June 2022 Paper 1 Q6

OCR MEICurrent spec9 marksMoments

6 A shelf consists of a horizontal uniform plank AB of length 0.8 m and mass 5 kg with light inextensible vertical strings attached at each end. A stack of bricks each of mass 2.3 kg is placed on the plank as shown in the diagram.

Horizontal plank AB hanging from vertical strings at A and B, with a stack of bricks standing on the plank to the right of its centre

(a) Explain the meaning of each of the following modelling assumptions.

The stack of bricks is modelled as a particle.
The plank is modelled as uniform.

[2]

Either of the strings will break if the tension exceeds 75 N.

(b) Find the greatest number of bricks that can be placed at the centre of the plank without breaking the strings. [2]

(c) Find an expression for the moment about A of the weight of a stack of \(n\) bricks when the stack is at a distance of \(x\) m from A. State the units for your answer. [2]

(d) Calculate the greatest distance from A that the largest stack of bricks can be placed without a string breaking. [3]

Mark scheme (a)

Scheme	Marks	AO
the bricks have negligible size so contact force with the plank acts at a point The mass of plank is evenly spread across its length the weight of plank acts at centre of plank.	B1 B1	3.3 3.3
	[2]

Notes

B1: Allow “no size” or “size doesn’t matter” or “shape is not relevant” etc
Allow “weight of bricks acts on the plank at point”

B1: Allow for either statement
Allow the plank is the same throughout or centre of mass at centre
Do not allow “mass acts” at the centre

Mark scheme (b)

Scheme	Marks	AO
when placed at the centre tensions are equal \(2 \times 75 = 2.3ng + 5g\)	M1	3.1b
\(n = \left[\dfrac{101}{2.3g}\right] = 4.48\ldots\) so 4 bricks	A1	1.1b
	[2]

Notes

M1: Using symmetry to establish equation for \(n\) soi. Allow if the weight of the plank or one of the tensions is missing
Trial and improvement may be used

A1: Final answer must be the integer
Allow if 4 seen www

Alternative using moments

Scheme	Marks
\(5g \times 0.4 + 2.3gn \times 0.4 = 75 \times 0.8\)	M1
\(n = \left[\dfrac{40.4}{9.016}\right] = 4.48\ldots\) so 4 bricks	A1

M1: Allow for missing moment of weight or one of the tensions
Every term must be a moment

A1: Final answer must be the integer
Allow if 4 seen www

Scheme	Marks	AO
\(2.3gnx\)	B1	1.1b
Nm	B1	1.2
	[2]

Mark scheme (d)

Scheme	Marks	AO
4 bricks on the point of breaking \(x\) m from A Taking moments about A \(5g \times 0.4 + 4 \times 2.3gx = 75 \times 0.8\)	M1	3.1b
\(9.2gx = 60 - 2g\)	A1	1.1b
\(x = 0.448\) [so 44.8 cm from A]	A1	1.1b
	[3]

Notes

M1: Taking moments about any point to form an equation. FT their \(n\)
All forces used in a moment. Allow sign errors. Allow an incorrect distance used. Could be an inequality

A1: Fully correct equation FT their \(n\). Allow corresponding inequality
Need not be simplified

A1: cao