Easy -1.8 This is a straightforward application of F=ma in vector form with only one unknown. Students simply need to recognize that the force and acceleration vectors are parallel (same direction), so the ratio of components must be equal: -2/-6 = 6/y, giving y=18. It's a 1-mark multiple-choice question requiring minimal calculation and no problem-solving insight.
A resultant force of \(\begin{bmatrix} -2 \\ 6 \end{bmatrix}\) N acts on a particle.
The acceleration of the particle is \(\begin{bmatrix} -6 \\ y \end{bmatrix} \text{ m s}^{-2}\)
Find the value of \(y\)
Circle your answer.
[1 mark]
\(2\) \quad \(3\) \quad \(10\) \quad \(18\)
A resultant force of $\begin{bmatrix} -2 \\ 6 \end{bmatrix}$ N acts on a particle.
The acceleration of the particle is $\begin{bmatrix} -6 \\ y \end{bmatrix} \text{ m s}^{-2}$
Find the value of $y$
Circle your answer.
[1 mark]
$2$ \quad $3$ \quad $10$ \quad $18$
\hfill \mbox{\textit{AQA AS Paper 1 2023 Q13 [1]}}