OCR Further Mechanics 2023 June — Question 4 9 marks

Exam BoardOCR
ModuleFurther Mechanics (Further Mechanics)
Year2023
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHooke's law and elastic energy
TypeElastic string equilibrium and statics
DifficultyStandard +0.3 This is a straightforward centre of mass problem requiring standard techniques: symmetry argument, decomposition into triangles with known centroids, and equilibrium of a suspended lamina. All methods are routine for Further Maths students with no novel insight required, making it slightly easier than average.
Spec6.04a Centre of mass: gravitational effect6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces
\(ABCD\) is a uniform lamina in the shape of a kite with \(BA = BC = 0.37\) m, \(DA = DC = 0.91\) m and \(AC = 0.7\) m (see diagram). The centre of mass of \(ABCD\) is \(G\). \includegraphics{figure_4}
  1. Explain why \(G\) lies on \(BD\). [1]
  2. Show that the distance of \(G\) from \(B\) is \(0.36\) m. [4]
The lamina \(ABCD\) is freely suspended from the point \(A\).
  1. Determine the acute angle that \(CD\) makes with the horizontal, stating which of \(C\) or \(D\) is higher. [4]
Question 4:
AnswerMarks Guidance
4(a) (BD is a line of) symmetry (of the kite)
[1]
AnswerMarks Guidance
(b)h = 0 .3 7 2 βˆ’ 0 .3 5 2 = 0 .1 2 so area = 0.042
A B CB1 3.1b
triangle ABD) with X as the point
where the diagonals meet:
β„Ž = 0.12 so area = 0.021
𝐴𝐡𝑋
h = 0 .9 1 2 βˆ’ 0 .3 5 2 = 0 .8 4 so area = 0.294
AnswerMarks Guidance
A D CB1 1.1
𝐴𝑋𝐢
Measuring from B:
( 0 .0 4 2 + 0 .2 9 4 ) x
AnswerMarks Guidance
= 0 .0 4 2 ο‚΄ 0 .0 8 + 0 .2 9 4 ο‚΄ ( 0 .1 2 + 0 .2 8 )M1 1.1
about any point with at least two
non-zero terms each comprising
the product of a force or mass
with an appropriate distance.
May be one error in the distance.
Measuring from AC
(0.042+0.294)π‘₯Μ„
= βˆ’0.04Γ—0.042
+0.28Γ—0.294
0.08064
Gives π‘₯Μ… = = 0.24
0.336
Final step: add 0.12 to measure
AnswerMarks
from BMeasuring from B:
(0.021+0.147)π‘₯Μ…
2
=0.021Γ— Γ—0.12
3
+0.147
1
Γ—(0.12+ Γ—0.84)
3
0 .1 2 0 9 6
x = = 0 .3 6
AnswerMarks Guidance
0 .3 3 6A1 1.1
working must be seen.0.06048
π‘₯Μ… = = 0.36 AG
0.168
[4]
AnswerMarks
(b)Alternative method
E.g. Taking B as (0,0)
h = 0 .3 7 2 βˆ’ 0 .3 5 2 = 0 .1 2 so A (0.12, 0.35)
AnswerMarks Guidance
A B CB1 1.1
may be on diagram. Award if
average of trianlge coords
method used
h = 0 .9 1 2 βˆ’ 0 .3 5 2 = 0 .8 4 so D (0.96,0)
AnswerMarks Guidance
A D CB1 1.1
1
From B, π‘₯ = (0+0.12+0.96)
AnswerMarks Guidance
3M1 1.1
1.08
= = 0.36 as triangle BAD and triangle
3
AnswerMarks
BCD are symmetrical1.08
= = 0.36 as triangle BAD and triangle
AnswerMarks Guidance
3A1 1.1
mention of symmetry along BD
BCD are symmetrical
must be seen.
[4]
AnswerMarks
(c)0 .2 4
t a n (  C A G ) =
AnswerMarks Guidance
0 .3 5B1 1.1
0 .8 4
1 8 0 9 0 " 3 4 .4 " t a n 1  = ο‚° βˆ’ ο‚° βˆ’ ο‚° βˆ’ βˆ’
AnswerMarks Guidance
0 .3 5M1 3.1a
∠𝐴𝐢𝐷 = =
0.35
π‘‘π‘Žπ‘›βˆ’112
,67.4Β° (3 sf) Allow a
5
AnswerMarks
sign error.22.6Β°=π‘‘π‘Žπ‘›βˆ’10.35
May see ∠𝐡𝐷𝐢=
0.84
Allow for method to gain 101.8
degrees with the vertical seen
AnswerMarks
180Β°βˆ’"55.6Β°"βˆ’π‘‘π‘Žπ‘›βˆ’10.35
0.84
Allow for combination of one of
(∠𝐢𝐴𝐺 or ∠𝐴𝐺𝐡) and one of
(∠𝐴𝐢𝐷 or ∠𝐡𝐷𝐢) to find angle
with vertical or horizontal.
