June 2023 Paper 2 Q16

AQACurrent spec4 marksForces & Newton's Laws Vectors

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]

Mark scheme

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)

Typical solution

\[\mathbf{F}_1 + \mathbf{F}_2 = \begin{bmatrix} 1.6 + k \\ 5k - 5 \end{bmatrix}\]\[\begin{bmatrix} 1.6 + k \\ 5k - 5 \end{bmatrix} = m\begin{bmatrix} 3.2 \\ 12 \end{bmatrix}\]\[1.6 + k = 3.2m\]\[5k - 5 = 12m\]\[k = 8.8\]