| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2021 |
| Session | June |
| Marks | 13 |
| Topic | Power and driving force |
| Type | Power from force and derived speed (non-equilibrium) |
| Difficulty | Standard +0.3 This is a straightforward vector mechanics question requiring standard applications of kinematic equations and power formulas (P = F·v). While it involves vectors and multiple parts, each step follows directly from given formulas with no novel problem-solving required. The calculations are routine for Further Maths students, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10d Vector operations: addition and scalar multiplication6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
1 A particle $Q$ of mass $m \mathrm {~kg}$ is acted on by a single force so that it moves with constant acceleration $\mathbf { a } = \binom { 1 } { 2 } \mathrm {~ms} ^ { - 2 }$. Initially $Q$ is at the point $O$ and is moving with velocity $\mathbf { u } = \binom { 2 } { - 5 } \mathrm {~ms} ^ { - 1 }$.
After $Q$ has been moving for 5 seconds it reaches the point $A$.
\begin{enumerate}[label=(\alph*)]
\item Use the equation $\mathbf { v } . \mathbf { v } = \mathbf { u } . \mathbf { u } + 2 \mathbf { a } . \mathbf { x }$ to show that at $A$ the kinetic energy of $Q$ is 37 m J .
\item \begin{enumerate}[label=(\roman*)]
\item Show that the power initially generated by the force is - 8 mW W.
\item The power in part (b)(i) is negative. Explain what this means about the initial motion of $Q$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Find the time at which the power generated by the force is instantaneously zero.
\item Find the minimum kinetic energy of $Q$ in terms of $m$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2021 Q1 [13]}}