M1 June 2017 Q1
1. Three forces, \((15\mathbf{i} + \mathbf{j})\) N, \((5q\mathbf{i} - p\mathbf{j})\) N and \((-3p\mathbf{i} - q\mathbf{j})\) N, where \(p\) and \(q\) are constants, act on a particle. Given that the particle is in equilibrium, find the value of \(p\) and the value of \(q\). (6)
| Scheme | Marks |
|---|---|
| \((15\mathbf{i} + \mathbf{j}) + (5q\mathbf{i} - p\mathbf{j}) + (-3p\mathbf{i} - q\mathbf{j}) = \mathbf{0}\) | M1 |
| \(3p - 5q = 15\) | M1 |
| \(p + q = 1\) | A1 |
| \(p = 2.5\ \ \ q = -1.5\) | M1 A1 A1 |
| (6 marks) |
Notes
First M1 for equating the sum of the three forces to zero (can be implied by subsequent working)
Second M1 for equating the sum of the \(\mathbf{i}\) components to zero AND the sum of the \(\mathbf{j}\) components to zero oe to produce TWO equations, each one being in \(p\) and \(q\) ONLY.
First A1 for TWO correct equations (in any form)
N.B. It is possible to obtain TWO equations by using \(\lambda(3p - 5q - 15) = \mu(p + q - 1)\) with TWO different pairs of values for \(\lambda\) and \(\mu\), with one pair not a multiple of the other e.g \(\lambda = 1\), \(\mu = 1\) AND \(\lambda = 1\), \(\mu = 2\).
Third M1(independent) for attempt (either by substitution or elimination) to produce an equation in either \(p\) ONLY or \(q\) ONLY.
Second A1 for \(p = 2.5\) (any equivalent form, fractions do not need to be in lowest terms)
Third A1 for \(q = -1.5\) (any equivalent form, fractions do not need to be in lowest terms)