| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: kinematics extension |
| Difficulty | Moderate -0.8 This is a straightforward M1 vector mechanics question requiring only routine procedures: vector addition, applying F=ma, using constant acceleration equations (v=u+at), and solving a simple quadratic. All steps are standard textbook exercises with no problem-solving insight needed, making it easier than average. |
| Spec | 1.10d Vector operations: addition and scalar multiplication3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration3.03a Force: vector nature and diagrams3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors |
5 The constant forces $\mathbf { F } _ { 1 } = ( 8 \mathbf { i } + 12 \mathbf { j } )$ newtons and $\mathbf { F } _ { 2 } = ( 4 \mathbf { i } - 4 \mathbf { j } )$ newtons act on a particle. No other forces act on the particle.
\begin{enumerate}[label=(\alph*)]
\item Find the resultant force acting on the particle.
\item Given that the mass of the particle is 4 kg , show that the acceleration of the particle is $( 3 \mathbf { i } + 2 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 2 }$.
\item At time $t$ seconds, the velocity of the particle is $\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }$.
\begin{enumerate}[label=(\roman*)]
\item When $t = 20 , \mathbf { v } = 40 \mathbf { i } + 32 \mathbf { j }$.
Show that $\mathbf { v } = - 20 \mathbf { i } - 8 \mathbf { j }$ when $t = 0$.
\item Write down an expression for $\mathbf { v }$ at time $t$.
\item Find the times when the speed of the particle is $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M1 2010 Q5 [14]}}