| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2024 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: tension at specific point |
| Difficulty | Standard +0.8 This is a comprehensive vertical circle problem requiring energy conservation, resolution of forces in tangential/radial directions, and analysis of conical pendulum motion. While the individual techniques are standard Further Maths content, the multi-part structure, need to distinguish tension/compression, and the conceptual reasoning in part (e) about physical impossibility elevate it above routine exercises. The calculations are straightforward but the question tests deeper understanding of circular motion mechanics. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.05b Circular motion: v=r*omega and a=v^2/r6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (a) | Tangential: ( − ) 6 g s i n 4 0 = 6 a |
| Answer | Marks |
|---|---|
| opposite to the direction of motion | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1b |
| 2.2a | Using NII with a component of |
| Answer | Marks |
|---|---|
| the direction of calculated value. | Condone e.g. “on |
| Answer | Marks |
|---|---|
| (b) | Initial KE = ½6122 (J) |
| Answer | Marks |
|---|---|
| awrt 55.4 (ms–2 ) towards O | B1 |
| Answer | Marks |
|---|---|
| [6] | 1.1 |
| Answer | Marks |
|---|---|
| 1.2 | DR |
| Answer | Marks |
|---|---|
| clear magnitude and direction. | Could see GPE term split if zero PE |
| Answer | Marks |
|---|---|
| (c) | 𝑇−6𝑔cos40° = 6𝑎 = 6×"55.41…" |
| Answer | Marks |
|---|---|
| is under a tension of awrt 378 (N) | M1FT |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 2.1 | Using NII in the radial direction |
| Answer | Marks |
|---|---|
| (d) | T c o s 4 0 = 6 g |
| Answer | Marks |
|---|---|
| so speed of P is awrt 3.56 (ms–1 ) | M1 |
| Answer | Marks |
|---|---|
| [4] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | Balancing vertical forces |
| Answer | Marks |
|---|---|
| 2.4cos(40) or r = sin(40) | If candidates combine and |
| Answer | Marks |
|---|---|
| (e) | (If θ = 90° then the tension in the rod is |
| Answer | Marks | Guidance |
|---|---|---|
| force.) | B1 | |
| [1] | 2.4 | They need to reference: |
| Answer | Marks |
|---|---|
| weight | Do not allow mention of air |
Question 4:
4 | (a) | Tangential: ( − ) 6 g s i n 4 0 = 6 a
t
a = (–)6.2993... so tangential acceleration is
t
awrt 6.30 (ms–2 ) (tangential) to the circle and
opposite to the direction of motion | M1
A1
[2] | 3.1b
2.2a | Using NII with a component of
the weight. May be implied by
( − ) g s i n 4 0 = a
t
Accept descriptions like 130
degrees anticlockwise from
vertical or 50 degrees to the
vertical or 40 degrees from the
leftward horizontal. Need to see
clear stated interpretation of
the direction of calculated value. | Condone e.g. “on
tangent/tangentially and
downwards and/or left”. Condone
“perpendicular to the rod and
down”.
Do not allow “clockwise
tangentially” as implies
turning/rotation (not a linear
direction). Do not allow e.g.
tangent to the rod/radius etc.
Do not allow 6.3
(b) | Initial KE = ½6122 (J)
GPE at A = 6g2.4(1 – cos40°) (J)
½6v2 + 6g2.4(1 – cos40°) = “432”
=> v2 = 132.99... (=> v = 11.532... )
𝑣2 "11.532…"2
Radial: 𝑎 = =
𝑟
𝑟 2.4
= 55.41... so radial acceleration at A is
awrt 55.4 (ms–2 ) towards O | B1
B1
M1*
A1
M1dep
A1
[6] | 1.1
1.1
1.1
1.1
3.1b
1.2 | DR
432 J
Finding expression for GPE
which is consistent with total
initial energy. Do not condone
missing brackets.
If expressions seen with m’s
then all m’s cancelled in energy
equation award B1 B1 at that
stage. Need to see correct
substitution of values into each
term.
Setting up CoE equation with
inital KE and GPE + final KE at
A. Correct signs. Do not allow
6g 2.4 cos 40 only.
soi
Using correct form for radial
acceleration soi. FT their v from
energy considerations.
