CAIE M2 2019 March — Question 2 6 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2019
SessionMarch
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeCentre of mass with variable parameter
DifficultyStandard +0.3 This is a straightforward centre of mass problem with standard techniques. Part (i) requires routine application of the centre of mass formula for composite bodies. Part (ii) involves setting up an equation with the new mass as a variable and solving algebraically. While it has multiple steps, it requires no novel insight—just systematic application of the standard formula with clear geometric setup.
Spec6.04c Composite bodies: centre of mass

2 \includegraphics[max width=\textwidth, alt={}, center]{b8e52188-f9a6-46fc-90bf-97965c6dd324-04_606_376_260_881} A uniform object is made by joining together three solid cubes with edges \(3 \mathrm {~m} , 2 \mathrm {~m}\) and 1 m . The object has an axis of symmetry, with the cubes stacked vertically and the cube of edge 2 m between the other two cubes (see diagram).
  1. Calculate the distance of the centre of mass of the object above the base of the largest cube.
    The smallest cube is now removed from the object. It is replaced by a heavier uniform cube with 1 m edges which is made of a different material. The centre of mass of the object is now at the base of the 2 m cube.
  2. Find the ratio of the masses of the two cubes of edge 1 m .

Question 2(i):
AnswerMarks Guidance
AnswerMarks Guidance
Total volume \(= 27 + 8 + 1 = 36\)B1
\(36x = 27 \times 1.5 + 8 \times 4 + 1 \times 5.5\)M1 Take moments about base of largest cube
\(x = \frac{13}{6} = 2.17 \text{ m}\)A1
Total: 3
Question 2(ii):
AnswerMarks Guidance
AnswerMarks Guidance
Mass of new cube \(= 35 + m\)B1 Where \(m\) is the mass of the new cube
\((35 + m) \times 3 = 27 \times 1.5 + 8 \times 4 + 5.5m\) (leads to \(m = 13\))M1 Take moments about base of largest cube
\(13:1\) or \(1:13\)A1 Accept 13
Total: 3
## Question 2(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Total volume $= 27 + 8 + 1 = 36$ | B1 | |
| $36x = 27 \times 1.5 + 8 \times 4 + 1 \times 5.5$ | M1 | Take moments about base of largest cube |
| $x = \frac{13}{6} = 2.17 \text{ m}$ | A1 | |
| **Total: 3** | | |

---

## Question 2(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Mass of new cube $= 35 + m$ | B1 | Where $m$ is the mass of the new cube |
| $(35 + m) \times 3 = 27 \times 1.5 + 8 \times 4 + 5.5m$ (leads to $m = 13$) | M1 | Take moments about base of largest cube |
| $13:1$ or $1:13$ | A1 | Accept 13 |
| **Total: 3** | | |
2\\
\includegraphics[max width=\textwidth, alt={}, center]{b8e52188-f9a6-46fc-90bf-97965c6dd324-04_606_376_260_881}

A uniform object is made by joining together three solid cubes with edges $3 \mathrm {~m} , 2 \mathrm {~m}$ and 1 m . The object has an axis of symmetry, with the cubes stacked vertically and the cube of edge 2 m between the other two cubes (see diagram).\\
(i) Calculate the distance of the centre of mass of the object above the base of the largest cube.\\

The smallest cube is now removed from the object. It is replaced by a heavier uniform cube with 1 m edges which is made of a different material. The centre of mass of the object is now at the base of the 2 m cube.\\
(ii) Find the ratio of the masses of the two cubes of edge 1 m .\\

\hfill \mbox{\textit{CAIE M2 2019 Q2 [6]}}