| Exam Board | OCR |
|---|---|
| Module | AS Pure (AS Pure Mathematics) |
| Year | 2017 |
| Session | Specimen |
| Marks | 10 |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: kinematics extension |
| Difficulty | Moderate -0.8 This is a straightforward mechanics question requiring vector addition, Newton's second law (F=ma), and constant acceleration kinematics (SUVAT). All steps are routine: add force vectors to show j-component cancels, calculate net force and acceleration, then apply standard kinematic equations. No novel problem-solving or geometric insight required—purely procedural application of standard AS-level mechanics techniques. |
| Spec | 1.10b Vectors in 3D: i,j,k notation3.02d Constant acceleration: SUVAT formulae3.03d Newton's second law: 2D vectors3.03p Resultant forces: using vectors |
In this question the unit vectors $\mathbf{i}$ and $\mathbf{j}$ are in the directions east and north respectively.
Distance is measured in metres and time in seconds.
A ship of mass 100 000 kg is being towed by two tug boats.
• The cables attaching each tug to the ship are horizontal.
• One tug produces a force of $(350\mathbf{i} + 400\mathbf{j})$ N.
• The other tug produces a force of $(250\mathbf{i} - 400\mathbf{j})$ N.
• The total resistance to motion is 200 N.
• At the instant when the tugs begin to tow the ship, it is moving east at a speed of 1.5 m s$^{-1}$.
\begin{enumerate}[label=(\alph*)]
\item Explain why the ship continues to move directly east. [2]
\item Find the acceleration of the ship. [2]
\item Find the time which the ship takes to move 400 m while it is being towed.
Find its speed after moving that distance. [6]
\end{enumerate}
\hfill \mbox{\textit{OCR AS Pure 2017 Q11 [10]}}