June 2023 Paper 1 Q13

OCR MEICurrent spec12 marksConnected Particles Resolving Forces

13 A block of mass 8 kg is placed on a rough plane inclined at \(15^\circ\) to the horizontal. The coefficient of friction between the block and the plane is 0.3.

One end of a light rope is attached to the block. The rope passes over a smooth pulley fixed at the top of the plane, and a sphere of mass 5 kg, attached to the other end of the rope, hangs vertically below the pulley. The part of the rope between the block and the pulley is parallel to the plane. The system is released from rest, and as the sphere falls the block moves directly up the plane with acceleration \(a\,\mathrm{m\,s^{-2}}\).

Block on a plane inclined at 15 degrees, connected by a rope parallel to the plane over a pulley at the top to a sphere hanging vertically

(a) On the diagram in the Printed Answer Booklet, show all the forces acting on the block and on the sphere. [4]

(b) Write down the equation of motion for the sphere. [2]

Force diagram: on the block, normal reaction R perpendicular to the plane, tension T up the plane, friction F down the plane, weight 8g N downwards; on the sphere, tension T upwards and weight 5g N downwards — Mark scheme (a)

Scheme	Marks	AO
	B1 B1 B1 B1	1.1b 1.1b 1.1b 1.1b
	[4]

Scheme	Marks	AO
Sphere (downwards positive) \(5g - T = 5a\)	M1 A1	1.1a 1.1b
	[2]

Mark scheme (c)

Scheme	Marks	AO
\(R = 8g\cos 15^\circ\)	B1	3.1b
\(F = 0.3R = 2.4g\cos 15^\circ\)	M1	3.4
Newton’s second law for the block \(T - 8g\sin 15^\circ - F = 8a\)	M1 A1	1.1a 1.1b
Add equations of motion \(5g - 8g\sin 15^\circ - 2.4g\cos 15^\circ = 13a\)	M1	1.1a
\(a = 0.461\)	A1	1.1b
	[6]

Notes

M1: FT their \(R\)

M1: Allow one missing or incorrect force, must be dimensionally correct. FT their \(F\)

A1: All correct FT their \(F\)

M1: Attempt to eliminate \(T\)
This equation with one missing or incorrect force implies the previous M1A0M1