June 2024 Paper 2 Q21

AQACurrent spec9 marksConnected Particles Resolving Forces

21 Two heavy boxes, \(M\) and \(N\), are connected securely by a length of rope.

The mass of \(M\) is 50 kilograms.
The mass of \(N\) is 80 kilograms.

\(M\) is placed near the bottom of a rough slope.
The slope is inclined at 60° above the horizontal.

The rope is passed over a smooth pulley at the top end of the slope so that \(N\) hangs with the rope vertical.

The boxes are initially held in this position, with the rope taut and running parallel to the line of greatest slope, as shown in the diagram below.

Box M on a slope inclined at 60° to the horizontal, connected by a rope parallel to the slope over a pulley at the top to box N hanging vertically

When the boxes are released, \(M\) moves up the slope as \(N\) descends, with acceleration \(a\) m s⁻²

The tension in the rope is \(T\) newtons.

(a) Explain why the equation of motion for \(N\) is\[80g - T = 80a\] [1 mark]

(b) Show that the normal reaction force between \(M\) and the slope is \(25g\) newtons. [1 mark]

Show that

\[a \geqslant \frac{(11 - 5\sqrt{3})g}{26}\] [6 marks]

(d) State one modelling assumption you have made throughout this question. [1 mark]

Scheme	Marks	AO
States that the resultant force is \(80g - T\) and states \(F = ma\)	E1	3.4
	(1)

Scheme	Marks	AO
States \(50g\cos 60 = 25g\) OE	B1	3.1b
	(1)

Mark scheme (c)

Scheme	Marks	AO
Resolves parallel to the slope to obtain \(50g\sin 60\) or better PI OE	B1	1.1b
Obtains \(\mu \times 25g\) for friction OE Or States \(\mu = 1\) and obtains \(25g\) for friction	B1	3.3
Forms a three or four-term equation of motion for \(M\) using \(F = ma\) parallel to slope Condone cos60 OE and sign errors	M1	3.3
Obtains single correct equation or inequality with \(\mu\) Or \(T - 50g\sin 60 - 25g = 50a\) where \(\mu = 1\) is stated Or \(T - 50g\sin 60 - 25g \leqslant 50a\) where \(\mu = 1\) is stated	A1	1.1b
Eliminates \(T\) using the equation in 21(a) and their equation or inequality with \(\mu\) Condone use of \(T = 80a + 80g\)	M1	1.1a
Completes reasoned argument to show \(a \geqslant \dfrac{(11 - 5\sqrt{3})g}{26}\) Must include clear reason for using \(\mu = 1\) or uses \(0 \leqslant \mu \leqslant 1\) or uses \(\mu \leqslant 1\) AG	R1	2.1
	(6)

Typical solution

parallel to slope

\[T - 50g\sin 60 - F = 50a\]\[F = 25g\mu\]\[T - 50g\sin 60 - 25g\mu = 50a\]

Using given equation for \(N\):

\[80g - T = 80a\]\[80g - 50g\sin 60 - 25g\mu = 130a\]\[a = \frac{80g - 50g\sin 60 - 25g\mu}{130}\]\[a = \frac{\left(16 - 5\sqrt{3} - 5\mu\right)g}{26}\]

Acceleration will be at its minimum when \(\mu = 1\)

Therefore

\[a \geqslant \frac{(11 - 5\sqrt{3})g}{26}\]

Scheme	Marks	AO
States any one of the following valid assumptions Rope is light rope inextensible pulley is fixed No air resistance	E1	3.5b
	(1)
	(9 marks)