OCR Further Mechanics 2019 June — Question 4 9 marks

Exam BoardOCR
ModuleFurther Mechanics (Further Mechanics)
Year2019
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeParticle on cone surface – with string attached to vertex or fixed point
DifficultyChallenging +1.2 This is a standard Further Mechanics circular motion problem requiring application of Hooke's law and resolution of forces in a conical pendulum setup. While it involves multiple steps (finding tension, resolving forces vertically and horizontally, applying circular motion equations), the approach is methodical and follows a well-established template for this topic type. The geometry is straightforward (3-4-5 triangle), and no novel insight is required beyond applying standard techniques.
Spec6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks
4 A right circular cone \(C\) of height 4 m and base radius 3 m has its base fixed to a horizontal plane. One end of a light elastic string of natural length 2 m and modulus of elasticity 32 N is fixed to the vertex of \(C\). The other end of the string is attached to a particle \(P\) of mass 2.5 kg . \(P\) moves in a horizontal circle with constant speed and in contact with the smooth curved surface of \(C\). The extension of the string is 1.5 m .
  1. Find the tension in the string.
  2. Find the speed of \(P\).
Question 4:
AnswerMarks Guidance
4(a) 321.5
T  24
AnswerMarks
2M1
A1
AnswerMarks
[2]3.4
1.1x
Use T 
l
Obtain 24
AnswerMarks
(b)𝜃 = 36.9 (𝑜𝑟 53.1)
𝑟(= 3.5sin𝜃)= 2.1
TcosRsinmg
TsinRcosma
v2
a
r
𝑚𝑣2
25𝑇 = 20𝑚𝑔+15
𝑟
AnswerMarks
v = 2.48B1
B1
*M1
*M1
B1
dep
*M1
AnswerMarks
A12.1
3.4
3.3
1.1
1.1
1.1
AnswerMarks
1.1Or correct trig ratio given
Candidate may give θ between
horizontal so cos and sin values will
be the other way round
soi
m, g, r and/or T may appear as
values throughout
Resolving tension and contact force
vertically and balancing with weight
NII with 3 terms, components of
tension and contact force
v2
Use a to eliminate a and
r
introduce v
Solve simultaneously to eliminate R
AnswerMarks
Substitute values and find vθ is the angle between the
string and the downwards
vertical
Must have 3 terms; allow
sign/trig errors
Must have 3 terms; allow
sign/trig errors
Could use 𝑎 = 𝑟𝜔2 but must
see 𝑣 = 𝑟𝜔 as well
53
May find 𝑅 = and sub in
6
√154
Or
5
Alternate method
AnswerMarks Guidance
𝜃 = 36.9 (𝑜𝑟 53.1)B1 Or correct trig ratio given
string and the downwards
vertical
Candidate may give θ between
horizontal so cos and sin values will
be the other way round
AnswerMarks Guidance
𝑟(= 3.5sin𝜃)= 2.1B1 soi
m, g, r and/or T may appear as
values throughout
AnswerMarks Guidance
Acc component parallel to string isM1 Attempt to resolve acceleration
asin
AnswerMarks Guidance
T mgcosmasinM1 NII with 3 terms, parallel to string
sign/trig errors
v2
= m sin
AnswerMarks Guidance
rM1 v2
Use a to eliminate a and
r
introduce v
rT mgcos
v2 
AnswerMarks Guidance
msinM1 Rearrange to make v2 the subject or
solve for v
AnswerMarks
v = 2.48A1
[7]
θ is the angle between the
string and the downwards
vertical
Rearrange to make v2 the subject or
solve for v
Question 4:
4 | (a) | 321.5
T  24
2 | M1
A1
[2] | 3.4
1.1 | x
Use T 
l
Obtain 24
(b) | 𝜃 = 36.9 (𝑜𝑟 53.1)
𝑟(= 3.5sin𝜃)= 2.1
TcosRsinmg
TsinRcosma
v2
a
r
𝑚𝑣2
25𝑇 = 20𝑚𝑔+15
𝑟
v = 2.48 | B1
B1
*M1
*M1
B1
dep
*M1
A1 | 2.1
3.4
3.3
1.1
1.1
1.1
1.1 | Or correct trig ratio given
Candidate may give θ between
horizontal so cos and sin values will
be the other way round
soi
m, g, r and/or T may appear as
values throughout
Resolving tension and contact force
vertically and balancing with weight
NII with 3 terms, components of
tension and contact force
v2
Use a to eliminate a and
r
introduce v
Solve simultaneously to eliminate R
Substitute values and find v | θ is the angle between the
string and the downwards
vertical
Must have 3 terms; allow
sign/trig errors
Must have 3 terms; allow
sign/trig errors
Could use 𝑎 = 𝑟𝜔2 but must
see 𝑣 = 𝑟𝜔 as well
53
May find 𝑅 = and sub in
6
√154
Or
5
Alternate method
𝜃 = 36.9 (𝑜𝑟 53.1) | B1 | Or correct trig ratio given | θ is the angle between the
string and the downwards
vertical
Candidate may give θ between
horizontal so cos and sin values will
be the other way round
𝑟(= 3.5sin𝜃)= 2.1 | B1 | soi
m, g, r and/or T may appear as
values throughout
Acc component parallel to string is | M1 | Attempt to resolve acceleration
asin
T mgcosmasin | M1 | NII with 3 terms, parallel to string | Must have 3 terms; allow
sign/trig errors
v2
= m sin
r | M1 | v2
Use a to eliminate a and
r
introduce v
rT mgcos
v2 
msin | M1 | Rearrange to make v2 the subject or
solve for v
v = 2.48 | A1
[7]
θ is the angle between the
string and the downwards
vertical
Rearrange to make v2 the subject or
solve for v
4 A right circular cone $C$ of height 4 m and base radius 3 m has its base fixed to a horizontal plane. One end of a light elastic string of natural length 2 m and modulus of elasticity 32 N is fixed to the vertex of $C$. The other end of the string is attached to a particle $P$ of mass 2.5 kg .\\
$P$ moves in a horizontal circle with constant speed and in contact with the smooth curved surface of $C$. The extension of the string is 1.5 m .
\begin{enumerate}[label=(\alph*)]
\item Find the tension in the string.
\item Find the speed of $P$.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Mechanics 2019 Q4 [9]}}
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