AnswerMarks Guidance
1 1 .8 ( 3 s f )  = ο‚°A1 1.1
C is above DA1 2.2a
working, may have found angle
from horizontal or vertical
[4]
E.g. Taking B as (0,0)
h = 0 .3 7 2 βˆ’ 0 .3 5 2 = 0 .1 2 so A (0.12, 0.35)
A B C
Question 4:
4 | (a) | (BD is a line of) symmetry (of the kite) | B1 | 2.1
[1]
(b) | h = 0 .3 7 2 βˆ’ 0 .3 5 2 = 0 .1 2 so area = 0.042
A B C | B1 | 3.1b | ALT (considering the CoM of
triangle ABD) with X as the point
where the diagonals meet:
β„Ž = 0.12 so area = 0.021
𝐴𝐡𝑋
h = 0 .9 1 2 βˆ’ 0 .3 5 2 = 0 .8 4 so area = 0.294
A D C | B1 | 1.1 | β„Ž = 0.84 so area = 0.147
𝐴𝑋𝐢
Measuring from B:
( 0 .0 4 2 + 0 .2 9 4 ) x
= 0 .0 4 2 ο‚΄ 0 .0 8 + 0 .2 9 4 ο‚΄ ( 0 .1 2 + 0 .2 8 ) | M1 | 1.1 | Attempt to balance moments
about any point with at least two
non-zero terms each comprising
the product of a force or mass
with an appropriate distance.
May be one error in the distance.
Measuring from AC
(0.042+0.294)π‘₯Μ„
= βˆ’0.04Γ—0.042
+0.28Γ—0.294
0.08064
Gives π‘₯Μ… = = 0.24
0.336
Final step: add 0.12 to measure
from B | Measuring from B:
(0.021+0.147)π‘₯Μ…
2
=0.021Γ— Γ—0.12
3
+0.147
1
Γ—(0.12+ Γ—0.84)
3
0 .1 2 0 9 6
x = = 0 .3 6
0 .3 3 6 | A1 | 1.1 | AG. Some intermediate
working must be seen. | 0.06048
π‘₯Μ… = = 0.36 AG
0.168
[4]
(b) | Alternative method
E.g. Taking B as (0,0)
h = 0 .3 7 2 βˆ’ 0 .3 5 2 = 0 .1 2 so A (0.12, 0.35)
A B C | B1 | 1.1 | Need to see clear coordinates,
may be on diagram. Award if
average of trianlge coords
method used
h = 0 .9 1 2 βˆ’ 0 .3 5 2 = 0 .8 4 so D (0.96,0)
A D C | B1 | 1.1
1
From B, π‘₯ = (0+0.12+0.96)
3 | M1 | 1.1
1.08
= = 0.36 as triangle BAD and triangle
3
BCD are symmetrical | 1.08
= = 0.36 as triangle BAD and triangle
3 | A1 | 1.1 | AG. Intermediate working and
mention of symmetry along BD
BCD are symmetrical
must be seen.
[4]
(c) | 0 .2 4
t a n (  C A G ) =
0 .3 5 | B1 | 1.1 | CAG=34.4ο‚° (3 sf) | ∠𝐴𝐺𝐡 = 55.6Β° (3 sf)
0 .8 4
1 8 0 9 0 " 3 4 .4 " t a n 1  = ο‚° βˆ’ ο‚° βˆ’ ο‚° βˆ’ βˆ’
0 .3 5 | M1 | 3.1a | π‘‘π‘Žπ‘›βˆ’10.84
∠𝐴𝐢𝐷 = =
0.35
π‘‘π‘Žπ‘›βˆ’112
,67.4Β° (3 sf) Allow a
5
sign error. | 22.6Β°=π‘‘π‘Žπ‘›βˆ’10.35
May see ∠𝐡𝐷𝐢=
0.84
Allow for method to gain 101.8
degrees with the vertical seen
|180Β°βˆ’"55.6Β°"βˆ’π‘‘π‘Žπ‘›βˆ’10.35
|
0.84
Allow for combination of one of
(∠𝐢𝐴𝐺 or ∠𝐴𝐺𝐡) and one of
(∠𝐴𝐢𝐷 or ∠𝐡𝐷𝐢) to find angle
with vertical or horizontal.
1 1 .8 ( 3 s f )  = ο‚° | A1 | 1.1 | or –11.8ο‚°
C is above D | A1 | 2.2a | Needs to come from correct
working, may have found angle
from horizontal or vertical
[4]
E.g. Taking B as (0,0)
h = 0 .3 7 2 βˆ’ 0 .3 5 2 = 0 .1 2 so A (0.12, 0.35)
A B C
$ABCD$ is a uniform lamina in the shape of a kite with $BA = BC = 0.37$ m, $DA = DC = 0.91$ m and $AC = 0.7$ m (see diagram). The centre of mass of $ABCD$ is $G$.

\includegraphics{figure_4}

\begin{enumerate}[label=(\alph*)]
\item Explain why $G$ lies on $BD$. [1]
\item Show that the distance of $G$ from $B$ is $0.36$ m. [4]
\end{enumerate}

The lamina $ABCD$ is freely suspended from the point $A$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Determine the acute angle that $CD$ makes with the horizontal, stating which of $C$ or $D$ is higher. [4]
\end{enumerate}

\hfill \mbox{\textit{OCR Further Mechanics 2023 Q4 [9]}}
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