Standard +0.3 This is a straightforward mechanics problem applying Newton's second law (F=ma) in vector form. Students must sum the forces, equate to mass times acceleration, and solve the resulting simultaneous equations. It requires only standard algebraic manipulation with no conceptual difficulty or novel insight, making it slightly easier than average.
Two forces, \(\mathbf{F_1}\) and \(\mathbf{F_2}\), are acting on a particle of mass 3 kilograms.
It is given that
$$\mathbf{F_1} = \begin{pmatrix} a \\ 23 \end{pmatrix} \text{ newtons and } \mathbf{F_2} = \begin{pmatrix} 4 \\ b \end{pmatrix} \text{ newtons}$$
where \(a\) and \(b\) are constants.
The particle has an acceleration of \(\begin{pmatrix} 4b \\ a \end{pmatrix}\) m s\(^{-2}\)
Find the value of \(a\) and the value of \(b\)
[4 marks]
Question 15:
15 | Obtains a + 4 and b + 23
OE | 1.1b | B1 | a+4
b+23
a+4 4b
=3
b+23 a
a + 4 = 12b
b + 23 = 3a
a = 8 and b = 1
Uses F = ma with m = 3 and
4b
a =
a | 3.3 | M1
Obtains two linear simultaneous
equations in a and b | 1.1a | M1
Obtains a = 8 and b = 1 | 1.1b | A1
Question 15 Total | 4
Q | Marking instructions | AO | Marks | Typical solution
Two forces, $\mathbf{F_1}$ and $\mathbf{F_2}$, are acting on a particle of mass 3 kilograms.
It is given that
$$\mathbf{F_1} = \begin{pmatrix} a \\ 23 \end{pmatrix} \text{ newtons and } \mathbf{F_2} = \begin{pmatrix} 4 \\ b \end{pmatrix} \text{ newtons}$$
where $a$ and $b$ are constants.
The particle has an acceleration of $\begin{pmatrix} 4b \\ a \end{pmatrix}$ m s$^{-2}$
Find the value of $a$ and the value of $b$
[4 marks]
\hfill \mbox{\textit{AQA Paper 2 2024 Q15 [4]}}