕ Tcos = 2g | B1 | 3.4 | Correctly balancing forces in the
vertical for B | T = 24.5 = 2.5g
4 3
𝑐𝑜𝑠𝜃 = ,𝑠𝑖𝑛𝜃 =
5 5
r = (1.26 – 0.9)sin
B | B1 | 2.2a | Correct expression or value for
the radius of B’s circle used | 0.36sin or 0.216
Tsin = 2r 2
B B | M1 | 3.4 | NII for B using a = r2 or v2/r or
v and correct value for m, 𝑠𝑖𝑛𝜃
and their calculated r
B, | T 𝑠𝑖𝑛𝜃=1.5g
May be seen as one equation for B
divided by the other in which case
give 1st B1 2nd M1 if correct
2𝑟 𝜔2
𝐵 𝐵
𝑡𝑎𝑛𝜃 =
2𝑔
T = 24.5 => = 35/6 or awrt 5.83
B
and = 14/3 or awrt 4.67
A | A1 | 1.1
[5]
(b) | for strings to realign need
(35/6)t – (14/3)t = k | M1FT | 3.1b | Condone 180 or any integer
multiple of rads or 180 .
2𝜋
Condone, 𝑇 = ,
Δ𝜔
Need to see a calculation of
difference of angular speed | Do not award for partially complete
common multiple methods.
SC2 for fully correct common
multiple or ratio method, finding 4
half turns for one object and 5 half
turns for the other leading to t = 6
/7 oe. Must come from correct .
7t/6 = => t = 6 /7 so time is awrt 2.69 s | A1 | 1.1 | cao
[2]Two particles $A$ and $B$ are connected by a light inextensible string of length $1.26$ m. Particle $A$ has a mass of $1.25$ kg and moves on a smooth horizontal table in a circular path of radius $0.9$ m and centre $O$. The string passes through a small smooth hole at $O$. Particle $B$ has a mass of $2$ kg and moves in a horizontal circle as shown in the diagram. The angle that the portion of string below the table makes with the downwards vertical through $O$ is $\theta$, where $\cos\theta = \frac{4}{5}$ (see diagram).
\includegraphics{figure_7}
\begin{enumerate}[label=(\alph*)]
\item Determine the angular speed of $A$ and the angular speed of $B$. [5]
\end{enumerate}
At the start of the motion, $A$, $O$ and $B$ all lie in the same vertical plane.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the first subsequent time when $A$, $O$ and $B$ all lie in the same vertical plane. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2023 Q7 [7]}}