June 2025 Paper 2 Q15

15 A particle moves under the actions of two forces, \(\mathbf{F}_1\) and \(\mathbf{F}_2\)

\(\mathbf{F}_1\) has magnitude 17 newtons and acts due East.
\(\mathbf{F}_2\) has magnitude 26 newtons and acts at a bearing of 310°

The resultant of \(\mathbf{F}_1\) and \(\mathbf{F}_2\) is \(\mathbf{R}\)

(a) Show that the magnitude of \(\mathbf{R}\) is 17.0 newtons, correct to three significant figures. [4 marks]

(b) Find the angle that \(\mathbf{R}\) makes with \(\mathbf{F}_1\)

Give your answer to the nearest degree.

[2 marks]

(i) State the magnitude of \(\mathbf{F}_3\) [1 mark]

(ii) State the bearing on which \(\mathbf{F}_3\) acts. [1 mark]

Mark scheme (a)

Scheme	Marks	AO
Resolves horizontally or vertically to find a correct expression for one of the i or j components of the resultant force. Or Uses the cosine rule with b = 17 and c = 26	M1	3.3
Resolves horizontally and vertically to find a correct expression for both i and j components of the resultant force PI by H: AWRT \(\pm\) 2.9 V: AWRT \(\pm\)16.7 Or Uses the cosine rule to obtain a² = AWRT 287.8	A1	1.1b
Finds the magnitude of the resultant vector of their horizontal and vertical components Or Finds the square root of their expression for b² + c² − 2bcCosA PI by AWFW [16.96, 16.97]	M1	1.1a
Completes reasoned argument showing a full method to obtain AWFW [16.96, 16.97] and concludes that the magnitude of R is 17.0 Condone 17 AG	R1	2.1
	(4)

Horizontally: \(17 - 26\cos(40)\)

\[= -2.917\ldots\]

Vertically: \(26\sin(40)\)

\[= 16.712\ldots\]\[\sqrt{(-2.917)^2 + 16.712^2} = 16.97\ \text{N}\]\[\approx 17.0\ \text{N}\]

Mark scheme (b)

Scheme	Marks	AO
Uses an appropriate trigonometric equation using their components for R Or Deduces that the two 17N forces make this an isosceles triangle, eg: 180 − (2 x 40) PI AWRT 79 or 80	M1	3.1a
Obtains any of the following angles AWRT 100, 101, \(\pm 259\), \(\pm 260\)	A1	1.1b
	(2)

\[\tan^{-1}\left(\frac{16.712}{2.917}\right) = 80^\circ\]\[180 - 80 = 100^\circ\]