M1 June 2014 Q6
6.

Two forces \(\mathbf{P}\) and \(\mathbf{Q}\) act on a particle at \(O\). The angle between the lines of action of \(\mathbf{P}\) and \(\mathbf{Q}\) is 120\(^\circ\) as shown in Figure 4. The force \(\mathbf{P}\) has magnitude 20 N and the force \(\mathbf{Q}\) has magnitude \(X\) newtons. The resultant of \(\mathbf{P}\) and \(\mathbf{Q}\) is the force \(\mathbf{R}\).
Given that the magnitude of \(\mathbf{R}\) is \(3X\) newtons, find, giving your answers to 3 significant figures
| Scheme | Marks |
|---|---|
![]() | |
| Resolve and use Pythagoras \((X - 20\cos 60)^2 + (20\cos 30)^2 = (3X)^2\) | M1 A1 |
| \(8X^2 + 20X - 400 = 0\) | A1 |
| \(X = \dfrac{-5 \pm \sqrt{25 + 800}}{4} = 5.93\) (3 SF) | M1A1 |
| (5) |
Notes
In this question a misquoted Cosine Rule is M0.
The question asks for both answers to 3 SF but only penalise under or over accuracy once in this question.
First M1 for a complete method to give an equation in \(X\) only i.e. producing two components and usually squaring and adding and equating to \((3X)^2\) (condone sign errors and consistent incorrect trig. in the components for this M mark BUT the \(x\)-component must be a difference)
First A1 for a correct unsimplified equation in \(X\) only
e.g, allow \((\pm(X - 20\cos 60^\circ))^2 + (\pm(20\cos 30^\circ))^2 = (3X)^2\)
Second A1 for any correct fully numerical 3 term quadratic = 0
Second M1(independent) for solving a 3 term quadratic
Third A1 for 5.93
Alternative using cosine rule:
First M1 for use of cosine rule with cos60\(^\circ\) (M0 if they use 120\(^\circ\))
First A1 for a correct equation unsimplified e.g, allow cos60\(^\circ\) and \((3X)^2\)
Second A1 for any correct fully numerical 3 term quadratic = 0
Second M1(independent) for solving a 3 term quadratic
Third A1 for 5.93
Alternative using 2 applications of the sine rule:
First M1 for using \(3X / \sin 60 = X / \sin a\) AND
Either: \(X / \sin a = 20 / \sin(120^\circ - a)\)
Or: \(3X / \sin 60^\circ = 20 / \sin(120^\circ - a)\)
(These could be in terms of \(b\) where \(b = (120^\circ - a)\))
First A1 for two correct equations
Second A1 for \(a = 16.778..^\circ\) (or \(b = 103.221..^\circ\))
Second M1 for solving: \(X / \sin a = 20 / \sin(120^\circ - a)\) or \(3X / \sin 60^\circ = 20 / \sin(120^\circ - a)\) with their \(a\) or \(b\), to find \(X\)
Third A1 for 5.93
6a alt
| Cosine rule \((3X)^2 = 20^2 + X^2 - 2.20X\cos 60\) \(8X^2 + 20X - 400 = 0\) | M1A1 A1 |
| \(X = \dfrac{-5 \pm \sqrt{25 + 800}}{4} = 5.93\) (3SF) | M1A1 |
| (5) |
| Scheme | Marks |
|---|---|
| \(|\mathbf{P} - \mathbf{Q}|^2 = 20^2 + X^2 - 2X \times 20 \times \cos 120\) | M1A1 |
| \(|\mathbf{P} - \mathbf{Q}| = 23.5\) (N) (3SF) | DM1 A1 |
| (4) | |
| (9 marks) |
Notes
First M1 for use of cosine rule unsimplified with cos120\(^\circ\) (M0 if they use 60\(^\circ\))
First A1 for a correct expression for \(|\mathbf{P} - \mathbf{Q}|\) in terms of \(X\) (does not need to be substituted)
Second M1, dependent on first M1, for substituting for their \(X\) and solving for \(|\mathbf{P} - \mathbf{Q}|\)
Second A1 for 23.5
Alternative using components:
First M1 for a complete method i.e. producing two components and squaring and adding (no square root needed) (condone sign errors and consistent incorrect trig. in the components for this M mark BUT the \(x\)-component must be a sum)
First A1 for a correct expression for \(|\mathbf{P} - \mathbf{Q}|\) (e.g, allow \((\pm(X + 20\cos 60^\circ))^2 + (\pm(20\cos 30^\circ))^2\)
Second M1, dependent on first M1, for substituting for their \(X\) and solving for \(|\mathbf{P} - \mathbf{Q}|\)
Second A1 for 23.5
6b alt
| \(|\mathbf{P} - \mathbf{Q}|^2 = (X + 20\cos 60)^2 + (20\cos 30)^2\) | M1A1 |
| \(|\mathbf{P} - \mathbf{Q}| = 23.5\) (N) (3SF) | DM1 A1 |
| (4) |
