OCR MEI Further Mechanics Major 2023 June — Question 8 8 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Major (Further Mechanics Major)
Year2023
SessionJune
Marks8
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TopicCentre of Mass 2
TypeCentre of mass of lamina by integration
DifficultyChallenging +1.8 This is a challenging Further Maths mechanics question requiring integration to find the centre of mass of a composite region, algebraic manipulation to express the answer in a specific form, and then proving an inequality. It involves multiple steps: finding the line equation, setting up double integrals for moments and area, evaluating them in terms of k, simplifying to the required form, and finally proving the y-coordinate cannot equal 3/2. The technical demands and multi-stage reasoning place it well above average difficulty.
Spec1.08e Area between curve and x-axis: using definite integrals6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids
\includegraphics{figure_8} The diagram shows the shaded region R bounded by the curve \(y = \sqrt{3x + 4}\), the \(x\)-axis, the \(y\)-axis, and the straight line that passes through the points \((k, 0)\) and \((4, 4)\), where \(0 < k < 4\). Region R is occupied by a uniform lamina.
  1. Determine, in terms of \(k\), an expression for the \(y\)-coordinate of the centre of mass of the lamina. Give your answer in the form \(\frac{a + bk}{c + dk}\), where \(a\), \(b\), \(c\) and \(d\) are integers to be determined. [6]
  2. Show that the \(y\)-coordinate of the centre of mass of the lamina cannot be \(\frac{3}{2}\). [2]
Question 8:
AnswerMarks Guidance
8(a) 4
Area below the curve =  3 x + 4 d x = 1 19 2
AnswerMarks Guidance
0B1 1.1
calculations)
4
A y = 12  3 x + 4 d x = 2 0
AnswerMarks Guidance
0B1 1.1
below the curve is 42 58 which implies B1B1
Might be seen in the evaluation of
 4  4  x − k  2
A y = 12 ( 3 x + 4 ) d x − 12 1 6 d x
4 − k
0 k
Possibly embedded in other calculations
4
Centre of mass of triangle is at a distance of from the x-axis
AnswerMarks Guidance
3B1 1.2
possibly implied in later working
If done by integration then this mark is awarded for
 4 x−k 2 8
correctly evaluating 1  16  dxas (4−k)
2  4−k 3
k
(or unsimplified)
AnswerMarks Guidance
M12.1 Table of values idea – correct number of terms
(dimensionally consistent) – so of the form
k f(k)=(k g(k))y where f(k) and g(k) are
1 2
equivalent to linear functions of k
1  4   1 1 2 1 
2 0 − ( 4 ) ( 4 − k ) = − ( 4 ) ( 4 − k ) y
AnswerMarks Guidance
2 3 9 2B1 1.1
For either the correct lhs or rhs seen (e.g. +2k or
9
2 8 8 k
+ or unsimplified equivalent expressions seen
3 3
anywhere in their solution so do not need to be seen
in a complete equation for y )
42+12k
y =
AnswerMarks Guidance
20+9kA1 2.2a
multiplies e.g. a = 84, b = 24, c = 40 and d = 18 etc.)
[6]
AnswerMarks
(b)4 2 + 1 2 k 3
=  k = ...
AnswerMarks Guidance
2 0 + 9 k 2M1 3.1a
fractions are cleared must be a linear equation in k)
and attempts to solve for k
AnswerMarks Guidance
k = 8 but 0 < k < 4 so y cannot be 1.5A1 2.3
[2]
Question 8:
8 | (a) | 4
Area below the curve =  3 x + 4 d x = 1 19 2
0 | B1 | 1.1 | BC or 1 29 8 − 1 69 (possibly embedded in other
calculations)
4
A y = 12  3 x + 4 d x = 2 0
0 | B1 | 1.1 | BC - y-coordinate of centre of mass of the lamina
below the curve is 42 58 which implies B1B1
Might be seen in the evaluation of
 4  4  x − k  2
A y = 12 ( 3 x + 4 ) d x − 12 1 6 d x
4 − k
0 k
Possibly embedded in other calculations
4
Centre of mass of triangle is at a distance of from the x-axis
3 | B1 | 1.2 | Used in a moment calculation (so not just stated) -
possibly implied in later working
If done by integration then this mark is awarded for
 4 x−k 2 8
correctly evaluating 1  16  dxas (4−k)
2  4−k 3
k
(or unsimplified)
M1 | 2.1 | Table of values idea – correct number of terms
(dimensionally consistent) – so of the form
k f(k)=(k g(k))y where f(k) and g(k) are
1 2
equivalent to linear functions of k
1  4   1 1 2 1 
2 0 − ( 4 ) ( 4 − k ) = − ( 4 ) ( 4 − k ) y
2 3 9 2 | B1 | 1.1 | 40
For either the correct lhs or rhs seen (e.g. +2k or
9
2 8 8 k
+ or unsimplified equivalent expressions seen
3 3
anywhere in their solution so do not need to be seen
in a complete equation for y )
42+12k
y =
20+9k | A1 | 2.2a | a = 42, b = 12, c = 20, d = 9 (or correct integer
multiplies e.g. a = 84, b = 24, c = 40 and d = 18 etc.)
[6]
(b) | 4 2 + 1 2 k 3
=  k = ...
2 0 + 9 k 2 | M1 | 3.1a | Sets their expression for y equal to 1.5 (which when
fractions are cleared must be a linear equation in k)
and attempts to solve for k
k = 8 but 0 < k < 4 so y cannot be 1.5 | A1 | 2.3 | k = 8 plus reason (so some mention of k < 4)
[2]
\includegraphics{figure_8}

The diagram shows the shaded region R bounded by the curve $y = \sqrt{3x + 4}$, the $x$-axis, the $y$-axis, and the straight line that passes through the points $(k, 0)$ and $(4, 4)$, where $0 < k < 4$.

Region R is occupied by a uniform lamina.

\begin{enumerate}[label=(\alph*)]
\item Determine, in terms of $k$, an expression for the $y$-coordinate of the centre of mass of the lamina. Give your answer in the form $\frac{a + bk}{c + dk}$, where $a$, $b$, $c$ and $d$ are integers to be determined. [6]
\item Show that the $y$-coordinate of the centre of mass of the lamina cannot be $\frac{3}{2}$. [2]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2023 Q8 [8]}}
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