| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2020 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Elastic string – horizontal circle on surface |
| Difficulty | Challenging +1.2 This is a structured Further Maths mechanics question with clear scaffolding through parts (a)-(e). Part (a) requires standard application of Hooke's law and circular motion (F=mv²/r), part (b) involves binomial approximation (A-level technique), and parts (c)-(e) are interpretative/numerical. While it's Further Maths content (inherently harder), the extensive guidance and routine application of formulas place it slightly above average difficulty rather than significantly challenging. |
| Spec | 4.04c Scalar product: calculate and use for angles6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.05a Angular velocity: definitions6.05b Circular motion: v=r*omega and a=v^2/r |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (a) | λx |
| Answer | Marks |
|---|---|
| ⇒λx2+λlx−lmv2 =0 | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.3 |
| Answer | Marks |
|---|---|
| 3.3 | Use of F = ma with |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (b) | −λl+ (λl)2−4λ(−lmv2) |
| Answer | Marks |
|---|---|
| 22 λl λ | M1 |
| Answer | Marks |
|---|---|
| [3] | 2.1 |
| Answer | Marks |
|---|---|
| 1.1 | Use of the quadratic equation |
| Answer | Marks |
|---|---|
| AG | 4mv2 |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (c) | λx mv2 |
| Answer | Marks | Guidance |
|---|---|---|
| and so the RHS would be the centripetal force | B1 | |
| [1] | 3.5a | |
| 8 | (d) | 260×0.03 |
| Answer | Marks | Guidance |
|---|---|---|
| 0.16 | B1 | |
| [1] | 3.4 | (6.982…) |
| 8 | (e) | While v in this situation is slightly below 7 |
| Answer | Marks |
|---|---|
| case the assumption would not be justified | B1 |
| [1] | 2.2b |
Question 8:
8 | (a) | λx
T = and r = l + x both used in solution
l
mv2
T =
l+x
λx mv2
= ⇒λx(l+x)=lmv2
l l+x
⇒λx2+λlx−lmv2 =0 | M1
M1
A1
[3] | 3.3
3.3
3.3 | Use of F = ma with
centripetal acceleration
AG
8 | (b) | −λl+ (λl)2−4λ(−lmv2)
x=
2λ
1
l 4mv2 2
x= 1+ −1
2 λl
l 1 4mv2
x= 1+ +...−1
2 2 λl
l 1 4mv2 mv2
x≈ = ⇒λx≈mv2
22 λl λ | M1
M1
A1
[3] | 2.1
3.1b
1.1 | Use of the quadratic equation
formula
Reject negative route and
rearranging to form with
1+
Use of binomial series
AG | 4mv2
No need to mention <1
λl
8 | (c) | λx mv2
λx≈mv2 ⇒ =T ≈ and if the string were
l l
inextensible, corresponding to an infinite value of λ
and x being 0, then l would be the radius of the motion
and so the RHS would be the centripetal force | B1
[1] | 3.5a
8 | (d) | 260×0.03
v≈ = awrt 7.0
0.16 | B1
[1] | 3.4 | (6.982…)
8 | (e) | While v in this situation is slightly below 7
nevertheless it is an estimate so we cannot be certain
that the modelled value does not exceed 7 in which
case the assumption would not be justified | B1
[1] | 2.2b
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8 One end of a light elastic string of natural length $l \mathrm {~m}$ and modulus of elasticity $\lambda \mathrm { N }$ is attached to a particle $A$ of mass $m \mathrm {~kg}$. The other end of the string is attached to a fixed point $O$ which is on a horizontal surface. The surface is modelled as being smooth and $A$ moves in a circular path around $O$ with constant speed $v \mathrm {~ms} ^ { - 1 }$. The extension of the string is denoted by $x \mathrm {~m}$.
\begin{enumerate}[label=(\alph*)]
\item Show that $x$ satisfies $\lambda x ^ { 2 } + \lambda | x - | m v ^ { 2 } = 0$.
\item By solving the equation in part (a) and using a binomial series, show that if $\lambda$ is very large then $\lambda \mathrm { x } \approx \mathrm { mv } ^ { 2 }$.
\item By considering the tension in the string, explain how the result obtained when $\lambda$ is very large relates to the situation when the string is inextensible.
The nature of the horizontal surface is such that the modelling assumption that it is smooth is justifiable provided that the speed of the particle does not exceed $7 \mathrm {~ms} ^ { - 1 }$.
In the case where $m = 0.16$ and $\lambda = 260$, the extension of the string is measured as being 3.0 cm .
\item Estimate the value of $v$.
\item Explain whether the value of $v$ means that the modelling assumption is necessarily justifiable in this situation.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2020 Q8 [9]}}