AQA M2 2006 January — Question 5 8 marks

Exam BoardAQA
ModuleM2 (Mechanics 2)
Year2006
SessionJanuary
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable acceleration (vectors)
TypeFind velocity by integrating acceleration
DifficultyModerate -0.3 This is a straightforward M2 mechanics question requiring direct application of F=ma for part (a) and integration of vector acceleration for part (b). Both parts are standard textbook exercises with no problem-solving insight needed—just routine manipulation of vectors and basic calculus, making it slightly easier than average.
Spec1.08h Integration by substitution1.10b Vectors in 3D: i,j,k notation3.02f Non-uniform acceleration: using differentiation and integration3.03d Newton's second law: 2D vectors

5 A particle moves such that at time \(t\) seconds its acceleration is given by $$( 2 \cos t \mathbf { i } - 5 \sin t \mathbf { j } ) \mathrm { m } \mathrm {~s} ^ { - 2 }$$
  1. The mass of the particle is 6 kg . Find the magnitude of the resultant force on the particle when \(t = 0\).
  2. When \(t = 0\), the velocity of the particle is \(( 2 \mathbf { i } + 10 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\). Find an expression for the velocity of the particle at time \(t\).

5 A particle moves such that at time $t$ seconds its acceleration is given by

$$( 2 \cos t \mathbf { i } - 5 \sin t \mathbf { j } ) \mathrm { m } \mathrm {~s} ^ { - 2 }$$
\begin{enumerate}[label=(\alph*)]
\item The mass of the particle is 6 kg . Find the magnitude of the resultant force on the particle when $t = 0$.
\item When $t = 0$, the velocity of the particle is $( 2 \mathbf { i } + 10 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$.

Find an expression for the velocity of the particle at time $t$.
\end{enumerate}

\hfill \mbox{\textit{AQA M2 2006 Q5 [8]}}