CAIE M2 2017 June — Question 3 7 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2017
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 2
TypeFolded lamina
DifficultyChallenging +1.2 This is a multi-part centre of mass problem requiring systematic application of the standard formula for composite bodies. Part (i) is straightforward calculation with 5 components. Part (ii) requires setting up an equation with the lid at angle θ, involving trigonometry (sin θ and cos θ terms) and algebraic manipulation, but follows a standard method without requiring novel geometric insight or proof techniques.
Spec6.04c Composite bodies: centre of mass

3 An open box in the shape of a cube with edges of length 0.2 m is placed with its base horizontal and its four sides vertical. The four sides and base are uniform laminas, each with weight 3 N .
  1. Calculate the height of the centre of mass of the box above its base.
    The box is now fitted with a thin uniform square lid of weight 3 N and with edges of length 0.2 m . The lid is attached to the box by a hinge of length 0.2 m and weight 2 N . The lid of the box is held partly open.
  2. Find the angle which the lid makes with the horizontal when the centre of mass of the box (including the lid and hinge) is 0.12 m above the base of the box.

Question 3:
Part (i)
AnswerMarks Guidance
AnswerMarks Notes
Height of C of M of each vertical face above the base \(= 0.1\) mB1
\(5 \times 3y = 4 \times 3 \times 0.1\)M1 Takes moments about the base; \(y\) is the height of C of M above the base
\(y = 0.08\) mA1
Total: 3
Part (ii)
AnswerMarks Guidance
AnswerMarks Notes
Moment of lid about the base \(= 3 \times (0.2 + 0.1\sin\theta)\)B1 \(\theta\) is the angle the lid makes with the horizontal
\((6 \times 3 + 2) \times 0.12 = 5 \times 3 \times 0.08 + 2 \times 0.2 + 3 \times (0.2 + 0.1\sin\theta)\)M1 Take moments about the base
A1
\(\theta = 41.8°\)A1
Total: 4
## Question 3:

**Part (i)**

| Answer | Marks | Notes |
|--------|-------|-------|
| Height of C of M of each vertical face above the base $= 0.1$ m | B1 | |
| $5 \times 3y = 4 \times 3 \times 0.1$ | M1 | Takes moments about the base; $y$ is the height of C of M above the base |
| $y = 0.08$ m | A1 | |
| **Total: 3** | | |

**Part (ii)**

| Answer | Marks | Notes |
|--------|-------|-------|
| Moment of lid about the base $= 3 \times (0.2 + 0.1\sin\theta)$ | B1 | $\theta$ is the angle the lid makes with the horizontal |
| $(6 \times 3 + 2) \times 0.12 = 5 \times 3 \times 0.08 + 2 \times 0.2 + 3 \times (0.2 + 0.1\sin\theta)$ | M1 | Take moments about the base |
| | A1 | |
| $\theta = 41.8°$ | A1 | |
| **Total: 4** | | |

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3 An open box in the shape of a cube with edges of length 0.2 m is placed with its base horizontal and its four sides vertical. The four sides and base are uniform laminas, each with weight 3 N .\\
(i) Calculate the height of the centre of mass of the box above its base.\\

The box is now fitted with a thin uniform square lid of weight 3 N and with edges of length 0.2 m . The lid is attached to the box by a hinge of length 0.2 m and weight 2 N . The lid of the box is held partly open.\\
(ii) Find the angle which the lid makes with the horizontal when the centre of mass of the box (including the lid and hinge) is 0.12 m above the base of the box.\\

\hfill \mbox{\textit{CAIE M2 2017 Q3 [7]}}