June 2023 Paper 2 Q16
16 A particle moves under the action of two forces, \(\mathbf{F}_1\) and \(\mathbf{F}_2\)
It is given that
\[\mathbf{F}_1 = (1.6\mathbf{i} - 5\mathbf{j})\text{ N}\]\[\mathbf{F}_2 = (k\mathbf{i} + 5k\mathbf{j})\text{ N}\]where \(k\) is a constant.
The acceleration of the particle is \((3.2\mathbf{i} + 12\mathbf{j})\) m s−2
Find \(k\) [4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Adds the two forces together. ACF | B1 | 1.1b |
| Uses \(\mathbf{F} = m\boldsymbol{a}\) and substitutes their \(\mathbf{F}_1 \pm \mathbf{F}_2\) and \(\boldsymbol{a} = \begin{bmatrix} 3.2 \\ 12 \end{bmatrix}\) PI by use of ratios For example: \(\dfrac{5k - 5}{12} = \dfrac{1.6 + k}{3.2}\) OE | M1 | 3.1a |
| Obtains two correct equations For example: \(1.6 + k = 3.2m\) and \(5k - 5 = 12m\) Only award if vectors are removed or Obtains a correct linear equation in \(k\) For example: \(\dfrac{5k - 5}{12} = \dfrac{1.6 + k}{3.2}\) OE PI by \(k = 8.8\) or \(m = 3.25\) | A1 | 1.1b |
| Obtains \(k = 8.8\) ACF | A1 | 1.1b |
| (4 marks) |