E.g. 50 degrees above negative
horizontal
If units are stated they must be
correct for A mark. Must give a
clear magnitude and direction. | Could see GPE term split if zero PE
not at initial level e.g. –
6g2.4cos40° J if total initial
energy given as
½6122 – 6g2.4 J
𝑣2 = 144−4.8𝑔(1−𝑐𝑜𝑠40°)
= 144−47.04(1
−𝑐𝑜𝑠40°)
Special case:
If candidates quote 𝑣2 = 𝑢2 −
2𝑔Δℎ without deriving it from
energy equation. B marks are not
awarded. Correct signs.
𝑣2 = 122 −2𝑔2.4(1−cos(40°))
gains M1 with values in
And then as main scheme.
(c) | 𝑇−6𝑔cos40° = 6𝑎 = 6×"55.41…"
𝑟
T = 377.53... so (because it is positive) the rod
is under a tension of awrt 378 (N) | M1FT
A1
[2] | 1.1
2.1 | Using NII in the radial direction
with a tension, a correct
component of the weight and
their a . (FT from part b)
r
Condone incorrect signs. Do not
allow sin/cos interchange.
(d) | T c o s 4 0 = 6 g
r = 2.4 sin 40°
6 v 2
T s i n 4 0 =
2 .4 s i n 4 0
v 2 = 2 .4 g s i n 4 0 t a n 4 0 = 12.685...
=> v = 3.56171...
so speed of P is awrt 3.56 (ms–1 ) | M1
B1
M1
A1
[4] | 1.1
1.1
3.1b
1.1 | Balancing vertical forces
soi T =76.8 N
Radius of horizontal circle soi (r
= 1.543)
Using NII in the horizontal with
a correct form for the centripetal
acceleration. Do not condone
r=2.4. Condone use of 𝑟 =
2.4cos(40) or r = sin(40) | If candidates combine and
rearrange equations first:
e.g.
𝑣2 = 𝑔𝑡𝑎𝑛(40)×2.4sin(40)
Gains B1M1M1 once values in.
𝑣2
Condone use of 𝑟𝜔2in place of
𝑟
for method mark. Must find v to
gain final mark.
(e) | (If θ = 90° then the tension in the rod is
horizontal because the acceleration must be
towards the centre of the circle and so there is)
no component of the tension to balance the
weight. (Thus, there would be an unbalanced
force.) | B1
[1] | 2.4 | They need to reference:
no *vertical* component of
tension
and
refer to unbalanced forces
or e.g
only *vertical* force acting is
weight | Do not allow mention of air
resistance
Do not allow ‘no tension in the rod
can counteract the weight’. Needs
to mention 0 component and be
clear about vertical.
Do not allow mass ‘acting’
Do not allow calculations that refer
to T being undefined.4 A particle, $P$, of mass 6 kg is attached to one end of a light inextensible rod of length 2.4 m . The other end of the rod is smoothly hinged at a fixed point $O$ and the rod is free to rotate in any direction.
Initially, $P$ is at rest, vertically below $O$, when it is projected horizontally with a speed of $12 \mathrm {~ms} ^ { - 1 }$. It subsequently describes complete vertical circles with $O$ as the centre.\\
\includegraphics[max width=\textwidth, alt={}, center]{05b479a4-4087-4332-924b-43b1aedbb4f2-3_611_517_536_246}
The angle that the rod makes with the downward vertical through $O$ at each instant is denoted by $\theta$ and $A$ is the point which $P$ passes through where $\theta = 40 ^ { \circ }$ (see diagram).
\begin{enumerate}[label=(\alph*)]
\item Find the tangential acceleration of $P$ at $A$, stating its direction.
\item Determine the radial acceleration of $P$ at $A$, stating its direction.
\item Find the magnitude of the force in the rod when $P$ is at $A$, stating whether the rod is in tension or compression.
The motion is now stopped when $P$ is at $A$, and $P$ is then projected in such a way that it now describes horizontal circles at a constant speed with $\theta = 40 ^ { \circ }$ (see diagram).\\
\includegraphics[max width=\textwidth, alt={}, center]{05b479a4-4087-4332-924b-43b1aedbb4f2-3_403_524_1877_242}
\item Find the speed of $P$.
\item Explain why, wherever $P$ 's motion is initiated from and whatever its initial velocity, it is not possible for $P$ to describe horizontal circles at constant speed with $\theta = 90 ^ { \circ }$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2024 Q4 [15]}}