CAIE Further Paper 3 2022 November — Question 2 6 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2022
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeLamina with removed triangle/rectangle/square
DifficultyChallenging +1.2 This is a standard centre of mass problem requiring students to find the centre of mass of a composite shape (triangle minus triangle), apply the equilibrium condition that the centre of mass must lie above the base, and solve an inequality. While it involves multiple steps (finding two centres of mass, using composite body formula, applying equilibrium condition), these are all routine techniques for Further Maths mechanics students with no novel insight required.
Spec6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces

\includegraphics{figure_2} A uniform lamina is in the form of a triangle \(ABC\) in which angle \(B\) is a right angle, \(AB = 9a\) and \(BC = 6a\). The point \(D\) is on \(BC\) such that \(BD = x\) (see diagram). The region \(ABD\) is removed from the lamina. The resulting shape \(ADC\) is placed with the edge \(DC\) on a horizontal surface and the plane \(ADC\) is vertical. Find the set of values of \(x\), in terms of \(a\), for which the shape is in equilibrium. [6]

Question 2:
AnswerMarks
2Area Distance from AB
ABC 27a2 2a
9 1
ABD ax x
2 3
9
Shape ADC 27a2 − ax x
2
Taking moments about AB
 9  9 1
x 27a2 − ax= 27a2 2a− ax  x
 2  2 3
 3 
54a3− ax2
 
2
x = 
9
 27a2 − ax 
AnswerMarks Guidance
 2 M1 Moments equation with 3 terms.
A1At least 2 terms correct.
A1All correct.
For equilibrium, x x,
3  9 
54a3− ax2 x27a2 − ax
AnswerMarks Guidance
2  2 B1 Use correct condition: allow strict inequality.
Can be implied by correct final answer x 3a.
54a2 −27ax+3x2 0
AnswerMarks Guidance
(x−3a)(x−6a) 0M1 Simplify and attempt to solve a quadratic inequality or equation.
(0 ) x 3a [only]A1 CAO
AreaDistance from AB
ABC27a2 2a
ABD9
ax
AnswerMarks
21
x
3
AnswerMarks
Shape ADC9
27a2 − ax
AnswerMarks Guidance
2x
QuestionAnswer Marks
2Alternative method for question 2
Taking moments with B as origin.M1
1 1
x = (0+x+6a)=2a+ x
AnswerMarks
3 3A2
1
For equilibrium, x x, so x 2a+ x
AnswerMarks Guidance
3B1 Allow strict inequality.
(0 ) x 3aM1
A1
6
AnswerMarks Guidance
QuestionAnswer Marks
Question 2:
2 | Area Distance from AB
ABC 27a2 2a
9 1
ABD ax x
2 3
9
Shape ADC 27a2 − ax x
2
Taking moments about AB
 9  9 1
x 27a2 − ax= 27a2 2a− ax  x
 2  2 3
 3 
54a3− ax2
 
2
x = 
9
 27a2 − ax 
 2  | M1 | Moments equation with 3 terms.
A1 | At least 2 terms correct.
A1 | All correct.
For equilibrium, x x,
3  9 
54a3− ax2 x27a2 − ax
2  2  | B1 | Use correct condition: allow strict inequality.
Can be implied by correct final answer x 3a.
54a2 −27ax+3x2 0
(x−3a)(x−6a) 0 | M1 | Simplify and attempt to solve a quadratic inequality or equation.
(0 ) x 3a [only] | A1 | CAO
Area | Distance from AB
ABC | 27a2 | 2a
ABD | 9
ax
2 | 1
x
3
Shape ADC | 9
27a2 − ax
2 | x
Question | Answer | Marks | Guidance
2 | Alternative method for question 2
Taking moments with B as origin. | M1
1 1
x = (0+x+6a)=2a+ x
3 3 | A2
1
For equilibrium, x x, so x 2a+ x
3 | B1 | Allow strict inequality.
(0 ) x 3a | M1
A1
6
Question | Answer | Marks | Guidance
\includegraphics{figure_2}

A uniform lamina is in the form of a triangle $ABC$ in which angle $B$ is a right angle, $AB = 9a$ and $BC = 6a$. The point $D$ is on $BC$ such that $BD = x$ (see diagram). The region $ABD$ is removed from the lamina. The resulting shape $ADC$ is placed with the edge $DC$ on a horizontal surface and the plane $ADC$ is vertical.

Find the set of values of $x$, in terms of $a$, for which the shape is in equilibrium.
[6]

\hfill \mbox{\textit{CAIE Further Paper 3 2022 Q2 [6]}}